Challenging Problem: $\int_{-\infty} ^{\infty} \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big)$ $$ \int_{-\infty} ^{\infty}  \frac{x \sin{3x} }{x^4 +1}dx =\pi^a e^{\frac{-b}{\sqrt{c}}}\sin \big({\frac {d}{\sqrt{e}}}\big) $$ where $a,b,c,d,e$ are positive integers and $c$ and $e$ are square free numbers. Find $a+b+c+d+e$
My Attempt
$$ \int_{-\infty} ^{\infty}  \frac{x \sin{3x} }{x^4 +1}dx = 
\int_{-\infty} ^{\infty}  \frac{16x \sin{3x} }{(2x-\sqrt{2}i-\sqrt{2})(2x-\sqrt{2}i+\sqrt{2})(2x+\sqrt{2}i-\sqrt{2})(2x+\sqrt{2}i+\sqrt{2})}dx =$$
by partial fractions
$$=\int  -\frac {i\sin(3x)}{2 \big(2x+\sqrt{2}+\sqrt{-2}\big) } +\frac {i\sin(3x)}{2 \big(2x+\sqrt{2}-\sqrt{-2}\big) }+\frac {i\sin(3x)}{2 \big(2x-\sqrt{2}+\sqrt{-2}\big) }-\frac {i\sin(3x)}{2 \big(2x-\sqrt{2}-\sqrt{-2}\big) }dx=\frac{i}{2}\int  \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }dx +\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x+\sqrt{2}-\sqrt{-2}\big) }dx+\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x-\sqrt{2}+\sqrt{-2}\big) }dx-\frac{i}{2}\int\frac {\sin(3x)}{ \big(2x-\sqrt{2}-\sqrt{-2}\big) }dx$$
Solving
$$\int  \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }dx$$ Substitute $u=2x+\sqrt{2}+\sqrt{-2}\longrightarrow \frac{du}{dx}=2$
$$\Rightarrow \int  \frac {\sin(3x)}{ \big(2x+\sqrt{2}+\sqrt{-2}\big) }=\frac{1}{2} \int  \frac {\sin(\frac{3u}{2}-\frac{3i}{\sqrt{2}}-\frac{3}{\sqrt{2}})}{ u }du$$
applying the addition formula
$$=\int\frac {\cos \big(\frac{3i}{\sqrt{2}}+\frac{3}{\sqrt{2}}\big) \sin \big(\frac{3u}{2}\big) -\sin \big(\frac{3i}{\sqrt{2}}+\frac{3}{\sqrt{2}}\big) \cos \big(\frac{3u}{2}\big) }{ u }du$$  I understand the integral will not have an antidervative with sin(x) and cos(x) over x but, how do you proceed from here?
 A: Since the $\Im z>0$ roots of $z^{4}+1$ are $\frac{\pm1+i}{\sqrt{2}}$ and $\lim_{z\to a}\frac{z-a}{f\left(z\right)-f\left(a\right)}=\frac{1}{f^{\prime}\left(a\right)}$,$$\begin{align}\int_{\mathbb{R}}\frac{x\sin3x}{x^{4}+1}dx&=\Im\int_{\mathbb{R}}\frac{xe^{3ix}}{x^{4}+1}dx\\&=\Im\left(2\pi i\sum_{\pm}\lim_{z\to\frac{\pm1+i}{\sqrt{2}}}\frac{\left(z-\frac{\pm1+i}{\sqrt{2}}\right)ze^{3iz}}{z^{4}+1}\right)\\&=\Re\left(\frac{\pi}{2}\sum_{\pm}\lim_{z\to\frac{\pm1+i}{\sqrt{2}}}\frac{e^{3iz}}{z^{2}}\right)\\&=\frac{\pi}{2}e^{-3/\sqrt{2}}\Re\sum_{\pm}e^{\pm\left(\frac{3}{\sqrt{2}}-\frac{\pi}{2}\right)i}\\&=\pi e^{-3/\sqrt{2}}\sin\frac{3}{\sqrt{2}}.\end{align}$$Although Wolfram Alpha doesn't get it in such an elegant form, it does agree with the above value of $\approx0.320952$. Compared with $\pi^ae^{-b/\sqrt{c}}\sin\frac{d}{\sqrt{E}}$ (I've capitalised the last unknown for disambiguation) gives$$a=1,\,b=3,\,c=2,\,d=3,\,E=2\implies a+b+c+d+E=11.$$
A: Without contour integration (the simplest, for sure)
This is just for your curiosity : you ended with a series of antiderivatives
$$I_k=\int \frac{\sin(nx) }{x+k}dx$$ where $k$ is a complex number.  Let $y=x+k$ to make
$$I_k=\int \frac{\sin (n (y-k))}{y}\,dy=\cos (k n)\int\frac{ \sin (n y)}{y}dy-\sin (k n)\int\frac{ \cos (n y)}{y} dy$$ Make $y=\frac t n$
$$I_k=\cos (k n)\int\frac{ \sin (t)}{t}dt-\sin (k n)\int\frac{ \cos (t)}{t} dt$$
$$I_k=\cos (k n)\,\text{Si}(t)-\sin (k n)\,\text{Ci}(t)$$ where appear the sine and cosine integrals (the are non-elementary functions). Now, back to $x$, let $k=a+ib$ and using trigonometric properties and bounds you should end with
$$J=\int_{-\infty}^{+\infty} \frac{\sin(nx) }{x+a+ib}dx=\pi  e^{n (b-i a)} $$
