Evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$. 
Evaluate $$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}.$$


My work so far and background to the problem.
This question was inspired by the second page of this paper. The author of the paper doesn't mention how he managed to prove the $4$ identities (shown below) at the top of the second page of the paper, so I tried to find my own method of doing so.
\begin{align}
\sum\limits_{n=1}^{\infty}\binom{2n}{n}^{-1} &= \frac{1}{3}+\frac{2\pi\sqrt3}{27} \\
\sum\limits_{n=1}^{\infty}\frac{1}{n}\binom{2n}{n}^{-1} &= \frac{\pi\sqrt3}{9} \\
\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\binom{2n}{n}^{-1} &= \frac{\pi^2}{18} \\
\sum\limits_{n=1}^{\infty}\frac{1}{n^4}\binom{2n}{n}^{-1} &= \frac{17\pi^4}{3240}
\end{align}
The method I used was similar to the method the author used in the rest of his paper: trying to find the generating functions for particular sequences. Using this method, like the author I was successful in proving the first $3$ identities, since I obtained the following:
$$\begin{align}\sum\limits_{n=1}^{\infty}\binom{2n}{n}^{-1}x^n&=4\sqrt\frac{x}{(4-x)^3}\arcsin\frac{\sqrt{x}}{2}+\frac{x}{4-x}\\
\sum\limits_{n=1}^{\infty}\frac{1}{n}\binom{2n}{n}^{-1}x^n&=2\sqrt{\frac{x}{4-x}}\arcsin\frac{\sqrt x}{2}\\
\sum\limits_{n=1}^{\infty}\frac{1}{n^2}\binom{2n}{n}^{-1}x^n&=2\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\end{align}$$
I obtained the second and third power series by dividing the power series above each one by $x$ and then integrating.
I then attempted to find a closed form for $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}x^n$ in order to evaluate $\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}$. The way I tried to do this was finding
$$\int\frac{2}{x}\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\mathrm{d}x$$
but according to WA this has no solution in terms of elementary functions and it involves some polylogarithms which I am not familiar with at all really, especially when they have a complex argument.
I then realized that
$$\sum\limits_{n=1}^{\infty}\frac{1}{n^3}\binom{2n}{n}^{-1}=\int_0^1\frac{2}{x}\left(\arcsin\frac{\sqrt{x}}{2}\right)^2\mathrm{d}x=\int_0^{1/2}\frac{4}{u}\left(\arcsin u\right)^2\mathrm{d}u$$
(where the second equality is found by using the substitution $u=\frac{\sqrt{x}}{2}$) which could be more helpful since definite integrals can often be evaluated in terms of elementary functions even if the indefinite integral cannot.
From there however I could not think of a way of continuing.

Thank you for your help.

Hint: We find in Evaluations of Binomial Series by J. M. Borwein and R. Girgensohn the formula (47)
\begin{align*}
\sum_{n\geq 1}\frac{1}{n^3\binom{2n}{n}}=\frac{2}{3}\pi\Im\left(L_2\left(e^{i\pi/3}\right)\right)
-\frac{4}{3}\zeta(3)
\end{align*}
with $L_p(z)=\sum_{n>0}\frac{z^n}{n^p}$ the polylogarithms.
