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
$$\int_0^{\infty } \left(\text{Li}_2\left(-\frac{1}{x^2}\right)\right)^n \, dx.$$ 

With a substitution and a step of integration by parts, the problems boils down to computing:
$$ I = -4\int_{0}^{+\infty}\frac{\log(1+x^2)\,\text{Li}_2(-x^2)}{x^2}\,dx. \tag{1}$$
Integrating by parts again, the problem boils down to computing:
$$ I_1 = \int_{0}^{+\infty}\frac{\log(1+x^2)^2}{x^2}\,dx,\qquad I_2=\int_{0}^{+\infty}\arctan\left(\frac{1}{x}\right)\frac{\log(1+x^2)}{x}\,dx.\tag{2} $$
The first integral is straightforward:
$$ I_1 = \frac{1}{2}\left.\frac{d^2}{d\alpha^2}\int_{0}^{+\infty}\frac{(1+x)^{\alpha}-1}{x^{3/2}}\,dx\,\right|_{\alpha=0} \tag{3}$$
since the innermost integral can be evaluated in terms of the beta function. 
That leads to $I_1=4\pi\log 2$. Now we just need to compute:
$$ J = \int_{1}^{+\infty}\int_{0}^{+\infty}\frac{\log(1+x^2)}{1+t^2 x^2}\,dx\,dt \tag{4}$$
to recover the value of $I_2=\frac{\pi^3}{8}$. That proves the conjecture:

$$ I = \frac{4\pi^3}{3}+32\pi\log 2.\tag{5}$$

