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?

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}
