integral of $\int_0^\infty \frac{ x^3 \cos(x)}{x^6 + 1} \mathrm{d}x $ How does someone solve
$$\int_0^\infty \frac{ x^3  \cos(x)}{x^6 + 1} \mathrm{d}x $$
Is this analytically possible? I have no idea and I think if it is, it must be probably be solved via the Fourier/residue theorem or something like that.

As Sangchul Lee commented, we could use partial fractions to get 
$$\frac{ x^3  \cos(x)}{x^6 + 1}=\sum_{i=1}^6 a_i \frac{\cos(x)}{x-b_i}$$ and then use $$\int_0^\infty \frac{\cos(x)}{x-c}\,dx=-\text{Ci}(-c) \cos (c)-\text{Si}(c) \sin (c)-\frac{1}{2} \pi  \sin (c)$$ (provided that $\Im(c)\neq 0\lor \Re(c)\leq 0$). This would be quite tedious but doable.
To my surprise, a CAS gave 
$$\int_0^\infty \frac{ x^3  \cos(x)}{x^6 + 1} \,dx=\frac{1}{2} \sqrt{\frac{\pi }{3}} G_{1,7}^{4,1}\left(\frac{1}{46656}|
\begin{array}{c}
 \frac{1}{3} \\
 0,\frac{1}{3},\frac{1}{3},\frac{2}{3},\frac{1}{6},\frac{1}{2},\frac{5}{6}
\end{array}
\right)$$ where appears the Meijer G function (see here and here).
Its numerical value is $0.118802427933651$.
