Compute the following sum in closed form : $\sum_{n=1}^{\infty}\frac{n\binom{2n}{n}}{4^{n}(2n+1)(2n-1)(4n+1)}$ $$\text{Find : }\sum_{n=1}^{\infty}\frac{n\binom{2n}{n}}{4^{n}(2n+1)(2n-1)(4n+1)}$$
I know that $\displaystyle\sum_{n=0}^{\infty}\binom{2n}{n}x^{2n}=\frac{1}{\sqrt{1-4x^{2}}}$ so $\displaystyle\sum_{n=1}^{\infty}n\binom{2n}{n}x^{2n}=\frac{2x^2}{\sqrt{1-4x^{2}}}$.
But I don't know he to complete this work because I find hypergeometric function.
 A: As remarked in the OP, 
\begin{equation}
\sum_{n=0}^{\infty}\binom{2n}{n}x^{2n}=\frac{1}{\sqrt{1-4x^{2}}}
\end{equation} 
or, by changing $x\to x/2$,
\begin{equation}
\sum_{n=0}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{2n}=\frac{1}{\sqrt{1-x^{2}}}
\end{equation} 
We use the decomposition
\begin{equation}
\frac{n}{\left( 2n-1 \right)\left( 2n+1 \right)\left( 4n+1 \right)}=\frac{1}{12}\frac{1}{2n-1}-\frac{1}{4}\frac{1}{2n+1}+\frac{1}{3}\frac{1}{4n+1}
\end{equation} 
Then, for $\left|x\right|<1$, as the three series converge absolutely
\begin{equation}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{nx^{2n}}{\left( 2n-1 \right)\left( 2n+1 \right)\left( 4n+1 \right)}=\frac{1}{12}S_--\frac{1}{4}S_++\frac{1}{3}S_2
\end{equation} 
where
\begin{align}
S_+&=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{ 2n+1}\\
S_-&=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{2n-1}\\
S_2&=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{4n+1}
\end{align} 
With the notation
\begin{equation}
f(x)=\frac{1}{\sqrt{1-x^{2}}}
\end{equation} 
by integrating the above series, we have
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{2n}&=f(x)-1\\
S_+=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{2n+1}&=\int_0^1 \left[f(x)-1\right]\,dx\\
&=\frac{\pi}{2}-1
\end{align}
A simple transformations is necessary forevaluating $S_-$:
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{2n-2}&=\frac{f(x)-1}{x^2}\\
S_-=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{2n-1}&=\int_0^1\frac{f(x)-1}{x^2}\,dx\\
&=1
\end{align} 
For the third term, changing $x\to x^2$ in the series before integrating 
\begin{align}
\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}x^{4n}&=f(x^2)-1\\
S_2=\sum_{n=1}^{\infty}\frac{1}{4^n}\binom{2n}{n}\frac{1}{4n+1}&=\int_0^1\frac{dx}{\sqrt{1-x^4}}-1\\
&=\frac{\Gamma(\frac{5}{4})\sqrt{\pi}}{\Gamma(\frac{3}{4})}-1\\
&=\frac{\left[\Gamma\left( 1/4\right)\right]^2 }{4\sqrt{2\pi}}-1
\end{align}
This classical integral can be found for example here. 
Putting all the results together gives
\begin{equation}
\sum_{n=1}^{\infty}\frac{n\binom{2n}{n}}{4^{n}(2n+1)(2n-1)(4n+1)}=-\frac{\pi}{8}+\frac{\left[\Gamma\left( 1/4\right)\right]^2 }{12\sqrt{2\pi}}
\end{equation} 
which is also identical to the result given by @ChipHurst in the comments.
