I am trying to evaluate $$\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(x\right)}{1+x}\:dx$$ But I am not sure what to do since integration by parts is not possible here.
I tried using a trilogarithm functional equation and got the following integrals. $$-\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(1-x\right)}{1+x}\:dx-\underbrace{\int _0^1\frac{\ln \left(1-x\right)\operatorname{Li}_3\left(\frac{x-1}{x}\right)}{1+x}\:dx}_{J}+\zeta (3)\int _0^1\frac{\ln \left(1-x\right)}{1+x}\:dx$$ $$+\frac{1}{6}\int _0^1\frac{\ln ^3\left(x\right)\ln \left(1-x\right)}{1+x}\:dx+\zeta (2)\int _0^1\frac{\ln \left(x\right)\ln \left(1-x\right)}{1+x}\:dx-\frac{1}{2}\int _0^1\frac{\ln ^2\left(x\right)\ln ^2\left(1-x\right)}{1+x}\:dx$$ All the integrals here are actually manageable except $J$.
Any idea on how to evaluate the main integral differently or the integral $J$? Thank you.