Closed-form of $\int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx $ I'm looking for a closed form of this integral.
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx ,$$
where $\operatorname{Li}_2$ is the dilogarithm function.
A numerical approximation of it is
$$ I \approx 1.39130720750676668181096483812551383015419528634319581297153...$$
As Lucian said $I$ has the following equivalent forms:
$$I = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x^2}} \,dx = \int_0^1 \frac{\operatorname{Li}_2\left( \sqrt{x} \right)}{2 \, \sqrt{x} \, \sqrt{1-x}} \,dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\sin x) \, dx = \int_0^{\frac{\pi}{2}} \operatorname{Li}_2(\cos x) \, dx$$
According to Mathematica it has a closed-form in terms of generalized hypergeometric function, Claude Leibovici has given us this form.
With Maple using Anastasiya-Romanova's form I could get a closed-form in term of Meijer G function. It was similar to Juan Ospina's answer, but it wasn't exactly that form. I also don't know that his form is correct, or not, because the numerical approximation has just $6$ correct digits. 
I'm looking for a closed form of $I$ without using generalized hypergeometric function, Meijer G function or $\operatorname{Li}_2$ or $\operatorname{Li}_3$.
I hope it exists. Similar integrals are the following.
$$\begin{align}
J_1 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{1+x} \,dx = \frac{\pi^2}{6} \ln 2 - \frac58 \zeta(3) \\
J_2 & = \int_0^1 \frac{\operatorname{Li}_2\left( x \right)}{\sqrt{1-x}} \,dx = \pi^2 - 8 \end{align}$$
Related techniques are in this or in this paper. This one also could be useful.

My attempt. This is by no means closer to the answer, but I want to address several equivalent forms that might be helpful for future calculations.
First, from Landen's identity of the following form
$$ \mathrm{Li}_2(z) = -\mathrm{Li}_2\left(-\frac{z}{1-z}\right) - \frac{1}{2}\log^{2}(1-z), \quad z \notin [1, \infty)$$
we observe that
\begin{align*}
I
&= -\int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \left\{ \mathrm{Li}_2\left(-\frac{x}{1-x}\right) - \frac{1}{2}\log^{2}(1-x) \right\} \, dx \\
&= -\int_{-\infty}^{0} \frac{2\mathrm{Li}_2 (t) + \log^{2}(1-t)}{2\sqrt{1-2t}(1-t)} \, dt
\tag{1}
\end{align*}
By noting that
$$ \frac{d}{dt} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) = \frac{1}{2\sqrt{1-2t}(1-t)}, $$
integration by parts and the substitution $x = (1-2t)^{-1/2}$ shows that (1) is equal to
\begin{align*}
I
&= \int_{-\infty}^{0} \arctan\left(\frac{1}{\sqrt{1-2t}}\right) \frac{2\log(1-t)}{t(t-1)}  \, dt \\
&= \int_{0}^{1} \frac{8x \arctan x}{1 - x^4} \log \left( \frac{1+x^2}{2x^2} \right) \, dx \tag{2}
\end{align*}
The following observation
$$ \Re \log \left(\frac{1+ix}{\sqrt{2}} \right) = \frac{1}{2}\log \left( \frac{1+x^2}{2} \right)
\quad \text{and} \quad
\Im \log \left(\frac{1+ix}{\sqrt{2}} \right) = \arctan x $$
somehow seems to suggest complex-analytic approach, but I have not been successful with such approaches so far. Next, from the following simple formula
$$ \log \left( \frac{1+x^2}{2x^2} \right) \, dx = \int_{0}^{1} \frac{d}{dy} \log \left( \frac{y^2+x^2}{y^2 + 1} \right) \, dy $$
the integral (2) can be further decomposed into the following form
$$ I = \int_{0}^{1}\int_{0}^{1} \frac{16xy \arctan x}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy. \tag{3} $$
Simple calculation shows that
$$ \int_{0}^{1}\int_{0}^{1} \frac{16xy}{(1+x^2)(1+y^2)(x^2+y^2)} \,dxdy = 2\zeta(2), $$
so I suspect that the situation in (3) is not that bad.
