Find $\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx$

Hint. A route.

One may recall the following result, which goes back at least to G. Boole (1857).

Proposition. Let $f \in L^1(\mathbb{R})$ and let $f$ be an even function. Then $$ \int_{-\infty}^{+\infty}x^{2n}f\left(x-\frac1x\right) dx=\sum_{k=0}^n \frac{(n+k)!}{(2k)!(n-k)!}\int_{-\infty}^{+\infty} x^{2k}f(x)\: dx. \tag1 $$

Then one may write $$ \begin{align} &\int_0^{+\infty}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{-16}}{\left(\left(x^4+\dfrac1{x^4}\right)+5\left(x^2+\dfrac1{x^2}\right)+14\right)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{-16}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \\\\&=\int_0^{+\infty}\frac{x^{14}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \qquad \left(x \to \dfrac1x \right) \\\\&=\frac12\int\limits^{\infty}_{-\infty}\frac{x^{14}}{\left(\left[\left(x-\dfrac1x\right)^2+2\right]^2+5\left(x-\dfrac1x\right)^2+22\right)^{4}}\:dx \\\\&=\frac12\sum_{k=0}^7 \frac{(7+k)!}{(2k)!(7-k)!}\int_{-\infty}^{+\infty} \frac{x^{2k}}{\left(\left(x^2+2\right)^2+5x^2+22\right)^{4}}\:dx \qquad (\text{using}\,\,(1)) \\\\&=\sum_{k=0}^7 \frac{(7+k)!}{(2k)!(7-k)!}\int_0^{+\infty} \frac{x^{2k}}{\left(x^4+6x^2+26\right)^{4}}\:dx \\\\&=\pi\:\frac{14325195794+(2815367209\sqrt{26})}{14623232(9+2\sqrt{26})^\frac{7}{2}} \end{align} $$ where we have concluded by using Theorem $3.1$ (p.$6$) here, in G. Boros and V. Moll's paper.