Is there a closed form for $\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\cdot\left({2n-1\over 2n+1}\right)^2?$

Let us denote: \begin{equation} S_p := \sum\limits_{n=0}^\infty \frac{2^{n+1}}{\binom{2 n}{n}} \cdot \frac{1}{(2n+1)^p} \end{equation} then the sum in question is just equal $S_0 - 4 S_1+4 S_2$. Now we have: \begin{eqnarray} S_0 &=& \sum\limits_{n=0}^\infty 2^{n+1} \cdot \underbrace{(2n+1)}_{\left. d_\theta \theta^{2n+1} \right|_{\theta=1}} \cdot \int\limits_0^1 t^n (1-t)^n dt \\ &=& \left. d_\theta 2 \theta \int\limits_0^1 \frac{1}{1-2 \theta^2 t (1-t)} dt \right|_{\theta=1} \\ &=& \left. d_\theta \frac{4 \arctan(\frac{\theta}{\sqrt{2-\theta^2}})}{\sqrt{2-\theta^2}} \right|_{\theta=1} = 4+\pi \end{eqnarray} We compute $S_1$ in exactly the same way. We have: \begin{eqnarray} S_1 &=& \sum\limits_{n=0}^\infty 2^{n+1} \int\limits_0^1 t^n (1-t)^n dt \\ &=& 2 \int\limits_0^1 \frac{1}{1-2 t(1-t) } dt = \frac{4 \arctan(1)}{1} = \pi \end{eqnarray} Now comes a little harder task. We have: \begin{eqnarray} S_2 &=& \sum\limits_{n=0}^\infty 2^{n+1} \cdot \underbrace{\frac{1}{(2n+1)}}_{\int\limits_0^1 \theta^{2 n} dt}\cdot \int\limits_0^1 t^n (1-t)^n dt \\ &=& 2 \int\limits_0^1 \int\limits_0^1 \frac{1}{1-2 \theta^2 t (1-t)} dt d\theta \\ &=& \int\limits_0^1 \frac{4 \arctan(\frac{\theta}{\sqrt{2-\theta^2}})}{\theta \sqrt{2-\theta^2}} d \theta \\ &\underbrace{=}_{u = \theta/\sqrt{2-\theta^2}}& 2 \sqrt{2} \int\limits_0^1 \frac{\arctan(u)}{u \sqrt{1+u^2}} du \\ &\underbrace{=}_{v=\arctan(u)}&2 \sqrt{2} \int\limits_0^{\frac{\pi}{4}} \frac{v}{\sin(v)} dv \\ &=& 2 \sqrt{2} \left.\left[v \left( \log(1-e^{\imath \cdot v}) - \log(1+e^{\imath \cdot v}) \right) + \imath \left( Li_2(-e^{\imath \cdot v}) - Li_2(e^{\imath \cdot v})\right)\right]\right|_{v=0}^{v=\pi/4} \\ &=& \sqrt{2} \frac{\pi}{2} \left( \left(\log(1-e^{\frac{\imath \pi}{4}}) - \log(1+e^{\frac{\imath \pi}{4}})\right) + \imath \frac{4}{\pi} \left( Li_2(-e^{\frac{\imath \pi}{4}}) - Li_2(e^{\frac{\imath \pi}{4}})\right)+\imath \pi\right)\\ &=& \frac{1}{16} \left( \zeta(2,\frac{1}{8})+\zeta(2,\frac{3}{8})-\zeta(2,\frac{5}{8})-\zeta(2,\frac{7}{8})\right) + \frac{\pi}{\sqrt{2}} \log(-1+\sqrt{2}) \end{eqnarray} In the last line we used the following identities: \begin{eqnarray} &&\log\left[ 1- e^{\imath \pi/4}\right] - \log\left[ 1+ e^{\imath \pi/4}\right] = \log\left[\sqrt{2}-1\right] - \imath \frac{\pi}{2}\\ &&Li_2(-e^{\frac{\imath \pi}{4}}) - Li_2(e^{\frac{\imath \pi}{4}}) = -\frac{1}{32 \sqrt{2}} \left( \right. \\ &&\zeta(2,\frac{1}{8}) - \zeta(2,\frac{3}{8}) - \zeta(2,\frac{5}{8})+\zeta(2,\frac{7}{8}) + \imath \left( \zeta(2,\frac{1}{8}) + \zeta(2,\frac{3}{8}) - \zeta(2,\frac{5}{8})-\zeta(2,\frac{7}{8})\right)\left.\right) \end{eqnarray} Here we only note that for generic $p\ge 1$ we have: \begin{eqnarray} &&S_{p+1} = -4 \imath \sqrt{2} \sum\limits_{t=1}^p \sum\limits_{s=0}^{p-t} \\ && \frac{(-1)^s}{2^{p-1} (p-s-t)! s! (t-1)!} (\log(-2))^{p-s-t} 2^{s+t-1} \int\limits_1^{\exp(\imath \pi/4)} \frac{[\log(z^2-1)]^s [\log(z)]^t}{z^2-1} d z \end{eqnarray} In general we also have: \begin{eqnarray} &&S_{p+1}(x) = -\imath 8 x \sum\limits_{t=1}^p \sum\limits_{s=0}^{p-t} \frac{(-1)^s}{(p-s-t)!s! (t-1)!} \cdot \left( \imath \frac{\pi}{2}+\log(2 x)\right)^{p-s-t} \cdot \\ &&\int\limits_1^{\exp(\imath \arcsin(x))} \frac{[\log(z^2-1)]^s [\log(z)]^t}{z^2-1} dz \end{eqnarray} where \begin{equation} S_{p+1}(x) := \sum\limits_{n=0}^\infty \frac{(2 x)^{2n+2}}{\binom{2n}{n}} \cdot \frac{1}{(2n+1)^{p+1}} \end{equation} It is still not clear whether the result reduces to polylogarithms only since for the time being we are unable to find the integrals in question.

This is not an answer at all.

Just out of curiosity, I had (using a CAS) a look at $$S_{a,b}=\sum_{n=0}^{\infty}{2^{n+1}\over {2n \choose n}}\,\frac{(2n-1)^a}{(2n+1)^b}\qquad \qquad a\geq 1\qquad b\geq 1$$ and what it seems is that, as soon as $b>1$, the result is given as linear combinations of hypergeometric functions as Brevan Ellefsen already commented.

However, what looks to be interesting is the case $b=1$ for which very simple expressions are obtained just as in $(1)$. $$\left( \begin{array}{cc} a & S_{a,1} \\ 1 & 4-\pi \\ 2 & 5 \pi \\ 3 & 68+11 \pi \\ 4 & 512+185 \pi \\ 5 & 7300+2279 \pi \\ 6 & 116224+37085 \pi \\ 7 & 2204868+701651 \pi \\ 8 & 48073728+15302705 \pi \\ 9 & 1186130180+377556239 \pi \\ 10 & 32669570048+10399048565 \pi \end{array} \right)$$

By the way, for the approximation of you result for $a=b=2$, the simple $\frac{5000}{1323}$, $\frac{43613}{11540}$, $\frac{92226}{24403}$ correspond to very small relative errors.

A somewhat easier way is the following:

Let's start with

$$\sum_{n=0}^{\infty}\frac{(2x)^{2n}}{\binom{2n}{n}}=\frac{1}{1-x^2}+\frac{x\arcsin x}{(1-x^2)^{\frac{3}{2}}}$$

Now, all we need is to integrate this expression twice and to differentiate it twice with respect to $x$.

I skip these simple procedures and write down only the end result.

$$\sum_{n=0}^{\infty}\frac{(2)^{2n}(x)^{2n-2}}{\binom{2n}{n}}\left (\frac{2n-1}{2n+1}\right )^2=$$

$$=\frac{1}{x^2(1-x^2)}-\frac{4-5x^2}{x^3(1-x^2)^{\frac{3}{2}}}\arcsin x+\frac{2\arcsin^2 x}{x^4}+\frac{2}{x^3}\int_{0}^{x}\left (\frac{\arcsin t}{t}\right )^2dt$$

To get the original sum, wich we denote as $S$, we evaluate this expression at $x=\frac{1}{\sqrt{2}}$

So the final result:

$$S=\frac{\pi^2}{2}-3\pi+4+4\sqrt{2}\int_{0}^{\frac{1}{\sqrt{2}}}\left (\frac{\arcsin t}{t}\right )^2dt$$