I was evaluating and integral involving iterated logarithms when the following integral appeared: $$\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \mathrm{d}x$$ which I have no idea how to tackle. I am sure it can be represented as an Euler sum, but do not know how to arrive at such a result. Wolfram gives the equality of $$\int_0^1\frac{\log^2x\log\left(1+\frac{1}{x}\right)\log^2\left(1+x\right)}{x\left(1+x\right)}\ \mathrm{d}x =0.305108.$$ Any hints would be greatly appreciated; thanks!
Related problem: Compute $\sum_{n=1}^\infty\frac{H_n^3}{n^4}-3\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}$
$$I=\int_0^1\frac{\ln^2x\log\left(1+\frac{1}{x}\right)\ln^2\left(1+x\right)}{x\left(1+x\right)}dx$$
$$\small{=\int_0^1\frac{\ln^2x\ln^3(1+x)}{x}dx-\int_0^1\frac{\ln^2x\ln^3(1+x)}{1+x}dx-\int_0^1\frac{\ln^3x\ln^2(1+x)}{x}dx+\int_0^1\frac{\ln^3x\ln^2(1+x)}{1+x}dx}$$
$$=I_1-I_2-I_3+I_4$$
By integration by parts
$$I_2=-\frac12\int_0^1\frac{\ln x\ln^4(1+x)}{x}dx$$
$$I_4=-\int_0^1\frac{\ln^2x\ln^3(1+x)}{x}dx=-I_1$$
Thus, the integral boils down to
$$I=\frac12\int_0^1\frac{\ln x\ln^4(1+x)}{x}dx-\int_0^1\frac{\ln^3x\ln^2(1+x)}{x}dx$$
$$=\frac12A-B$$
The first integral is calculated here
$$A\small{=-120\operatorname{Li}_6\left(\frac12\right)-72\ln2\operatorname{Li}_5\left(\frac12\right)-24\ln^22\operatorname{Li}_4\left(\frac12\right)+78\zeta(6)+\frac34\ln2\zeta(5)-\frac32\ln^22\zeta(4)}$$ $$\small{-3\ln^32\zeta(3)+2\ln^42\zeta(2)+12\zeta^2(3)-12\ln2\zeta(2)\zeta(3)-\frac{17}{30}\ln^62+24\sum_{n=1}^\infty\frac{H_n}{n^52^n}}$$
For the second integral, use $\ln^2(1+x)=2\sum_{n=1}^\infty\frac{(-1)^n}{n}H_{n-1}x^n$
$$B=2\sum_{n=1}^\infty\frac{(-1)^n}{n}H_{n-1}\int_0^1 x^{n-1}\ln^3xdx=-12\sum_{n=1}^\infty\frac{(-1)^n}{n^5}H_{n-1}$$
$$=-12\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}+12\sum_{n=1}^\infty\frac{(-1)^n}{n^6}=-12\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}-\frac{93}{8}\zeta(6)$$
Collecting the two integrals we have
$$I=\small{-60\operatorname{Li}_6\left(\frac12\right)-36\ln2\operatorname{Li}_5\left(\frac12\right)-12\ln^22\operatorname{Li}_4\left(\frac12\right)+\frac{405}{8}\zeta(6)+\frac38\ln2\zeta(5)-\frac34\ln^22\zeta(4)}$$ $$\small{-\frac32\ln^32\zeta(3)+\ln^42\zeta(2)+6\zeta^2(3)-6\ln2\zeta(2)\zeta(3)-\frac{17}{60}\ln^62+12\sum_{n=1}^\infty\frac{H_n}{n^52^n}+12\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^5}}$$
Unfortunately, there is no closed form for your integral as the latter two sums have no known closed forms.