$\int_{0}^{\infty}\frac{x^{s}}{ax^{3}+bx^{2}+cx+d}dx$ I know that the following formulas are true:
$$\int_{0}^{\infty}\frac{x^{s}}{ax+b}dx=\frac{-\pi}{a\sin\left(\pi s\right)}\left(\frac{b}{a}\right)^{s}$$
and
$$\int_{0}^{\infty}\frac{x^{s}}{ax^{2}+bx+c}dx=\frac{\pi\sin\left(s\cos^{-1}\left(\frac{b}{2\sqrt{ac}}\right)\right)}{c\sin\left(\pi s\right)\sqrt{1-\frac{b^{2}}{4ac}}}\left(\frac{c}{a}\right)^{\frac{s+1}{2}}$$
And so I was wondering if there is a generalisation of this integral to the ratio of $x^s$ and a polynomial of general degree. More specificaly, is there is closed form for the below integral?
$$\int_{0}^{\infty}\frac{x^{s}}{ax^{3}+bx^{2}+cx+d}dx$$
 A: Problems of the form
$$\int_0^\infty\frac{x^s}{\prod_i(x-a_i)}~\mathrm dx$$
can all be handled by taking a keyhole/pacman contour, where we define $x^s$ to have a branch cut along the positive real line. Supposing that $a_i$ are distinct non-positive numbers, and that $s$ is chosen so that it converges, the above integral can easily be given by
$$\int_0^\infty\frac{x^s}{\prod_i(x-a_i)}~\mathrm dx=\frac{2\pi i}{1-e^{2\pi is}}\sum_j\frac{a_j^s}{\prod_{i\ne j}(a_j-a_i)}$$
so it boils down to simply factoring the denominator.

Intuitively, one could also simply apply partial fractions and get a similar result, using the first of your listed results.
A: Write $1/(ax^3 + b x^2 + c x + d)$ in partial fractions:
$$ \frac{1}{a x^3 + b x^2 + c x + d} = \sum_\alpha \frac{1}{(3 \alpha^2 a + 2\alpha b + c)(x - \alpha)} $$
where the sum is over the roots of $a x^3 + b x^2 + c x + d$ (assuming these are distinct).  Of course you'll also want to assume none of these roots are positive reals.  Then, if $s < -1$, you can use your first formula on each of the summands.  
