Integrals of the form ${\large\int}_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx$

Question

I'm interested in integrals of the form $$I(a,b)=\int_0^\infty\operatorname{arccot}(x)\cdot\operatorname{arccot}(a\,x)\cdot\operatorname{arccot}(b\,x)\ dx,\color{#808080}{\text{ for }a>0,\,b>0}\tag1$$ It's known$\require{action}\require{enclose}\texttip{{}^\dagger}{Gradshteyn & Ryzhik, Table of Integrals, Series, and Products, 7th edition, page 599, (4.511)}$ that $$I(a,0)=\frac{\pi^2}4\left[\ln\left(1+\frac1a\right)+\frac{\ln(1+a)}a\right].\tag2$$ Maple and Mathematica are also able to evaluate $$I(1,1)=\frac{3\pi^2}4\ln2-\frac{21}8\zeta(3).\tag3$$

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Is it possible to find a general closed form for $I(a,1)$? Or, at least, for $I(2,1)$ or $I(3,1)$?

Answer 1

I finally managed to find a general solution to this problem. Although I put a lot of effort into simplification of the result, it is still not as pretty and symmetric as I would like it to be. I hope to improve it later. Sorry for poor typesetting.

Assuming $0<a<b<c,$ $$\int_0^\infty\operatorname{arccot}(ax)\,\operatorname{arccot}(bx)\,\operatorname{arccot}(cx)\,dx=\\ \frac1{24 a b c}\left(2 a (b+2 c) \ln ^3(a)-3 \left(2 b (a+c) \ln (b)+(4 a c-2 b c) \ln (c)+\\ b \left(-2 c \operatorname{artanh}\left(\frac{b}{c}\right)+(c-a) \ln (c-a)-(a+c) \ln (a+c)+a \ln \left(c^2-b^2\right)\right)\right) \ln ^2(a)\\ +3 \left(b c \left(\ln ^2(b)+2 \left(2 \ln (a+c)+\ln \left(-\frac{c (b+c)}{(a-b) (a-c) (b-c)}\right)\right) \ln (b)+\ln ^2(b-a)\\ -3 \ln ^2(c)+\ln ^2(c-a)-\ln ^2(c-b)+\ln ^2(b+c)-4 \operatorname{artanh}\left(\frac{a}{c}\right) \ln (a+b)+\\ 2 \ln (b-a) \ln \left(\frac{c}{c-a}\right)-6 \ln (c) \ln (a+c)+2 \ln (a+c) \ln (c-b)\\ +4 \ln (c) (\ln (c-a)+\ln (c-b))-4 \ln (c) \ln (b+c)-2 \ln (c-a) \ln (b+c)\right)\\ +a \left(b \left(2 \ln ^2(b)-2 \ln \left(\frac{a^2-c^2}{b^2-c^2}\right) \ln (b)+\ln ^2(a+b)-\ln ^2(c-a)-\ln ^2(a+c)\\ -3 \ln ^2(b+c)-2 \ln (b-a) \ln (c-b)+2 \ln (c-a) (\ln (b-a)+\ln (b+c))\\ +2 \ln (a+b) \ln \left(\frac{b+c}{c^2-a^2}\right)+2 \ln (a+c) \ln \left(c^2-b^2\right)\right)-c \left(\ln ^2(b)-2 (\ln ((b-a) c)-\ln (a+c)) \ln (b)+\ln ^2(b-a)+\ln ^2(a+b)\\ -3 \ln ^2(c)+\ln ^2(a+c)+\ln ^2(c-b)+2 \ln ^2(b+c)+2 \ln (b-a) \ln \left(\frac{c}{c-b}\right)\\ -2 \ln (a+b) \ln (b+c)-2 \ln (a+c) \ln (c (b+c))\right)\right)\right) \ln (a)\\ -2 (a (b-c)+b c) \ln ^3(b)+a c \ln ^3(b-a)-b c \ln ^3(b-a)-3 a c \ln ^3(c)+5 b c \ln ^3(c)\\ +2 a b \ln ^3(c-a)-2 b c \ln ^3(c-a)+2 a c \ln ^3(a+c)-2 b c \ln ^3(a+c)-a b \ln ^3(c-b)\\ -b c \ln ^3(c-b)-a b \ln ^3(b+c)-3 b c \ln ^3(b+c)-3 b c \ln (b-a) \ln ^2(c)\\ -3 a b \ln ((a-b) (b-c)) \ln ^2(c-a)+3 b c \ln ((a-b) (b-c)) \ln ^2(c-a)\\ +3 b c \ln \left(\left(b^2-a^2\right) c\right) \ln ^2(a+c)-3 a c \ln (a+b) \ln ^2(c-b)+3 a c \ln ((b-a) c) \ln ^2(c-b)\\ -3 a b (\ln (b-a)-\ln ((a+b) (a+c))) \ln ^2(c-b)+3 b c \ln (c (a+c)) \ln ^2(c-b)\\ -3 a c \ln \left(\frac{b-a}{a+b}\right) \ln ^2(b+c)+6 a c \ln (c) \ln ^2(b+c)-12 b c \ln (c) \ln ^2(b+c)\\ +3 a b \ln (a+c) \ln ^2(b+c)+3 b c \left(\ln \left(b^2-a^2\right)+\ln (a+c)\right) \ln ^2(b+c)\\ -3 a c \ln ^2(b-a) \ln \left(1-\frac{b}{c}\right)-b c \pi ^2 \left(\ln (b-a)+\ln \left(\frac{b-c}{a-c}\right)\right)-5 a c \pi ^2 \ln (c)-5 b c \pi ^2 \ln (c)\\ -3 b c \ln ^2(b-a) \ln \left(-\frac{c}{b-c}\right)-12 b c \ln ^2(c) \ln \left(\frac{c-a}{a+c}\right)-12 b c \ln (a+b) \ln (c) \ln \left(\frac{c-a}{a+c}\right)\\ +\pi ^2 a c (\ln (b-a)-\ln (a+c))+3 a b \pi ^2 \ln (a+c)+4 b c \pi ^2 \ln (a+c)-6 b c \ln ^2(c) \ln (c-b)\\ +6 a b \ln (b-a) \ln (c-a) \ln (c-b)-6 b c \ln (b-a) \ln (c-a) \ln (c-b)\\ -6 a b \ln (a+b) \ln (a+c) \ln (c-b)-6 b c \ln (a+b) \ln (a+c) \ln (c-b)\\ -12 b c \ln (c) \ln (a+c) \ln (c-b)-a b \pi ^2 (\ln (c-a)+\ln (c-b))\\ +3 a c \ln ^2(a+b) \ln \left(\frac{c (c-b)}{b+c}\right)+3 a c \ln ^2(c) (\ln (b-a)-\ln (b+c))+9 b c \ln ^2(c) \ln (b+c)\\ -6 a c \ln (a+b) \ln (c) \ln (b+c)+6 b c \ln (a+b) \ln (c) \ln (b+c)+12 b c \ln (c) \ln (c-a) \ln (b+c)\\ +5 \pi ^2 a b \ln (b+c)+6 a c \pi ^2 \ln (b+c)+3 b c \pi ^2 \ln (b+c)-3 a b \ln ^2(a+c) \ln \left(\frac{b+c}{a+b}\right)\\ +6 a c \ln (b-a) \ln (a+c) \ln (c (b+c))-6 b c \ln (b-a) \ln (a+c) \ln (c (b+c))\\ -3 a c \ln ^2(a+c) (\ln (b-a)+\ln (c (b+c)))-3 b c \ln ^2(a+b) \ln \left(-\frac{c (b+c)}{b-c}\right)\\ -3 \ln ^2(b) \left(a \left(c (\ln (b-a)+\ln (c)-2 \ln (a+c)+\ln (b+c))\\ +b \ln \left(\frac{b^2-c^2}{a^2-c^2}\right)\right)-b c \ln \left(\frac{(a-b) c (c-b)}{a^2-c^2}\right)\right)-6 a c \ln (b-a) \ln (c) \ln \left(c^2-b^2\right)\\ +6 b c \ln (b-a) \ln (c) \ln \left(c^2-b^2\right)\\ -\ln (b) \left(b c \left(3 \left(-\ln ^2(a+b)+2 (\ln ((a+c) (b+c))-\ln (c-a)) \ln (a+b)\\ +2 \ln ^2(c)+\ln ^2(c-a)+\ln ^2(a+c)-3 \ln ^2(b+c)-2 \ln (a+c) \ln (c-b)\\ +2 \ln (c-a) \ln (b+c)+2 \ln (c) (\ln ((a+c) (b+c))-2 \ln (c-a))\\ -2 \ln (b-a) \ln \left(c^2-a^2\right)+2 \ln (b-a) \ln \left(c^2-b^2\right)\right)+\pi ^2\right)\\ +a \left(c \left(6 (\ln (b-a)+\ln (c)-\ln (a+c)) \ln \left(\frac{a+c}{b+c}\right)+\pi ^2\right)\\ +3 b \left(\ln ^2(a+b)+2 \ln \left(\frac{b+c}{c^2-a^2}\right) \ln (a+b)-\ln ^2(c-a)-\ln ^2(a+c)-3 \ln ^2(b+c)\\ +2 \ln (b-a) \ln \left(\frac{a-c}{b-c}\right)+2 \ln (c-a) \ln (b+c)+2 \ln (a+c) \ln \left(c^2-b^2\right)+2 \pi ^2\right)\right)\right)\\ +6 \left(2 \left(a \ln \left(\frac{a}{c}\right)+b \ln \left(\frac{c}{b}\right)\right) \operatorname{Li}_2\left(\frac{a}{c}\right) c+2 \left(a \ln \left(\frac{a}{c}\right)+b \ln \left(\frac{c}{b}\right)\right) \operatorname{Li}_2\left(-\frac{c}{a}\right) c-2 b \ln \left(\frac{b}{c}\right) \operatorname{Li}_2\left(\frac{b (a+c)}{a (b-c)}\right) c+2 b \ln \left(\frac{b}{c}\right) \operatorname{Li}_2\left(\frac{b (a-c)}{a (b+c)}\right) c-2 a \operatorname{Li}_3\left(\frac{a}{c}\right) c+2 b \operatorname{Li}_3\left(-\frac{b}{c}\right) c-2 b \operatorname{Li}_3\left(\frac{b}{c}\right) c+2 a \operatorname{Li}_3\left(-\frac{c}{a}\right) c-(a-b) \operatorname{Li}_3\left(\frac{(b-a) c}{b (a+c)}\right) c-(a-b) \operatorname{Li}_3\left(\frac{a (c-b)}{(a-b) c}\right) c+(a+b) \operatorname{Li}_3\left(\frac{a (c-b)}{(a+b) c}\right) c-(a-b) \operatorname{Li}_3\left(\frac{b-a}{b+c}\right) c+(a+b) \operatorname{Li}_3\left(\frac{a+b}{b+c}\right) c-(a+b) \operatorname{Li}_3\left(\frac{a (b+c)}{(a+b) c}\right) c+(a-3 b) \zeta (3) c+2 a b \ln \left(\frac{a}{b}\right) \operatorname{Li}_2\left(\frac{a}{b}\right)+2 a b \ln \left(\frac{a}{b}\right) \operatorname{Li}_2\left(-\frac{b}{a}\right)+\left(a b \ln \left(\frac{a}{b}\right)+b c \ln \left(\frac{b}{c}\right)+a c \ln \left(\frac{c}{a}\right)\right) \operatorname{Li}_2\left(\frac{b-a}{b-c}\right)+\left(a b \ln \left(\frac{a}{b}\right)+b c \ln \left(\frac{b}{c}\right)+a c \ln \left(\frac{c}{a}\right)\right) \operatorname{Li}_2\left(\frac{b-c}{a+b}\right)-\left(a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)-b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{a+b}{b+c}\right)-\left(a b \ln \left(\frac{a}{b}\right)+a c \ln \left(\frac{a}{c}\right)-b c \ln \left(\frac{b}{c}\right)\right) \operatorname{Li}_2\left(\frac{a+c}{b+c}\right)-2 a b \operatorname{Li}_3\left(\frac{a}{b}\right)+2 a b \operatorname{Li}_3\left(-\frac{b}{a}\right)-b (a-c) \operatorname{Li}_3\left(\frac{a-b}{a-c}\right)-a (b-c) \operatorname{Li}_3\left(\frac{b-a}{b-c}\right)+a (b-c) \operatorname{Li}_3\left(\frac{a+b}{b-c}\right)-b (a-c) \operatorname{Li}_3\left(\frac{a (b-c)}{b (a-c)}\right)+b (a+c) \operatorname{Li}_3\left(\frac{a+b}{a+c}\right)+b (a+c) \operatorname{Li}_3\left(\frac{b (a+c)}{a (b-c)}\right)+b (a-c) \operatorname{Li}_3\left(\frac{b (a-c)}{a (b+c)}\right)-b (a-c) \operatorname{Li}_3\left(\frac{c-a}{b+c}\right)+b (a+c) \operatorname{Li}_3\left(\frac{a+c}{b+c}\right)-a (b+c) \operatorname{Li}_3\left(\frac{a (b+c)}{b (a+c)}\right)\right)\right)$$

Here is the equivalent Mathematica expression. The formula can be proved using differentiation by parameters $a,b,c.$

In fact, the integrand even has a closed-form antiderivative in terms of elementary functions, dilogarithms and trilogarithms, but it is too large to put it here.

Answer 2

$$\begin{align}I(2,1)&=\frac{\pi^2}3\ln2-\frac{\pi^2}6\ln3+2\ln^22\cdot\ln3-3\ln2\cdot\ln^23+\frac{29}{24}\ln^33\\&+\frac{73}{16}\zeta(3)-2\ln2\cdot\operatorname{Li}_2\left(\tfrac13\right)-\frac{13}4\operatorname{Li}_3\left(\tfrac13\right)-4\operatorname{Li}_3\left(\tfrac23\right)\end{align}$$

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$$\begin{align}I(3,1)&=\frac{13\,\pi^2}{12}\ln2-\frac{4\,\pi^2}9\ln3-\frac13\ln2\cdot\ln^23+\frac7{18}\ln^33\\&-\frac{13}8\zeta(3)+\ln3\cdot\operatorname{Li}_2\left(\tfrac13\right)+\frac43\operatorname{Li}_3\left(\tfrac13\right)-\frac23\operatorname{Li}_3\left(\tfrac23\right)\end{align}$$

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Update (in response to a comment): $$\begin{align}&I(\phi,1)=\frac32\ln^32+\frac{\pi^2}{12}\Big[\left(6-3\sqrt5\right)\ln2+\left(3\sqrt5-4\right)\ln\left(1+\sqrt5\right)\Big]+\frac{51-21\sqrt5}{48}\zeta(3)\\&-\frac{\ln\left(1+\sqrt5\right)}2\Bigg[15\ln^22-15\ln\left(1+\sqrt5\right)\ln2+4\ln^2\left(1+\sqrt5\right)+2\operatorname{Li}_2\left(\frac{1-\sqrt5}4\right)\Bigg]\\&-\ln\left(3+\sqrt5\right)\operatorname{Li}_2\left(\sqrt5-2\right)+\frac{11+3\sqrt5}{48}\operatorname{Li}_3\left(9-4\sqrt5\right)-\frac{13+3\sqrt5}6\operatorname{Li}_3\left(\sqrt5-2\right)\end{align}$$

Answer 3

Some further raw results (largely inspired by Cleo's answer :+1) for the impatient :

$$I\left(4,1\right)=- \frac{1915}{128}\zeta(3)+\left(\frac {359}{192}\ln(2) + 2\ln(3) + \frac{175}{96}\ln(5)\right)\pi^2 +\frac{513}{32}\ln(2)^2\ln(5) - \frac{1535}{96}\ln(2)^3 - \frac {15}4\ln(3)^3 - \frac{1031}{96}\ln(5)^3 - \frac{31}8\ln(2)\ln(5)^2 + \frac{93}8\ln(3)\ln(5)^2 + \frac {45}4\operatorname{Li}_3(1/3) +\frac{93}8\operatorname{Li}_3(2/3) - \frac{81}{32}\operatorname{Li}_3(1/5) - \operatorname{Li}_3(2/5) - \frac 38\operatorname{Li}_3(3/5) - \frac {91}{32}\operatorname{Li}_3(4/5)- \frac 38\operatorname{Li}_3(1/6) + \frac 5{16}\operatorname{Li}_3(1/10) \\- \ln(2)\left(12\operatorname{Li}_2(1/3) + 8\operatorname{Li}_2(1/4) + 2\operatorname{Li}_2(1/5)+\frac {163}{16}\operatorname{Li}_2(2/5) + \frac {195}{16}\operatorname{Li}_2(3/5)\right) \\- \frac{189}{16}\,\ln(5)\;(\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5))$$

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$$I\left(5,1\right)=- \frac{577}{160}\zeta(3)+\left(\frac {11}{15}\ln(3) + \frac{191}{480}\ln(4) - \frac{11}{60}\ln(5)\right)\pi^2 + \frac{17}{10}\ln(3)\ln(5)^2 - \frac{17}{30}\ln(3)^3 - \frac{173}{960}\ln(4)^3 - \frac{181}{120}\ln(5)^3 - \frac 1{10}\ln(4)\ln(5)^2 + \frac{51}{160}\ln(4)^2\ln(5) + 2\operatorname{Li}_3(1/3) + \frac 75\operatorname{Li}_3(2/3) - \frac 38\operatorname{Li}_3(1/5) + \frac 9{10}\operatorname{Li}_3(2/5) - \frac 35\operatorname{Li}_3(3/5) -\frac 18\operatorname{Li}_3(4/5) + \frac 14\operatorname{Li}_3(1/10) - \frac 75\ln(2)\,\left(\operatorname{Li}_2(1/3)+\frac{\operatorname{Li}_2(1/4)}2 +\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5)\right) - \ln(5)\,\left(\operatorname{Li}_2(2/5)+2\operatorname{Li}_2(3/5)\right)$$

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$$I\left(\frac 12,1\right)=-\frac{39}8\zeta(3)+\left(\ln(2)+5\ln(3)+2\ln(5)\right)\zeta(2)-\frac 5{12}\ln(3)^3+\ln(2)^2\ln(25/8)-\ln(25/9)\ln(5)^2 -\ln(4)\left(\operatorname{Li}_2(1/3)+\frac 34 \operatorname{Li}_2(1/4)\right)-(\ln(4)+2\ln(5))\;(\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5))+\frac 12\operatorname{Li}_3(1/3)+2\operatorname{Li}_3(2/3)$$

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$$I\left(\frac 13,1\right)=-\frac{139}{24}\zeta(3)+\left(\frac 94\ln(2)-\frac 5{12}\ln(3)+\frac 13\ln(5)\right)\pi^2 + 2\ln(2)^2\ln(5) - 2\ln(2)^3 -\frac 23\ln(3)^3 - 2\ln(5)^3 + 2\ln(3)\ln(5)^2 + 2\operatorname{Li}_3(1/3) + 2\operatorname{Li}_3(2/3) - 2\ln(2)\,\left(\operatorname{Li}_2(1/3)+\frac 12\operatorname{Li}_2(1/4)+\operatorname{Li}_2(2/5) +\operatorname{Li}_2(3/5)\right) - 2\ln(5)\;(\operatorname{Li}_2(2/5)+\operatorname{Li}_2(3/5))$$

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$$I\left(\frac 23,1\right)=\frac{573}{128}\zeta(3)+\left(\frac{31}{64}\ln(2)-\frac 23\ln(3)-\frac 5{48}\ln(5)\right)\pi^2 + \frac {47}{32}\ln(2)^3 + \frac 34\ln(3)^3 + \frac {79}{32}\ln(5)^3 - 3\ln(2)\ln(5)^2 + \frac{45}{32}\ln(2)^2\ln(5) - \frac 74\ln(3)\ln(5)^2 - \frac 54\operatorname{Li}_3(1/3) - \frac 94\operatorname{Li}_3(2/3) - \frac {17}{32}\operatorname{Li}_3(1/5) - \operatorname{Li}_3(1/6) - \frac {13}8\operatorname{Li}_3(2/5) - \operatorname{Li}_3(3/5) - \frac {27}{32}\operatorname{Li}_3(4/5) - \frac 5{16}\operatorname{Li}_3(1/10) + \ln(2)\,\left(\frac 54\operatorname{Li}_2(1/3) + \frac 58\operatorname{Li}_2(1/4)\right) + (\ln(3)-\ln(2))\operatorname{Li}_2(1/5) + \frac{3\ln(2)+7\ln(5)}4\,(\operatorname{Li}_2(2/5) + \operatorname{Li}_2(3/5))$$

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$$I\left(\frac 14,1\right)=\frac{335}{32}\zeta(3)-\left(\frac{95}{48}\ln(2)+\frac 23\ln(3)+\frac 1{24}\ln(5)\right)\pi^2 + \frac {107}{24}\ln(2)^3 + \frac 73\ln(3)^3 + \frac {41}8\ln(5)^3 - \ln(2)\ln(5)^2 - \frac{21}8\ln(2)^2\ln(5) - \frac {11}2\ln(3)\ln(5)^2 - 7\operatorname{Li}_3(1/3) - \frac{11}2\operatorname{Li}_3(2/3) - \frac 78\operatorname{Li}_3(1/5) + \operatorname{Li}_3(2/5) - \frac 32\operatorname{Li}_3(3/5) + \frac 38\operatorname{Li}_3(4/5) - \frac 32\operatorname{Li}_3(1/6)- \frac 54\operatorname{Li}_3(1/10) + \ln(2)\;\left(4\operatorname{Li}_2(1/3) + 2\operatorname{Li}_2(1/4) + \frac 54\operatorname{Li}_2(2/5) + \frac {13}4\operatorname{Li}_2(3/5)\right) + \frac {19}4\ln(5)\;\left(\operatorname{Li}_2(2/5) + \operatorname{Li}_2(3/5)\right)$$

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Raw expressions for (the much needed) simplifications and manipulations
($I(4,1),I(5,1),I(1/2,1),I(1/3,1),I(2/3,1),I(1/4,1)$ respectively)

- 1915/128*zeta(3)+(359/192*ln(2) + 2*ln(3) + 175/96*ln(5))*PI^2 +(513/32*ln(2)^2*ln(5) - 1535/96*ln(2)^3 - 15/4*ln(3)^3 - 1031/96*ln(5)^3 - 31/8*ln(2)*ln(5)^2  + 93/8*ln(3)*ln(5)^2 + 45/4*polylog(3, 1/3) +93/8*polylog(3, 2/3) - 81/32*polylog(3, 1/5) - 3/8*polylog(3, 1/6) - polylog(3, 2/5) - 3/8*polylog(3, 3/5) - 91/32*polylog(3, 4/5) + 5/16*polylog(3, 1/10) - 12*ln(2)*polylog(2, 1/3) - 8*ln(2)*polylog(2, 1/4) - 2*ln(2)*polylog(2, 1/5) - 163/16*ln(2)*polylog(2, 2/5) - 195/16*ln(2)*polylog(2, 3/5) - 189/16*ln(5)*polylog(2, 2/5) - 189/16*ln(5)*polylog(2, 3/5))

- 577/160*zeta(3) +(191/240*ln(2)+11/15*ln(3)-11/60*ln(5))*PI^2 + 17/10*ln(3)*ln(5)^2 - 17/30*ln(3)^3 - 173/960*ln(4)^3 - 181/120*ln(5)^3  - 1/10*ln(4)*ln(5)^2 + 51/160*ln(4)^2*ln(5) + 2*polylog(3, 1/3) + 7/5*polylog(3, 2/3) - 3/8*polylog(3, 1/5) + 9/10*polylog(3, 2/5) - 3/5*polylog(3, 3/5) - 1/8*polylog(3, 4/5) + 1/4*polylog(3, 1/10) - 7/10*ln(4)*(polylog(2,1/3)+polylog(2,1/4)/2+polylog(2,2/5)+polylog(2,3/5)) - ln(5)*polylog(2, 2/5) - 2*ln(5)*polylog(2, 3/5)

-39/8*zeta(3)+(ln(2)+5*ln(3)+2*ln(5))*zeta(2)-5/12*ln(3)^3+ln(2)^2*ln(25/8)-ln(25/9)*ln(5)^2 -ln(4)*(polylog(2,1/3)+3/4*polylog(2,1/4))- (ln(4)+2*ln(5))*(polylog(2,2/5)+polylog(2,3/5))+polylog(3,1/3)/2+2*polylog(3,2/3)

-139/24*zeta(3)+(9/4*ln(2)-5/12*ln(3)+1/3*ln(5))*PI^2 + 2*ln(2)^2*ln(5) - 2*ln(2)^3 -2/3*ln(3)^3 - 2*ln(5)^3  + 2*ln(3)*ln(5)^2 + 2*polylog(3,1/3) + 2*polylog(3, 2/3) - 2*ln(2)*(polylog(2, 1/3)+polylog(2, 1/4)/2+polylog(2, 2/5) +polylog(2, 3/5)) - 2*ln(5)*(polylog(2,2/5)+polylog(2, 3/5))

573/128*zeta(3)+(31/64*ln(2)-2/3*ln(3)-5/48*ln(5))*PI^2 + 47/32*ln(2)^3 + 3/4*ln(3)^3 + 79/32*ln(5)^3 - 3*ln(2)*ln(5)^2 + 45/32*ln(2)^2*ln(5) - 7/4*ln(3)*ln(5)^2 - 5/4*polylog(3, 1/3) - 9/4*polylog(3, 2/3) - 17/32*polylog(3, 1/5) - polylog(3, 1/6) - 13/8*polylog(3, 2/5) - polylog(3, 3/5) - 27/32*polylog(3, 4/5) - 5/16*polylog(3, 1/10) + ln(2)*(5/4*polylog(2,1/3) + 5/8*polylog(2, 1/4)) + (ln(3)-ln(2))*polylog(2, 1/5) + (3*ln(2)+7*ln(5))/4*(polylog(2, 2/5) + polylog(2, 3/5))

335/32*zeta(3)-(95/48*ln(2)+2/3*ln(3)+1/24*ln(5))*PI^2 + 107/24*ln(2)^3 + 7/3*ln(3)^3 + 41/8*ln(5)^3 - ln(2)*ln(5)^2 - 21/8*ln(2)^2*ln(5) - 11/2*ln(3)*ln(5)^2 - 7*polylog(3, 1/3) - 11/2*polylog(3, 2/3) - 7/8*polylog(3, 1/5) + polylog(3, 2/5) - 3/2*polylog(3, 3/5) + 3/8*polylog(3, 4/5) - 3/2*polylog(3, 1/6)- 5/4*polylog(3, 1/10) + ln(2)*(4*polylog(2, 1/3) + 2*polylog(2, 1/4) + 5/4*polylog(2, 2/5) + 13/4*polylog(2,3/5)) + 19/4*ln(5)*(polylog(2, 2/5) + polylog(2, 3/5))

Answer 4

By the substitution $x\mapsto x^{-1}$ it is not hard to see that $$I(\nu^{-1},1)=\int^\infty_0\frac{\arctan^2{x}\arctan(\nu x)}{x^2}{\rm d}x$$

First, start off by re-expressing the integral \begin{align} \int^\pi_0x^2\cos(nx)\cot\left(\frac{x}{2}\right)\ {\rm d}x &=-{\rm Re}\int_C\frac{z+1}{z-1}z^{n-1}\ln^2{z}\ {\rm d}z\\ &=(-1)^{n}\int^1_0\frac{1-x}{1+x}x^{n-1}(\ln^2{x}-\pi^2)\ {\rm d}x-\int^1_0\frac{1+x}{1-x}x^{n-1}\ln^2{x}\ {\rm d}x\\ \end{align} where $C$ is the arc joining $z=1$ to $z=-1$. (The first equality follows from $z=e^{ix}$, whereas the second follows from the fact that the integral along $[-1,\epsilon]\cup\epsilon\exp(i[\pi,0])\cup[\epsilon,1]\cup C$ is $0$ since $z=1$ is a removable singularity and the indent around $z=0$ vanishes.)

Next, note that \begin{align} I_\nu(\nu^{-1},1) &=\int^\frac{\pi}{2}_0\frac{x^2}{\tan{x}(\cos^2{x}+\nu^2\sin^2{x})}{\rm d}x\\ &=\frac{1}{4}\int^\pi_0\frac{x^2}{\tan\left(\frac{x}{2}\right)(1+(1-\nu^2)\cos{x}+\nu^2)}{\rm d}x\\ &=\frac{1}{8\nu}\int^\pi_0\left(x^2\cot\left(\frac{x}{2}\right)+2\sum^\infty_{n=1}\left(\frac{\nu-1}{\nu+1}\right)^nx^2\cos(nx)\cot\left(\frac{x}{2}\right)\right){\rm d}x\\ &=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\int^1_0\frac{1-x}{(1+x)(1+\xi x)}(\ln^2{x}-\pi^2)+\frac{1+x}{(1-x)(1-\xi x)}\ln^2{x}\ {\rm d}x\right) \end{align} Here $\xi=\dfrac{\nu-1}{\nu+1}$. Utilising the partial fraction decompositions $$\frac{1-x}{(1+x)(1+\xi x)}=\frac{\nu+1}{1+x}-\frac{\nu}{1+\xi x}$$ $$\frac{1+x}{(1-x)(1-\xi x)}=\frac{\nu+1}{1-x}-\frac{\nu}{1-\xi x}$$ in tandem with the easily verifiable fact $$\int^1_0\frac{\ln^2{x}}{1+\lambda x}{\rm d}x=-\frac{2{\rm Li}_3(-\lambda)}{\lambda}$$ yields \begin{align} I_\nu(\nu^{-1},1) &=\frac{\pi^2}{4\nu}\ln{2}-\frac{7\zeta(3)}{8\nu}-\frac{\xi}{4\nu}\left(\frac{\pi^2\nu}{\xi}\ln(1+\xi)-\pi^2(\nu+1)\ln{2}+\frac{7\zeta(3)}{2}(\nu+1)-\frac{4\nu}{\xi}\chi_3(\xi)\right)\\ &=\chi_3\left(\frac{\nu-1}{\nu+1}\right)-\frac{7\zeta(3)}{8}+\frac{\pi^2}{4}\ln\left(1+\frac{1}{\nu}\right) \end{align} where $\displaystyle\chi_s(z)=\sum_{n\ \text{odd}}\frac{z^n}{n^s}=\frac{1}{2}\left({\rm Li}_s(z)-{\rm Li}_s(-z)\right)$ is the Legendre-chi function.

Integrating back, \begin{align} I_\nu(\nu^{-1},1) &=\underbrace{\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)-\frac{7\zeta(3)}{8}\nu}_{\text{Let this be C}}+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_3(v)}{(1+v)^2}{\rm d}v\\ &=C-\left.\frac{2\chi_3(v)}{1+v}\right|^\frac{1-\nu}{1+\nu}_1+2\int^\frac{1-\nu}{1+\nu}_1\frac{\chi_2(v)}{v(1+v)}{\rm d}v\\ &=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}-\left.\color{white}{\frac{1}{1}}2\chi_2(v)\ln(1+v)\right|^\frac{1-\nu}{1+\nu}_1+\int^\frac{1-\nu}{1+\nu}_1\frac{\ln(1+v)\ln\left(\frac{1+v}{1-v}\right)}{v}{\rm d}v\\ &=C+(1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-\frac{7\zeta(3)}{8}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln{2}\\ &\ \ \ \ +\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)-\ln^2(1-v)+\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v \end{align} Repeatedly integrating by parts, it is not hard to derive that \begin{align} \frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1+v)}{v}{\rm d}v =&\ -\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)+\frac{7\zeta(3)}{8}-{\rm Li}_3\left(\frac{1+\nu}{2}\right)+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)\\ &\ +\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)\\ -\frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2(1-v)}{v}{\rm d}v =&\ {\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)\\ \frac{1}{2}\int^\frac{1-\nu}{1+\nu}_1\frac{\ln^2\left(\frac{1-v}{1+v}\right)}{v}{\rm d}v =&\ -2\chi_3(\nu)+2\chi_2(\nu)\ln{\nu}+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right) \end{align} Therefore, we get, for $I(\nu^{-1},1)$, \begin{align} I(\nu^{-1},1) =&\color{brown}{\ (1-\nu)\chi_3\left(\frac{1-\nu}{1+\nu}\right)-2\chi_3(\nu)+{\rm Li}_3\left(\frac{2\nu}{1+\nu}\right)-{\rm Li}_3\left(\frac{1+\nu}{2}\right)-\frac{7\zeta(3)}{8}\nu}\\ &\ \color{brown}{+2\chi_2(\nu)\ln{\nu}+2\chi_2\left(\frac{1-\nu}{1+\nu}\right)\ln\left(\frac{1+\nu}{2}\right)-{\rm Li}_2\left(\frac{2\nu}{1+\nu}\right)\ln\left(\frac{2\nu}{1+\nu}\right)}\\ &\ \color{brown}{+{\rm Li}_2\left(\frac{1+\nu}{2}\right)\ln\left(\frac{1+\nu}{2}\right)+\frac{\pi^2}{4}\ln\left(\frac{(1+\nu)^{1+\nu}}{\nu^\nu}\right)+\frac{\pi^2}{4}\ln{2}}\\ &\ \color{brown}{+\frac{1}{2}\ln^2\nu\ln\left(\frac{1-\nu}{1+\nu}\right)+\frac{1}{2}\ln\left(\frac{1-\nu}{2}\right)\ln^2\left(\frac{1+\nu}{2}\right)-\frac{1}{6}\ln^3\left(\frac{1+\nu}{2}\right)}\\ &\ \color{brown}{-\frac{1}{2}\ln\left(\frac{1-\nu}{1+\nu}\right)\ln^2\left(\frac{2\nu}{1+\nu}\right)} \end{align} If no mistakes were made, this formula should hold for $0<\nu\le1$. Further simplifications to the formula may be possible through some polylogarithm identities.