Prove $\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$ $$S=\sum_{n=0}^{\infty}{2^n(n^2-n\pi+1)(n^2+n-1)\over (2n+1)(2n+3){2n\choose n}}=1$$
Beta function
$${1\over {2n\choose n}}=n\int_{0}^{1}x^{n-1}(1-x)^ndx$$
$$\sum_{n=0}^{\infty}{2^n\over {2n\choose n}}=\int_{0}^{1}{2(1-t)\over [1-2t(1-t)]^2}dt$$
We can split out the sum
$$\sum_{n=0}^{\infty}{2^{n-1}(n^2-n\pi+1)\over (2n+1){2n\choose n}}-\sum_{n=0}^{\infty}{2^{n-1}(n^2+n-1)\over (2n+3){2n\choose n}}=1$$
With out the denominator $2n+1$ and $2n+3$ We can apply the beta function directly and integrate. I don't how to get the factors $2n+1$ and $2n+3$ with the beta function. Any hints on how to prove S?

Through Euler's beta function you may prove that:
$$ \forall x\in(0,4), \qquad f(x)=\sum_{n\geq 0}\frac{x^n}{(2n+1)\binom{2n}{n}}=\frac{4\arcsin\left(\frac{\sqrt{x}}{2}\right)}{\sqrt{x(4-x)}} \tag{1}$$
(have also a look at this question), then by computing $\int x^2 f(x^2)\,dx$ we get:
$$ \forall x\in(0,2),\quad g(x)=\sum_{n\geq 0}\frac{x^{2n}}{(2n+1)(2n+3)\binom{2n}{n}}=\frac{4 \left(x \sqrt{4-x^2}-(4-x^2)\arcsin\frac{x}{2}\right)}{x^3 \sqrt{4-x^2}}\tag{2}$$
that is more than enough to prove your claim.
