Evaluate $\int_0^1 \frac{x \operatorname{Li}_2(x) \log (1+x)}{x^2+1} \, dx$

Question

$$\int_0^1 \frac{x \operatorname{Li}_2(x) \log (1+x)}{x^2+1} \, dx=-\frac{3\pi }{4} \Im(\operatorname{Li}_3(1+i))+\frac{189}{128} \zeta (3) \log (2)+\frac{C^2}{2}-\frac{1}{4} \pi C \log (2)+\frac{ \operatorname{Li}_4\left(\frac{1}{2}\right)}{8}+\frac{89 \pi ^4}{5760}+\frac{ \log ^4(2)}{192}+\frac{7}{96} \pi ^2 \log ^2(2)+\frac{1}{4}\int_0^1\frac{\ln(1-x)\operatorname{Li}_2(-x^2)}{1+x}dx$$ I was inspired by the method used by Ali Shather (see here) to calculate this integral. But here, without going into the details of the long and tedious calculations, I could only deduce this relation. I believe that the last integral has never been calculated, is therefore a closed form for the proposed integral?$$$$

Answer 1

$$I=\int_0^1\frac{x\operatorname{Li}_2(x)\ln(1+x)}{1+x^2}\ dx=-\int_0^1\ln(y)\left(\int_0^1\frac{x^2\ln(1+x)}{(1+x^2)(1-xy)}dx\right)dy$$

$$=-\int_0^1\ln(y)\left(-\frac{1}{1+y^2}\frac{\pi\ln(2)}{8}-\frac{y}{1+y^2}\left(\frac{\pi^2}{96}+\frac{\ln^2(2)}{8}\right)-\frac{\ln2\ln(1-y)}{(1+y^2)y}+\frac{\ln2\ln(1+y)}{(1+y^2)y}+\frac{\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{y(1+y^2)}-\frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{y(1+y^2)}\right)\ dy$$ $$\int_0^1\ln(y)\frac{\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{y(1+y^2)} \, dy = \int_0^1 \ln(y) \frac{\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{y}dy-\int_0^1{y\ln(y)}\frac{\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{(1+y^2)}\,dy$$

$$\int_0^1{y\ln(y)}\frac{\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{(1+y^2)}dy=\frac{C^2}{2}+\frac{15 \operatorname{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{701 \pi ^4}{46080}+\frac{7}8\ln2\zeta(3)+\frac{5 \log ^4(2)}{128}-\frac{3}{64} \pi ^2 \log ^2(2)$$ $$\int_0^1 \ln(y) \frac{\operatorname{Li}_2\left(\frac{y}{1+y}\right)}{y(1+y^2)}dy=-\frac{C^2}{2}+\frac{17 \operatorname{Li}_4\left(\frac{1}{2}\right)}{16}-\frac{707 \pi ^4}{46080}+\frac{7}8\ln2\zeta(3)+\frac{17 \log ^4(2)}{384}-\frac{7}{192} \pi ^2 \log ^2(2)$$ $$I=\int_0^1 \ln(y) \frac{\operatorname{Li}_2 \left(\frac{2y}{1+y}\right)}{y(1+y^2)}\,dy+\frac{C^2}{2}-\frac{1}{8} \pi C \log (2)-\frac{17 \operatorname{Li}_4 \left(\frac{1}{2}\right)}{16}+\frac{697 \pi ^4}{46080}-\frac{17 \log ^4(2)}{384}+\frac{61}{384} \pi ^2 \log ^2(2)$$ $$\int_0^1\ln(y)\frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{y(1+y^2)}dy=\int_0^1\ln(y)\frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{y}dy-\int_0^1 y\ln(y) \frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{(1+y^2)} \,dy$$ $$\int_0^1\ln(y)\frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{y}\,dy={4 \operatorname{Li}_4\left(\frac{1}{2}\right)}+\frac{7}4\ln2\zeta(3)-\frac{17 \pi ^4}{360}+\frac{ \log ^4(2)}{6}-\frac{1}{6} \pi ^2 \log ^2(2)$$ $$\int_0^1 y\ln(y) \frac{\operatorname{Li}_2 \left(\frac{2y}{1+y}\right)}{(1+y^2)}dy=-\frac{1}{2} \int_0^1\ln(1+y^2)\frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{y}dy+\frac{1}{2} \int_0^1\log y \log(1+y^2) \left[\log\left(\frac{1-y}{1+y}\right)\right]\operatorname{d}\!y\frac{1}{y(1+y}$$ $$\int_0^1\log y \log (1+y^2) \left[\log\left(\frac{1-y}{1+y}\right)\right] \, \mathrm{d}y\frac{1}{y(1+y}=2 \pi \Im(\operatorname{Li}_3(1+i))+ {10\operatorname{Li}_4 \left(\frac{1}{2}\right)}+\frac{35}{8} \zeta (3) \log (2)-\frac{89 \pi ^4}{576}+\frac{5}{12} \log ^4(2)-\frac{17}{48} \pi^2 \log^2(2)$$ $$\int_0^1 y\ln(y) \frac{\operatorname{Li}_2 \left(\frac{2y}{1+y}\right)}{(1+y^2)}\,dy=-\frac{1}{2}\int_0^1\ln(1+y^2)\frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{y}dy+\frac{1}{2} \int_0^1\log y \log (1+y^2) \left[\log\left(\frac{1-y}{1+y}\right)\right] \operatorname{d}\!y\frac{1}{y(1+y}$$ $$\int_0^1\log y \log \left(1+y^2\right) \left[\log\left(\frac{1-y}{1+y}\right) \right] \, \mathrm{d}y\frac{1}{y(1+y}=2 \pi \Im(\operatorname{Li}_3(1+i))+ {10\operatorname{Li}_4 \left(\frac{1}{2}\right)}+\frac{35}{8} \zeta (3) \log (2)-\frac{89 \pi^4}{576}+\frac{5}{12} \log^4(2)-\frac{17}{48} \pi^2 \log^2(2)$$ $$K=\int_0^1\ln(1+y^2)\frac{\operatorname{Li}_2\left(\frac{2y}{1+y}\right)}{y}dy=\frac{\pi^4}{144}(IBP)-\frac{1}{2}\int_0^1\ln\left(\frac{1-y}{1+y} \right) \frac{\operatorname{Li}_2(-y^2)}{y} \, dy+\frac{1}{2} \int_0^1\ln\left(\frac{1-y}{1+y}\right) \frac{\operatorname{Li}_2(-y^2)}{1+y}\, dy$$ $$\int_0^1\ln\left(\frac{1-y}{1+y}\right) \frac{\operatorname{Li}_2 (-y^2)}{y} \, dy = -4 \operatorname{Li}_4 \left(\frac{1}{2}\right) -\frac{7}2 \ln2 \zeta(3)+\frac{151 \pi^4}{2880}-\frac{ \log^4(2)}{6} + \frac{1}{6} \pi ^2 \log^2(2)$$ $$K= 2 \operatorname{Li}_4\left(\frac{1}{2} \right) + \frac{7}4\ln2\zeta(3)-\frac{37 \pi ^4}{1920}+\frac{ \log ^4(2)}{12}-\frac{1}{12} \pi ^2 \log ^2(2)+\frac{1}{2}\int_0^1\ln\left(\frac{1-y}{1+y}\right) \frac{\operatorname{Li}_2(-y^2)}{1+y} \, dy$$ $$\int_0^1 \ln(1+y) \frac{\operatorname{Li}_2(-y^2)}{1+y} \, dy = -\pi \Im(\operatorname{Li}_3(1+i))-\frac{19 \operatorname{Li}_4\left(\frac{1}{2}\right)}{4}-\frac{133}{32} \zeta (3) \log (2)+\frac{185 \pi^4}{2304}-\frac{19}{96} \log^4(2)+\frac{7}{32} \pi ^2 \log^2(2)+\frac{1}{2} \pi C \log (2)$$