Formula 20 on page 9 of this text has that for
$$f(x) = \frac{1}{x^4 + 2a^2x^2\cos(2\theta)+a^4}\qquad a>0, \quad|\theta|<\frac{\pi}{2},$$ we have $$\hat f(y) = \frac{\pi}{2a^3}e^{-a y \cos\theta}\sin(\theta+a y \sin\theta)\csc(2\theta)$$
where
$$\hat f(y) \equiv \int_0^{\infty}f(x)\cos(yx)dx,\qquad y>0$$
is the Fourier transform for even functions $f(x)$. I would appreciate it if someone could provide a derivation.
The function $g(z)=z^4+2a^2\cos(2\theta)z^2+a^4$ has zeroes for the four values of $z$ such that
$$z=\pm ia e^{\ i\theta}, \pm iae^{-i\theta}$$
For $a>0$, the zeroes of $g(z)$ that are in the upper-half plane are at $iae^{\pm i\theta}$. Hence, the residue theorem in conjunction with Jordan's Lemma guarantee that
$$\begin{align} \int_{-\infty}^\infty \frac{e^{iyx}}{x^4+2a^2\cos(2\theta)x^2+a^4}\,dx&=2\pi i \text{Res}\left(\frac{e^{iyz}}{z^4+2a^2\cos(2\theta)z^2+a^4}, z=ia e^{\pm i\theta}\right)\\\\ &=2\pi i \left(\frac{e^{-aye^{i\theta}}}{-4ia^3e^{i3\theta}+i2a^3e^{i\theta}\cos(2\theta)}+\frac{e^{-aye^{-i\theta}}}{-4ia^3e^{-i3\theta}+i2a^3e^{-i\theta}\cos(2\theta)}\right)\\\\ &=(\pi/2a^3)\left(i\csc(2\theta)e^{-(i\theta+ a y e^{i\theta})}-i\csc(2\theta)e^{(i\theta-a y e^{-i\theta})}\right)\\\\ &=-\frac{\pi \csc(2\theta)}{a^3} \text{Im}\left(e^{-aye^{i\theta}-i\theta}\right)\\\\ &=\frac{\pi\csc(2\theta)}{a^3}e^{-ay\cos(\theta)}\sin(\theta+ay\sin(\theta)) \end{align}$$
as was to be shown!