Is there a closed form for the integral $ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $ Is there a closed form for the integral 
$$ \int_{0}^{1}\frac{\mathrm{Li}_2(-x^2)}{x^2+1} \mathrm{d}x $$
where $ \displaystyle \mathrm{Li}_2(x) \;\;=\;\; \sum_{n=1}^\infty \frac{x^n}{n^2}  $
a related sum is $\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n H_n^{(2)}}{2n+1} $
 A: \begin{align}
I&=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\
&=-\frac{\pi^3}{48}+2\int_0^1\frac{\arctan x\ln(1+x^2)}{x}\ dx\tag1\\
&=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}\int_0^1 x^{2n}\ dx\tag2\\
&=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{(2n+1)^2}\tag3\\
&=-\frac{\pi^3}{48}-4\sum_{n=0}^\infty\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+4\underbrace{\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^3}}_{\beta(3)=\frac{\pi^3}{32}}\tag4\\
&=\frac{5\pi^3}{48}-4\Im\sum_{n=1}^\infty\frac{(i)^nH_n}{n^2}\tag5\\
&=\frac{5\pi^3}{48}-4\left(-\frac{\pi}{16}\ln^22-\frac12\ln2\ G-\Im\operatorname{Li}_3(1-i)\right)\tag6\\
&=\frac{5\pi^3}{48}+\frac{\pi}{4}\ln^22+2\ln2\ G+4\Im\operatorname{Li}_3(1-i)
\end{align}

The related sum can be obtained from the the generating function
$$\sum_{n=1}^\infty H_n^{(q)}x^n=\frac{\operatorname{Li}_q(x)}{1-x}$$
set $q=2$ and replace $x$ with $-x^2$ we get
$$\sum_{n=1}^\infty (-1)^nH_n^{(2)}x^{2n}=\frac{\operatorname{Li}_2(-x^2)}{1+x^2}$$
now integrate both sides from $x=0$ to $1$
$$\sum_{n=1}^\infty \frac{(-1)^nH_n^{(2)}}{2n+1}=\int_0^1\frac{\operatorname{Li}_2(-x^2)}{1+x^2}\ dx$$
The index $n$ can start from $0$ as $H_{0}^{(2)}=0$
.

Explanation:
$(1)$ Integration by parts.
$(2)$ Using the identity $\arctan x\ln(1+x^2)=-2\sum_{n=0}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}$.
$(3)$ $\int_0^1 x^{a-1}\ dx=\frac1a$
$(4)$ Using $H_n=H_{n+1}-\frac1{n+1}$
$(5)$ Using $\sum_{n=0}^\infty (-1)^n a_{2n+1}=\Im\sum_{n=1}^\infty i ^n a_n$
$(6)$ Using the generating function  $\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$
