Integrate it!

We can consider $\displaystyle \int_{-\infty}^\infty \frac {x\sin 3x}{x^4+1}dx = \Im \left(\int_{-\infty}^\infty \frac {xe^{3ix}}{x^4+1}dx \right)$ and apply contour integration using a semicircle with radius $R \to \infty$ in the upper complex plane as the contour as follows:

$\oint \frac {ze^{3iz}}{z^4+1} dz = \int_{-\infty}^\infty \frac {xe^{3ix}}{x^4+1}dx + \int_{\text{arc}} \frac {ze^{3iz}}{z^4+1} dz$

The imaginary part of first integral on the right hand side is twice the integral we are solving for. The last integral on the right and side approaches to 0 when $R \to \infty$ by Jordan's lemma. Then we have:

$\begin{aligned} I & = \int_{-\infty}^\infty \frac {xe^{3ix}}{x^4+1}dx = \oint \frac {ze^{3iz}}{z^4+1} dz \\ & = \oint \frac {ze^{3iz}}{\blue{(z-e^{\frac \pi 4i})(z-e^{\frac {3\pi} 4i})}(z+e^{\frac \pi 4i})(z+e^{\frac {3\pi} 4i})} dz & \small \blue{\text{Only 2 of the 4 poles are in the contour}} \\ & = \oint \blue{\frac 1{\sqrt 2}\left(\frac 1{z-e^{\frac \pi 4i}} - \frac 1{z-e^{\frac {3\pi} 4i}}\right)}\frac {ze^{3iz}}{(z+e^{\frac \pi 4i})(z+e^{\frac {3\pi} 4i})} dz & \small \blue{\text{By Cauchy integral formula}} \\ & = \frac {2\pi i ze^{3iz}}{\sqrt 2(z+e^{\frac \pi 4i})(z+e^{\frac {3\pi} 4i})} \bigg|_{e^{\frac {3\pi}4i}}^{e^{\frac \pi 4i}} \\ & = \frac \pi 2 \left(e^{-\frac 3{\sqrt 2}(1-i)} - e^{-\frac 3{\sqrt 2}(1+i)} \right) \\ & = i \pi e^{-\frac 3{\sqrt 2}} \sin \left(\frac 3{\sqrt 2} \right) \end{aligned}$

Therefore, $\Im(I) = \pi e^{-\frac 3{\sqrt 2}} \sin \left(\dfrac 3{\sqrt 2} \right)$ and $a+b+c+d+e = 1+3+2+3+2 = \boxed{11}$ .

Define the contour $\mathscr{C}_R$ given by the upper semicircle of radius $R > 1$ oriented counterclockwise and consider the contour integral $\oint_{\mathscr{C}_R} \frac{ze^{3iz}}{z^4+1}\mathrm{d}z = \int_{-R}^R \frac{xe^{3ix}}{x^4+1}\mathrm{d}x + \int_{\mathcal{C}_R} \frac{ze^{3iz}}{z^4+1}\mathrm{d}z,$ where $\mathcal{C}_R$ is the upper half circle of radius $R$ . We will prove that $\lim_{R\to\infty} \int_{\mathcal{C}_R} \frac{ze^{3iz}}{z^4+1}\mathrm{d}z = 0.$ Note that by the reverse triangle inequality, $\left|\int_{\mathcal{C}_R} \frac{ze^{3iz}}{z^4+1}\mathrm{d}z \right| \leq \pi R \cdot \max\limits_{z \in \mathcal{C}_R} \left|\frac{ze^{3iz}}{z^4+1}\right| \leq \pi R \cdot \max\limits_{z \in \mathcal{C}_R} \frac{|z||e^{3iz}|}{|z|^4-1} = \frac{\pi R^2}{R^4-1} \cdot \max\limits_{z \in \mathcal{C}_R} |e^{3iz}| \xrightarrow{R\to\infty} 0,$ as $|e^{3iz}|$ is bounded by 1 in the upper half plane. Therefore, by the Residue theorem, $\int_{-\infty}^\infty \frac{xe^{3ix}}{x^4+1}\mathrm{d}x = \lim_{R\to\infty} \oint_{\mathscr{C}_R} \frac{ze^{3iz}}{z^4+1}\mathrm{d}z = 2\pi i \left(\text{Res}_{z=e^{\pi i/4}} \frac{ze^{3iz}}{z^4+1} + \text{Res}_{z=e^{3\pi i/4}} \frac{ze^{3iz}}{z^4+1}\right).$ Since the poles are simple, for a residue at $z=z_0$ we have that $\text{Res}_{z=z_0} \frac{ze^{3iz}}{z^4+1}= \lim_{z\to z_0} \frac{(z-z_0)ze^{3iz}}{z^4+1} = \frac{z_0e^{3iz_0}}{4z_0^3} = \frac{e^{3iz_0}}{4z_0^2},$ by l'Hôpital's rule. By doing some algebraic stuff we find that $\int_{-\infty}^\infty \frac{x\sin(3x)}{x^4+1}\mathrm{d}x = \mathfrak{Im}\left(\int_{-\infty}^\infty \frac{xe^{3ix}}{x^4+1}\mathrm{d}x\right) = \boxed{\pi e^{-3/\sqrt{2}}\sin(3/\sqrt{2})}.$ Therefore $a = 1, b = 3, c = 2, d = 3$ and $e = 2$ and $a+b+c+d+e = 11$ .