Elegant proof of $\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$? Let $a, b > 0$ satisfy $a^2-4b^2 \geq 0$. Then:
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}$$
One way to calculate this is by computing the residues at the poles in the upper half-plane and integrating around the standard semicircle. However, the sum of the two residues becomes a complicated expression involving nested square roots, which magically simplifies to the concise expression above.
Sometimes such 'magical' cancellations indicate that there is a faster, more elegant method to reach the same result. 
Is there a faster or more insightful way to compute the above integral?
 A: It's time to return the favor from here Ruben. (>‿◠)✌
First, we will prove
$$\int_{-\infty}^\infty \frac{dx}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}=\frac{\pi}{2\beta\cosh\alpha}$$
Note that
$$\frac{1}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\frac{1}{\beta^2 x^2+e^{-2\alpha}}-\frac{1}{\beta^2 x^2+e^{2\alpha}}\right]$$
Hence
\begin{align}
\int_{-\infty}^\infty \frac{dx}{\beta^4x^4+2\beta^2\cosh(2\alpha)\,x^2+1}&=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\int_0^\infty\frac{dx}{\beta^2 x^2+e^{-2\alpha}}-\int_0^\infty\frac{dx}{\beta^2 x^2+e^{2\alpha}}\right]\\
&=\frac{1}{e^{2\alpha}-e^{-2\alpha}}\left[\frac{e^{\alpha}}{\beta}\arctan\left(e^{\alpha}x\right)-\frac{e^{-\alpha}}{\beta}\arctan\left(e^{-\alpha}x\right)\right]_{x=-\infty}^\infty\\
&=\frac{\pi}{\beta}\left[\frac{e^{\alpha}-e^{-\alpha}}{\left(e^{\alpha}-e^{-\alpha}\right)\left(e^{\alpha}+e^{-\alpha}\right)}\right]\\
&=\frac{\pi}{2\beta\cosh\alpha}\qquad\qquad\square
\end{align}
Now
$$\int_{-\infty}^\infty \frac{dx}{x^4+\frac{2\cosh(2\alpha)}{\beta^2}\,x^2+\frac{1}{\beta^4}}=\frac{\beta^3\pi}{2\cosh\alpha}$$
Setting $a=\frac{2\cosh(2\alpha)}{\beta^2}$ and $b^2=\frac{1}{\beta^4}$, then using $\cosh\alpha=\sqrt{\frac{\cosh(2\alpha)+1}{2}}$ will give
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}\qquad\qquad\square$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{-\infty}^{\infty}{\dd x \over x^{4} + ax^{2} + b^{2}}
    ={\pi \over b\root{2b + a}}:\ {\large ?}.\qquad a, b\ >\ 0\,,\quad
    a^{2} - 4b^{2}\ \geq\ 0}$.

Indeed, this is essentially the @user111187 proof with the addition of some omitted details:

\begin{align}{\cal I}&\equiv\color{#66f}{\large
\int_{-\infty}^{\infty}{\dd x \over x^{4} + ax^{2} + b^{2}}}
=2\int_{0}^{\infty}{1 \over x^{2}}\,{\dd x \over x^{2} + a + b^{2}/x^{2}}
\\[5mm]&\imp\quad \half\,{\cal I}=\int_{0}^{\infty}{1 \over x^{2}}\,
{\dd x \over \pars{x - b/x}^{2} + 2b + a}\tag{1}
\end{align}

With $\ds{{b \over x}\equiv t\ \imp\ x = {b \over t}}$ we'll have:

\begin{align}\half\,{\cal I}&
=\int_{\infty}^{0}{t^{2} \over b^{2}}\,
{-b\,\dd t/t^{2} \over \pars{b/t - t}^{2} + 2b + a}
={1 \over b}\int_{0}^{\infty}{\dd t \over \pars{b/t - t}^{2} + 2b + a}\tag{2}
\end{align}

With $\pars{1}$ and $\pars{2}$:

\begin{align}
\half\,b{\cal I}&=\int_{0}^{\infty}{b \over x^{2}}\,
{\dd x \over \pars{x - b/x}^{2} + 2b + a}
\\[5mm]\half\,b{\cal I}&
=\int_{0}^{\infty}{\dd x \over \pars{x - b/x}^{2} + 2b + a}
\\[5mm]\mbox{and}\ 
b{\cal I}&=\half\,b{\cal I} + \half\,b{\cal I}
=\int_{0}^{\infty}{\pars{b/x^{2} + 1}\,\dd x \over \pars{x - b/x}^{2} + 2b + a}
\end{align}

With $\ds{u \equiv x - {b \over x}\ \imp\ \dd u = \pars{1 + {b \over x^{2}}}
\,\dd x}$ we'll get

\begin{align}
b{\cal I}&
=\int_{-\infty}^{\infty}{\dd u \over u^{2} + 2b + a}
={2 \over \root{2b + a}}\
\overbrace{\int_{0}^{\infty}{\dd u \over u^{2} + 1}}
^{\ds{\color{#c00000}{\pi \over 2}}}\ =\
{\pi \over \root{2b + a}}
\\[5mm]\imp{\cal I}&\equiv\color{#66f}{\large
\int_{-\infty}^{\infty}{\dd x \over x^{4} + ax^{2} + b^{2}}
={\pi \over b\root{2b + a}}}
\end{align}
A: Let $b=\frac{a}{2}\sin2\theta$ ($0<\theta<\frac{\pi}{2}$), then
\begin{eqnarray}
\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + \frac14a^2\sin^22\theta}&=&\int_{-\infty}^{\infty}\frac{dx}{(x^2+a\sin^2\theta)(x^2+a\cos^2\theta)}\\
&=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\int_{-\infty}^{\infty}\left(\frac{1}{x^2+a\sin^2\theta}-\frac{1}{x^2+a\cos^2\theta}\right)dx\\
&=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\left(\frac{1}{\sqrt{a}\sin\theta}\arctan\frac{1}{\sqrt{a}\sin\theta}x\right.\\
&&-\left.\frac{1}{\sqrt{a}\cos\theta}\arctan\frac{1}{\sqrt{a}\cos\theta}x\right)\bigg|_{-\infty}^\infty\\
&=&\frac{1}{a(\cos^2\theta-\sin^2\theta)}\left(\frac{1}{\sqrt{a}\sin\theta}-\frac{1}{\sqrt{a}\cos\theta}\right)\pi\\
&=&\frac{\pi}{a\sqrt{a}(\cos\theta+\sin\theta)\sin\theta\cos\theta}\\
&=&\frac{\pi}{b\sqrt{2b+a}}.
\end{eqnarray}
A: Ok, I finally found a nice method.
We have $$ \begin{align} \int_0^{\infty} \frac{dx}{x^4+ax^2+b^2} &= \int_0^{\infty} \frac{dx}{x^2}\frac{1}{(x-b/x)^2+2b+a} \\&=  \frac{1}{b}\int_0^{\infty} \frac{dx}{(x-b/x)^2+2b+a} \\&= \frac{1}{b}\int_0^{\infty} \frac{dx}{x^2+2b+a}  \\&=  \frac{
\pi}{2b\sqrt{2b+a}}   \end{align}$$
and the desired integral follows by symmetry.
Here the nontrivial step made use of the Cauchy-Schlömilch transformation (see e.g. here): if the integrals exist and $b > 0$, then $$\int_0^{\infty} f\left((x-b/x)^2\right)\, dx = \int_0^{\infty} f(x^2) \, dx$$
It is quite interesting that the above proof doesn't make use of the assumption that $a^2-4b^2 \geq 0$.
A: Note
\begin{align}  \int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} 
 &= 2\int_{0}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} \overset{x\to\frac b x}= 
 2\int_{0}^{\infty} \frac{\frac{x^2}b\ dx}{x^4 + a x^2 + b ^2}\\
&= \int_{0}^{\infty} \frac{1+ \frac{x^2}b}{x^4 + a x^2 + b ^2}dx=\frac1b \int_{0}^{\infty} \frac{d(x-\frac b x)}{(x-\frac b x)^2 +a+2b}\\
&= \frac1{b\sqrt{a+2b}}\tan^{-1}\frac{x-\frac b x}{\sqrt{a+2b}}\bigg|_0^\infty= \frac{\pi}{b\sqrt{a+2b}}
\end{align}
