Calculating $\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$ Do you see any fast way of calculating this one? $$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^2 \, dx$$
Numerically, it's about 
$$\approx 111.024457130115028409990464833072173251135063166330638343951498907293$$
or in a predicted closed form 
$$\frac{4 }{3}\pi ^3+32 \pi  \log (2).$$
Ideas, suggestions, opinions are welcome, and the solutions are optionals.
Supplementary question for the integrals lovers: calculate in closed form
$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^3 \, dx.$$ 
As a note, it would be remarkable to be able to find a solution for the generalization below
$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^n \, dx.$$ 

$$I=\int_0^\infty\operatorname{Li}_2^2\left(-\frac1{x^2}\right)\ dx\overset{IBP}{=}-4\int_0^\infty\operatorname{Li}_2\left(-\frac1{x^2}\right)\ln\left(1+\frac1{x^2}\right)\ dx$$
By using $$\int_0^1\frac{x\ln^n(u)}{1-xu}\ du=(-1)^n n!\operatorname{Li}_{n+1}(x)$$  setting $n=1$ and replacing $x$ with $-\frac1{x^2}$ we can write
$$\operatorname{Li}_2\left(-\frac1{x^2}\right)=\int_0^1\frac{\ln u}{u+x^2}\ du$$
Then
\begin{align}
I&=-4\int_0^1\ln u\left(\int_0^\infty\frac{\ln\left(1+
\frac1{x^2}\right)}{u+x^2}\ dx\right)\ du\\
&=-4\int_0^1\ln u\left(\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)\ du, \quad \color{red}{\sqrt{u}=x}\\
&=-16\pi\int_0^1\ln x\left(\ln(1+x)-\ln x\right)\ dx\\
&=-16\pi\left(2-\frac{\pi^2}{12}-2\ln2-2\right)\\
&=\boxed{\frac43\pi^3+32\pi\ln2}
\end{align}


Proof of $\ \displaystyle\int_0^\infty\frac{\ln\left(1+
\frac1{x^2}\right)}{u+x^2}\ dx=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$

Let  $$J=\int_0^\infty\frac{\ln\left(1+\frac1{x^2}\right)}{u+x^2}\ dx\overset{x\mapsto\ 1/x}{=}\int_0^\infty\frac{\ln(1+x^2)}{1+ux^2}\ dx$$
and $$J(a)=\int_0^\infty\frac{\ln(1+ax^2)}{1+ux^2}\ dx, \quad J(0)=0,\quad J(1)=J$$
\begin{align}
J'(a)&=\int_0^\infty\frac{x^2}{(1+ux^2)(1+ax^2)}\ dx\\
&=\frac1{a-u}\int_0^\infty\left(\frac1{1+ux^2}-\frac1{1+ax^2}\right)\ dx\\
&=\frac1{a-u}*\frac{\pi}{2}\left(\frac1{\sqrt{u}}-\frac1{\sqrt{a}}\right)\\
&=\frac{\pi}{2}\frac{1}{\sqrt{u}a+u\sqrt{a}}
\end{align}
Then $$J=\int_0^1 J'(a)\ da=\frac{\pi}{2}\int_0^1\frac{da}{\sqrt{u}a+u\sqrt{a}}\\=\frac{\pi}{2}\left(\frac{2}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)\right)=\frac{\pi}{\sqrt{u}}\ln\left(\frac{1+\sqrt{u}}{\sqrt{u}}\right)$$
