$\int\limits_0^\infty {x^4 \over (x^4-x^2+1)^4}\ dx$ I want to calculate
$$\int\limits_0^\infty \frac{x^4}{(x^4-x^2+1)^4}dx$$
I have searched with keywords "\frac{x^4}{(x^4-x^2+1)^4}" and "x^4/(x^4-x^2+1)^4". But there are no results

For $b > 0$, define $I_b(a)$ by
$$ I_b(a)
= \int_{0}^{\infty} \frac{x^{a}}{((x - x^{-1})^2 + 1)^b} \, \mathrm{d}x
= \int_{0}^{\infty} \frac{x^{a+2b}}{(x^4 - x^2 + 1)^b} \, \mathrm{d}x. $$
This integral converges if $|a+1| < 2b$. We can also prove that $I_b(a) = I_b(-a-2)$ holds, by using the substitution $x \mapsto 1/x$. Then
\begin{align*}
\int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x
&= I_4(-4)
 = I_4(2) \\
&= I_4(2) - I_4(0) + I_4(-2) \\
&= \int_{0}^{\infty} \frac{1}{((x - x^{-1})^2 + 1)^3} \, \mathrm{d}x
\end{align*}
So, by the Glasser's master theorem,
\begin{align*}
\int_{0}^{\infty} \frac{x^4}{(x^4 - x^2 + 1)^4} \, \mathrm{d}x
&= \int_{0}^{\infty} \frac{1}{(u^2 + 1)^3} \, \mathrm{d}u \\
&= \int_{0}^{\frac{\pi}{2}} \cos^4\theta \, \mathrm{d}\theta \tag{$(u=\tan\theta)$} \\
&= \frac{3\pi}{16}
 \approx 0.58904862254808623221 \cdots.
\end{align*}
