computing $\int_{-\infty}^{\infty}\frac{x^3\sin(ax)}{x^4+b^4}$ Find the value of the integral
$$\int_{-\infty}^{\infty}\frac{x^3\sin(ax)}{x^4+b^4}$$
assuming $a,b>0$.
I tried using complex analysis by defining the function 
$$f(z)=\frac{z^3e^{azi}}{z^4+b^4}$$
$f$ has simple poles in $b\{e^{\frac{i\pi}{4}},e^{\frac{3i\pi}{4}},e^{\frac{5i\pi}{4}},e^{\frac{7i\pi}{4}}\}$.
So, I took $\forall t\in[-R,R]:\gamma_1(t)=t$, $\forall t\in[0,\pi]:\gamma_2(t)=Re^{it}$ and $\gamma=\gamma_1+\gamma_2$.
Now, let aside the computation of the integrals on $\gamma_1,\gamma_2$, I find it very inconvenient to compute the residues
$$Res_{be^{\frac{i\pi}{4}}}(f(z)),\ Res_{be^{\frac{3i\pi}{4}}}(f(z))$$ 
 A: Enforcing the substitution $x\mapsto x/a$ reveals
$$\int_{-\infty}^\infty \frac{x^3\sin(ax)}{x^4+b^4}\,dx=\int_{-\infty}^\infty \frac{x^3\sin(x)}{x^4+(ab)^4}\,dx$$
Let $f(z)=\frac{z^3e^{iz}}{z^4+(ab)^4}$ and let $C$ be the contour comprised of $(1)$ the straight line segment from $-R$ to $R$ and $(2)$ the circular arc from $R$ to $-R$ in upper-half plane.  Then, we have
$$\begin{align}
\oint_C f(z)\,dz&=\int_{-R}^R \frac{x^3e^{ix}}{x^4+(ab)^4}\,dx+\int_0^{\pi}\frac{(Re^{i\phi})^3}{(Re^{i\phi})^4+(ab)^4}\,iRe^{i\phi}\,d\phi\\\\
&=2\pi i \text{Res}\left(\frac{z^3e^{iz}}{z^4+(ab)^4},z=abe^{i\pi/4}, z=abe^{i3\pi/4}\right)\\\\
&=\frac{\pi i}2 \left( e^{iabe^{i\pi/4}}+e^{iabe^{i3\pi/4}}\right)
\end{align}$$
Letting $R\to \infty$ and taking the imaginary part, we find that 
$$\int_{-\infty}^\infty \frac{x^3\sin(ax)}{x^4+b^4}\,dx=\pi e^{-ab/\sqrt2}\cos(ab/\sqrt2)$$ 

NOTE:
For general $a$ and $b$, we have
$$\int_{-\infty}^\infty \frac{x^3\sin(ax)}{x^4+b^4}\,dx=\pi e^{-|ab|/\sqrt2}\cos(ab/\sqrt2)\text{sgn}(a)$$
where $\text{sgn}(a)$ is given by
$$\text{sgn}(a)=\begin{cases}1&,a>0\\\\0&,a=0\\\\-1&,a<0\end{cases}$$
