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?

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
$$\int_{-\infty}^{\infty} \frac{dx}{x^4 + a x^2 + b ^2} =\frac {\pi} {b \sqrt{2b+a}}\qquad\qquad\square$$
