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.
 A: 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
$$\sum_{n=1}^\infty \frac{(-1)^nH_nH_n^{(2)}}{2n+1}$$
$$=\int_0^1
\frac{\operatorname{Li}_3(-x^2)+\operatorname{Li}_3(1+x^2)+\frac12\ln(-x^2)\ln^2(1+x^2)-\zeta(2)\ln(1+x^2)-\zeta(3)}{1+x^2}dx$$
A: Using double integration we have:
$$\scriptsize I=\frac{\pi ^2 C}{12}+2 C \log ^2(2)-16 \Im(\text{Li}_4(1+i))-\frac{21 \pi  \zeta (3)}{8}+\frac{1}{6} \pi  \log ^3(2)+\frac{5}{24} \pi ^3 \log (2)+\frac{11 \psi ^{(3)}\left(\frac{1}{4}\right)}{768}-\frac{11 \psi ^{(3)}\left(\frac{3}{4}\right)}{768}$$
And a corollary (thank to Shadhar)
$$\scriptsize \sum_{n=1}^\infty \frac{(-1)^nH_nH_n^{(2)}}{2n+1}=-\frac{\pi ^2 C}{12}+C \log ^2(2)-24 \Im(\text{Li}_4(1+i))+4 \log (2) \Im(\text{Li}_3(1+i))+\frac{5}{24} \pi ^3 \log (2)+\frac{5 \psi ^{(3)}\left(\frac{1}{4}\right)}{384}-\frac{5 \psi ^{(3)}\left(\frac{3}{4}\right)}{384}$$
