Integrating $\int_{0}^{\infty} \frac{p^6 dp }{1 + a p^4 + b p^6 } \int_{0}^{\pi}\frac{\sin^5 \theta \,d\theta}{1 + a |p-k|^4 + b |p-k|^6 }$

Question

This is my first question here, so I hope I'm not giving too little/too much information. I need some help calculating (or even approximating) an integral which I've been wrestling with for a while.

As part of my internship, I need to calculate or even just approximate a power spectrum which boils down to the following integral:

$$I(k)=\int \frac{d^3p}{(2\pi)^3} (1 - \kappa^2)(1-\lambda^2) \frac{\left\lvert p \right\rvert^2}{(1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6)} \frac{\left\lvert p-k \right\rvert^2}{(1 + a \left\lvert p-k \right\rvert^4 + b \left\lvert p-k \right\rvert^6)}$$

where $\kappa = \hat{k}.\hat{p}$ and $\lambda = \hat{k}.\widehat{k-p}$ (Obviously $\left\lvert p-k \right\rvert^2 = p^2 + k^2 - 2 p k \cos \theta$). I need the answer in terms of $a,b$, two positive constants that I'd like to vary to describe different physical situations.

First attempt at a solution:

I moved into spherical coordinates and (I'm reasonably sure that) the integral reduces to:

$$\int \frac{d^3p}{(2\pi)^3} \frac{\left\lvert p \right\rvert^2}{1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6 } \frac{\left\lvert p \right\rvert^2\sin^4 \theta}{1 + a \left\lvert p-k \right\rvert^4 + b \left\lvert p-k \right\rvert^6 }$$

So naturally I considered performing the integration

$$\frac{1}{2\pi^2}\int_{0}^{\infty} \frac{\left\lvert p \right\rvert^6 dp }{1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6 } \int_{0}^{\pi}\frac{\sin^5 \theta \,d\theta}{1 + a \left\lvert p-k \right\rvert^4 + b \left\lvert p-k \right\rvert^6 }$$

Sadly, this doesn't really help me too much since these integrals are still pretty hard. Mathematica does manage to calculate them, but the results are really far too cumbersome to handle.

Further attempts:

As an approximation, I considered sloppily breaking the integral into $p<k$ and $p>k$, which leads me to something of the form (modulo constants):

$$\frac{1}{1 + a \left\lvert k \right\rvert^4 + b \left\lvert k \right\rvert^6 }\int_{0}^{k} \frac{\left\lvert p \right\rvert^6 dp}{1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6} +\int_{k}^{\infty} \frac{\left\lvert p \right\rvert^6 dp}{(1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6)^2 } $$

But I found even these "simpler" integrals too hard to calculate, and I'm not particularly fond of the approximation either.

Questions:

1) Does anyone know of a way I could solve or even approximate any of the above integrals as a function of the parameters $a$ and $b$?

2) It reminds me slightly on a convolution product, though not exactly since it's the norm of the vector $\left\lvert p-k \right\rvert$, which means I can't use the convolution theorem (can I?).

3) I know how to use the Feynman Parametrization for terms that are quadratic in the denominator. But I don't think it would work with $\left\lvert p\right\rvert^6$ (would it?).

3) The parameters $a$ and $b$ are quite small ($a$ is of the order of $10^{-2}$ and $b$ around the same), I was wondering if I could use that to my advantage, maybe perform some sort of Taylor expansion in either of them, reducing the integral to something more manageable, but I'm not sure that's allowed, and at any rate such integrals seem to diverge.

Thanks!