Compute $\int_0^1 \frac{\text{Li}_2(-x^2)\log (x^2+1)}{x^2+1} \, dx$ How can we evaluate: $$\int_0^1 \frac{\text{Li}_2\left(-x^2\right) \log \left(x^2+1\right)}{x^2+1} \, dx$$
Any help will be appreciated.

Here is a different transformation that might be easier than the integral in the question. 
Recall the identity
$$\sum_{n=1}^\infty H_nH_n^{(2)}x^n=
\frac{\operatorname{Li}_3(x)+\operatorname{Li}_3(1-x)+\frac12\ln x\ln^2(1-x)-\zeta(2)\ln(1-x)-\zeta(3)}{1-x}$$
where if we replace $x$ by $-x^2$ then $\int_0^1$ we have
$$\sum_{n=1}^\infty \frac{(-1)^nH_nH_n^{(2)}}{2n+1}$$
$$=\int_0^1
\frac{\operatorname{Li}_3(-x^2)+\operatorname{Li}_3(1+x^2)+\frac12\ln(-x^2)\ln^2(1+x^2)-\zeta(2)\ln(1+x^2)-\zeta(3)}{1+x^2}dx$$
