Can you generalize it?

Useful identities:

Generalized geometric series: Let $n\in\mathbb{N}_0$ and $q\in\mathbb{C}$ . Then $\begin{aligned} \sum_{k=0}^\infty \binom{k+n}{n}q^k&=\frac{1}{(1-q)^{k+1}}, &&& |q|&<1 &&&&&(*) \end{aligned}$

In the link about central binomial coefficients, there is a useful identity to rewrite them as an integral: $\begin{aligned} n&\in\mathbb{N}_0:&\binom{2n}{n}&=\frac{4^n}{\pi}\int_{\mathbb{R}}\frac{dx}{(1+x^2)^{n+1}}&&&&&(**) \end{aligned}$

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First steps

Let $S$ be the value of the series and $P(n)$ the numerator polynomial. Rewrite the central binomial coefficient using $(**)$ : $S=\sum_{n=0}^\infty (-1)^n \frac{P(n)}{2^{12n}} \cdot\frac{4^{5n}}{\pi^5}\int_{\mathbb{R}^5} \frac{dV}{Q(\vec{x})^{n+1}}=\frac{1}{\pi^5}\sum_{n=0}^\infty P(n)\cdot\frac{(-1)^n}{4^n}\int_{\mathbb{R}^5} \frac{dV}{Q(\vec{x})^{n+1}},\qquad Q(\vec{x})=\prod_{k=1}^5(1+x_k^2)$ By dominated convergence, we interchange summation and integration to get $S=\frac{1}{\pi^5}\int_{\mathbb{R}^5}\sum_{n=0}^\infty P(n)\cdot\frac{(-1)^n}{4^nQ(\vec{x})^{n+1}}\:dV$ Rewrite the numerator in terms of binomial coefficients in preparation for $(*)$ : $20n^2+8 n+1=20(n+1)(n+2) - 52(n+1) + 13 = 40\binom{n+2}{2} - 52\binom{n+1}{1} + 13\binom{n+0}{0}$ The integrand consists of three generalized geometric series: $S=\frac{1}{\pi^5}\int_{\mathbb{R}^5} \frac{40}{(1 + f(\vec{x}))^3}-\frac{52}{(1 + f(\vec{x}))^2}+\frac{13}{1 + f(\vec{x})}\:\frac{dV}{Q(\vec{x})},\qquad f(\vec{x}):=\frac{1}{4Q(\vec{x})}$ These integrals seem to have an analytic solution - the first component is easy, but afterwards you get radicals. At that point, I used numeric integration and manually checked low powers of pi. If anyone knows of a nice way to analytically solve these integrals (or another way altogether), please let me know!