Integration of $\frac{1}{u^4 + (4\zeta^2-2)u^2 + 1}$ I am trying to compute
$$I(\zeta) = \int_{-\infty}^{\infty} \frac{1}{u^{4} +  \left(4 \zeta^{2} - 2\right)u^{2} + 1}\, du$$
for positive real $\zeta$. Can anyone help?
I'm way out of practice for integrals except for simple stuff like $\int 1/(1+u^2)\, du = \tan^{-1} u + C$.
Sympy fails on the definite integral and gives me this weird RootSum expression for the indefinite integral:
$$\operatorname{RootSum} {\left(t^{4} \left(4096 \zeta^{8} - 8192 \zeta^{6} + 4096 \zeta^{4}\right) + t^{2} \left(256 \zeta^{6} - 384 \zeta^{4} + 128 \zeta^{2}\right) + 1, \left( t \mapsto t \log{\left (- 512 t^{3} \zeta^{6} + 768 t^{3} \zeta^{4} - 256 t^{3} \zeta^{2} - 32 t \zeta^{4} + 32 t \zeta^{2} - 4 t + u \right )} \right)\right)}$$
Wolfram Alpha gives me the following for the indefinite integral :
$$\begin{align}
& \frac{\frac{1}{a_1}\tan^{-1} \frac{u}{a_1} - \frac{1}{a_2}\tan^{-1} \frac{u}{a_2}}{4\zeta\sqrt{\zeta^2-1}} + C \\
\\
a_1 &= \sqrt{2\zeta^2-2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b-c}\\
a_2 &= \sqrt{2\zeta^2+2\zeta\sqrt{\zeta^2-1}-1} = \sqrt{b+c}\\
\end{align}$$
(with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$) but I'm a bit lost how it got there, and then I'm not exactly sure what to do if $\zeta \le 1$ (is the formula still valid?!)
edit: OK, partial fraction expansion is sloooowwwwly coming back to me. It looks like $a_1a_2 = 1$ and $a_1{}^2 + a_2{}^2 = 4\zeta^2-2$, so I guess they used the expansion
$$
\frac{1}{u^{4} +  \left(4 \zeta^{2} - 2\right)u^{2} + 1} = \frac{1}{4\zeta\sqrt{\zeta^2-1}}\left(\frac{1}{u^2+a_1{}^2} - \frac{1}{u^2+a_2{}^2}\right)
$$

Let $I$ denote the integral. Then using the identity
$$ u^4 + (4\zeta^2-2)u^2 + 1 = u^2 \left( ( u - u^{-1} )^2 + 4\zeta^2 \right), $$
we may write
$$ I = 2 \int_{0}^{\infty} \frac{u^{-2}}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u. \tag{1} $$
Now applying the substitution $u \mapsto u^{-1}$,
$$ I = 2 \int_{0}^{\infty} \frac{1}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u. \tag{2} $$
Averaging $\text{(1)}$ and $\text{(2)}$, we get
\begin{align*}
I
&= \int_{0}^{\infty} \frac{1 + u^{-2}}{( u - u^{-1} )^2 + 4\zeta^2} \, \mathrm{d}u \\
&= \int_{-\infty}^{\infty} \frac{1}{t^2 + 4\zeta^2} \, \mathrm{d}t \tag{$t=u-u^{-1}$} \\
&= \frac{\pi}{2|\zeta|}.
\end{align*}
