Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$ We have
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot{2n-1\over 2n+1}=4-\pi\tag1$$
I would like to know if there exist a closed form for
$$\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2 =\,??\tag2$$
I was able to roughly estimate it as $\approx\sqrt{8+2\pi}$ but it is not the closed form.
How can we find the closd form for $(2)?$

This is not an answer to this question but an attempt to answer a generalized question. Our generalization consists in replacing the power two in the last term in the parentheses by a power three.Now, by starting from the (corrected--the term $(1-x^2)$ in the denominator in front of the first power of the arc sine should be raised to the power $3/2$ and not as it is now to the power one) second formula given in the answer by Martin Gales and then by multiplying the right hand side of that formula by $x^2$ and then integrating over $x$ we have derived the following formula:
\begin{eqnarray}
&&\sum\limits_{n=0}^\infty \frac{2^{2 n} x^{2n+1}}{\binom{2 n}{n}} \cdot \frac{(2n-1)^2}{(2n+1)^3}=\\
&&\left(4 \log(1+\sqrt{1-x^2})-4 \log(x)+ \frac{1}{\sqrt{1-x^2}}\right) \arcsin(x) + \\
&&4 \imath \left( \log\left[\frac{\left(x-\imath(1-\sqrt{1-x^2})\right)^2}{2-2 \sqrt{1-x^2}}\right] \log\left[ \frac{1-\sqrt{1-x^2}}{x}\right]-\right.\\
&&\left.Li_2[(-\imath) \frac{1-\sqrt{1-x^2}}{x} ]+Li_2[(\imath) \frac{1-\sqrt{1-x^2}}{x})]\right)+\\
&&2 \int\limits_0^x \left(\frac{\arcsin(t)}{t}\right)^2 \left(1+\log(\frac{x}{t})\right)dt =\\
&&-4 i \text{Li}_2\left(-\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+4 i \text{Li}_2\left(\frac{i \left(1-\sqrt{1-x^2}\right)}{x}\right)+\\
&&-8 i \text{Li}_3\left(-e^{i \sin ^{-1}(x)}\right)+8 i \text{Li}_3\left(e^{i \sin
   ^{-1}(x)}\right)+\\
&&\left(\text{Li}_2\left(e^{i \sin ^{-1}(x)}\right)-\text{Li}_2\left(-e^{i \sin ^{-1}(x)}\right)\right) \left(8 \sin ^{-1}(x)-4 i \log (2 i x)\right)+\\
&&\sin ^{-1}(x) \left(\frac{1}{\sqrt{1-x^2}}+4 i \log (2
   i x) \cos ^{-1}\left(\frac{1}{x}\right)\right)+\\
&&4 \sin ^{-1}\left(\frac{1}{x}\right) \sin ^{-1}(x)^2+2 i \pi  \log (2 i x) \cos ^{-1}(x)-14 i \zeta (3)+\\
&&8 \imath \int\limits_1^{\exp(\imath \arcsin(x))} \frac{\log(z)}{z^2-1} \log(z^2-1) dz
\end{eqnarray} 
We will evaluate the result further and simplify it later on.
