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I am looking for whether the integral $$\int_{-\infty}^{\infty}\frac{x^i}{((x^2+a)^2+b^2)^{\frac{3}{2}}}dx$$ where $i=0,\ldots, 4$ and $a,b>0$ are parameters can be calculated in an elementary way. I myself got stuck and online calculators gave quite horribly looking answers. It would be especially nice to see a connection to probability (i.e. interpreting the above as moments of some known distribution), as the background of the question, which is a PDE problem, would actually suggest there might be one.

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    For $i=1,3$, the integrand is an odd function, so the integral of course vanishes identically. For the remaining cases, the integrals are general elliptic integrals. So in principle, they can definitely be solved, though it may end up being a pain in arse. – 2017-01-26
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    Have you tried contour integration? – 2017-02-03

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Cases $i=1,3$ are considered in comments, the interal is zero.

If $$i=2k,\quad k=0,1,2,$$ then $$J_{2k} = 2\int_0^{\infty}\frac{x^{2k}}{((x^2+a)^2+b^2)^{3/2}}\,dx,$$ $$\frac14J_{2k}^2 = \int_0^{\infty}\frac{x^{2k}}{((x^2+a)^2+b^2)^{3/2}}\,dx\cdot\int_0^{\infty}\frac{y^{2k}}{((y^2+a)^2+b^2)^{3/2}}\,dy$$ $$= \int_0^{\infty}\int_0^{\infty}\frac{(xy)^{2k}}{(((x^2+a)^2+b^2)((y^2+a)^2+b^2))^{3/2}}\,dxdy.$$ Using polar coordnates for the first quadrant: $$x=\rho\cos\varphi, y=\rho\sin\phi, dxdx=\rho\,d\rho\,d\varphi,$$ $$\frac14 J_{2k}^2 = \frac1{2^k}\int_0^\infty\int_0^{\frac\pi2}\frac{\rho^{4k}\sin^k{2\varphi}d\varphi}{\left(\left(\left(\rho^2\cos^2\varphi+a\right)^2+b^2\right)\left(\left(\rho^2\sin^2\varphi+a\right)^2+b^2\right)\right)^{3/2}}\,d{\frac{\rho^2}2}.$$ Using substitution $$r=\frac{\rho^2}2:$$ $$\frac14J_{2k}^2 = \int_0^\infty\int_0^{\frac\pi2}\frac{r^k\sin^k{2\varphi}d\varphi}{\left(\left(\left(r+a+r\cos{2\varphi}\right)^2+b^2\right)\left(\left(r+a-r\cos{2\varphi}\right)^2+b^2\right)\right)^{3/2}}\,dr.$$ There are many ways for the further elementary calculations. However, I do not see how to make effective progress.