How can i evaluate $\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{x}\:dx$ I want to evaluate $$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(1-x\right)\ln \left(1+x\right)}{x}\:dx$$
I tried integration by parts and shape it in a way that i could expand either $\ln$ terms.
$$-\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1-x\right)}{1+x}dx+\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1+x\right)}{1-x}\:dx$$
After this i tried expand the terms but i still couldnt go through, any different approaches are welcome.
 A: To evaluate this you can make use of the following identity
$$\ln \left(1-x\right)\ln \left(1+x\right)=-\sum _{k=1}^{\infty }x^{2k}\frac{H_{2k}-H_k}{k}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{x^{2k}}{k^2}$$
Resuming on your integral,
$$\int _0^1\frac{\ln \left(1-x\right)\ln ^2\left(x\right)\ln \left(1+x\right)}{x}\:dx$$
$$=-\sum _{k=1}^{\infty }\frac{H_{2k}-H_k}{k}\int _0^1x^{2k-1}\ln ^2\left(x\right)\:dx-\frac{1}{2}\sum _{k=1}^{\infty }\frac{1}{k^2}\int _0^1x^{2k-1}\ln ^2\left(x\right)\:dx$$
$$=-\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_{2k}}{k^4}+\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_k}{k^4}-\frac{1}{8}\sum _{k=1}^{\infty }\frac{1}{k^5}$$
$$=-\frac{7}{4}\sum _{k=1}^{\infty }\frac{H_k}{k^4}-2\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k^4}-\frac{1}{8}\zeta \left(5\right)$$
$$=-\frac{21}{4}\zeta \left(5\right)+\frac{7}{4}\zeta \left(2\right)\zeta \left(3\right)-\zeta \left(2\right)\zeta \left(3\right)+\frac{59}{16}\zeta \left(5\right)-\frac{1}{8}\zeta \left(5\right)$$
$$=\frac{3}{4}\zeta \left(2\right)\zeta \left(3\right)-\frac{27}{16}\zeta \left(5\right)$$
Those sums are evaluated here.
