I'd like to evaluate and find the closed form for $$\int _0^1\frac{\ln \left(x\right)\ln ^2\left(\frac{1-x}{1+x}\right)\operatorname{Li}_2\left(x\right)}{x}\:dx$$ But it's very difficult since its related to alternating harmonic series of weight 6.
To show so one can make use of $\frac{1}{4}\ln ^2\left(\frac{1-x}{1+x}\right)=-\frac{1}{2}\sum _{n=1}^{\infty }\frac{\left(H_n-2H_{2n}\right)}{n}x^{2n}$ and obtain: $$-2\sum _{n=1}^{\infty }\frac{\left(H_n-2H_{2n}\right)}{n}\int _0^1x^{2n-1}\ln \left(x\right)\operatorname{Li}_2\left(x\right)\:dx$$ $$=-\frac{1}{2}\underbrace{\sum _{n=1}^{\infty }\frac{H_nH_{2n}}{n^4}}_{S_1}-\frac{1}{2}\underbrace{\sum _{n=1}^{\infty }\frac{H_nH_{2n}^{\left(2\right)}}{n^3}}_{S_2}+\zeta \left(2\right)\sum _{n=1}^{\infty }\frac{H_n}{n^3}+\underbrace{\sum _{n=1}^{\infty }\frac{H_{2n}^2}{n^4}}_{S_3}+\underbrace{\sum _{n=1}^{\infty }\frac{H_{2n}H_{2n}^{\left(2\right)}}{n^3}}_{S_4}-2\zeta \left(2\right)\sum _{n=1}^{\infty }\frac{H_{2n}}{n^3}$$ From this it can be seen that some of the series like $S_1, S_2, S_3$ and $S_4$ are very difficult yet one can find relations involving them such as $$5\sum _{n=1}^{\infty }\frac{\left(-1\right)^nH_n}{n^5}-\sum _{n=1}^{\infty }\frac{\left(-1\right)^nH_n^2}{n^4}+\sum _{n=1}^{\infty }\frac{\left(-1\right)^nH_n^{\left(2\right)}}{n^4}=-\frac{175}{32}\zeta \left(6\right)+\frac{1}{2}\zeta ^2\left(3\right)$$ Yet it's still quite hard, I'd appreciate any insight, thanks.