Okay, I am going to be very specific. I have the following integral $$\int_{-1}^1 \mathop{dx}\frac{x^{n-2m}(a^2+x^2)^{(k-2)n/2+(3-k)m/2}}{(c_1 +c_2(a^2+x^2)^{k/2}+c_3(a^2+x^2)^{k/2-1}x^2)^{n-m+1/2}}\left(c_4+c_5\frac{x^2}{a^2+x^2}\right)^{n-2m}\left(c_6\frac{x^2}{a^2+x^2}+c_7\right)^m.$$ where all $c_i$ are real constants, $n$ and $m$ are non-negative integers such that $n\geq 2m$. $k$ is a real number which could be both positive or negative.
I am interested in the scaling of the above integral when $a\ll 1$. I believe the scaling will be different when $k>0$ in comparison to when $k<0$.
My approach is the following
As $a\ll1$, $a^2$ is negligible in comparison to $x^2$, except in the region $x\sim 0$. But, in that region, our integrand contains $x^{n-2m}$, and therefore, the contribution of $a$ is negligible even in the region $x\sim 0$. Therefore, my integral should be of the following from $$ \mathcal{I}\sim \int_{-1}^1 \mathop{dx}\frac{x^{n-2m}x^{(k-2)n+(3-k)m}}{(c_1 +(c_2+c_3)x^{k})^{n-m+1/2}}\left(c_4+c_5\right)^{n-2m}\left(c_6+c_7\right)^m.$$ It is quite clear from the above equation that the integral is independent of $a$. Therefore the scaling of the integral is $$\mathcal{I}\sim a^0.$$ The unfortunate part of my answer is that the information about $k$ and $a$ is completely lost in my crude approximation. I do not want this to happen. Do you guys have any idea how to improve this approximation? I am solely interested in the dependence of integral on $a$.
I write here because this stupid site allows me to give an answer but not a simple comment because I don't have 50 reputations yet. NONSENSE.
Anyway: think about the power:
$$(x^2 + a^2)^k$$
where $k$ is the whole stuff you have as exponent. Now:
$$\left(x^2 \left(1 + \frac{a^2}{x^2}\right)\right)^k = x^{2k}\cdot (1+ X)^k$$
and then use the binomial expansion for small $X$