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:

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

Another way to evaluate the integral $$I_{2} = \int_{0}^{\infty} \frac{\arctan (\frac{1}{x}) \log(1+x^{2})}{x} \, dx$$ in Jack D'Aurizio's answer (which according to Wolfram Alpha evaluates to $\frac{\pi^{3}}{12}$) is to consider the complex function $$f(z) = \frac{\arctan \left(\frac{1}{z} \right) \log(1-iz)}{z} = \frac{\text{arccot}(z) \log(1-iz)}{z}.$$

Using the principal branch of the logarithm, $\text{arccot}(z)$ has a branch cut on $[-i, i]$ and $\log(1-iz)$ has a branch cut on $(-i\infty, -i]$.

So integrating around a semicircle in the upper half-plane deformed around the branch cut and using the fact $\text{arccot}(z) \sim \frac{1}{z}$ when $z$ is large in magnitude, we get

$$\int_{-\infty}^{\infty} \frac{\arctan(\frac{1}{x}) \log(1-ix)}{x} \, dx + \int_{0}^{1} \frac{\frac{i}{2} (2 \pi i) \log(1+t)}{it} \, i \, dt =0.$$

Then equating the real parts on both sides of the equation, we get

$$\int_{0}^{\infty} \frac{\arctan(\frac{1}{x}) \log(1+x^{2})}{x} \, dx = \pi \int_{0}^{1} \frac{\log(1+t)}{t} \, dt = - \pi \, \text{Li}_{2}(-1) = \frac{\pi^{3}}{12}.$$