Calculating $\int_{-\infty}^{\infty}\frac{x^i}{((x^2+a)^2+b^2)^{\frac{3}{2}}}dx$ 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.
 A: 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 $$r=\frac{\rho^2}2:$$
$$\frac14J_{2k}^2 = \int_0^\infty\int_0^{\frac\pi2}\frac{r^{2k}\sin^k{2\varphi}d\varphi}{\left(\left(\left(r+a+r\cos{2\varphi}\right)^2+b^2\right)\left(\left(r+a-r\cos{2\varphi}\right)^2+b^2\right)\right)^{3/2}}\,dr.$$
There are many ways for the further elementary calculations. However, I do not see how to  make effective progress.
