Find $\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx$ I was asked to prove that
$$\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^{4}}dx=\pi\frac{14325195794+(2815367209\sqrt{26})}{14623232(9+2\sqrt{26})^\frac{7}{2}}$$
I checked the result numerically and the first digits correct using W|F
$$\int\limits^{\infty}_{0}\frac{1}{(x^8+5x^6+14x^4+5x^2+1)^4}dx\approx 0.19874620328$$
I tried to start with trig substitution but the high power in the integral make it more complicated. Is there any way to evaluate this integral?
 A: 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.
