While attempting to solve this problem I ran into a nasty limit.
Mathematica claims that the indefinite integral
$$\int\left[\frac{1}{\sqrt{a+b}(z^2)^{n/4}}-\frac{1}{\sqrt{a+b\cos^2\theta}(R^2+z^2)^{n/4}}\right]\,dz $$
where $$\cos\theta = \frac{z}{\sqrt{z^2+R^2}}$$
is equal to
$$-\frac{z \left(R^2+z^2\right)^{-\frac{n}{4}} \left(\frac{z^2}{R^2}+1\right)^{\frac{n+2}{4}} \sqrt{\frac{a \left(R^2+z^2\right)+b z^2}{R^2+z^2}}}{a (a+b) \sqrt{\frac{z^2 (a+b)}{a R^2}+1}}-\frac{2 z \left(z^2\right)^{-\frac{n}{4}}}{(n-2) \sqrt{a+b}}\cdot\left[b F_1\left(\frac{1}{2};\frac{n+2}{4},\frac{1}{2};\frac{3}{2};-\frac{z^2}{R^2},-\frac{(a+b) z^2}{a R^2}\right)+a F_1\left(\frac{1}{2};\frac{n+2}{4},-\frac{1}{2};\frac{3}{2};-\frac{z^2}{R^2},-\frac{(a+b) z^2}{a R^2}\right)\right]$$
With the restrictions $a>0$, $b>0$, $0<n<2$ and $R\ll 1$.
We want the definite integral from $-\infty $ to $\infty$ and because the function is even we can take the difference of the limits at infinity and $0$. Mathematica evaluates to limit at $0$ to be equal to zero but isn't able to evaluate the upper limit.
In an effort to make the arguments of the Appell functions converge to something I applied the following transformation (citation)
$$F_1\left(a;b_1,b_2;c;z_1,z_2\right)\to\left(1-z_1\right){}^{-b_1} \left(1-z_2\right){}^{-b_2} F_1\left(c-a;b_1,b_2;c;\frac{z_1}{z_1-1},\frac{z_2}{z_2-1}\right)$$
Which simplifies to
$$\lim_{z\to\infty}z\left(\frac{\left(R^2+z^2\right)^{-\frac{n}{4}} \sqrt{a+\frac{b z^2}{R^2+z^2}}}{(a+b) \left(z^2 (a+b)+a R^2\right)}+\frac{2 \left(z^2\right)^{-\frac{n}{4}}}{(n-2) \sqrt{a+b}}\left[b R^2 F_1\left(1;\frac{n+2}{4},\frac{1}{2};\frac{3}{2};\frac{z^2}{R^2+z^2},\frac{(a+b) z^2}{a R^2+(a+b) z^2}\right)+\left(z^2 (a+b)+a R^2\right) F_1\left(1;\frac{n+2}{4},-\frac{1}{2};\frac{3}{2};\frac{z^2}{R^2+z^2},\frac{(a+b) z^2}{a R^2+(a+b) z^2}\right)\right]\right)$$
The arguments now converge to 1, but mathematica claims that the limit is complex infinity which I know not to be true because the integral converges for the special cases that I have tried numerically. Where is my mistake? The transformation seems valid because the only restriction is that $$z_1, z_2\not\in(1, \infty)$$ which is clearly the case. Do I need more restrictions to ensure convergence?
Edit: Using the comment from Pierpaolo Vivo the limit in question becomes
$$\lim_{\zeta\to\infty}\left[\frac{\zeta \sqrt{\frac{\zeta ^2}{\alpha ^2}+1}}{\sqrt{\alpha ^2+\zeta ^2}}+\frac{2 \zeta ^{1-\frac{n}{2}}}{n-2}F_1\left(\frac{1}{2};\frac{n-2}{4},\frac{1}{2};\frac{3}{2};-\zeta ^2,-\frac{\zeta ^2}{\alpha ^2}\right)\right]$$
When I apply the same transformation as above we obtain
$$F_1\left(\frac{1}{2};\frac{n-2}{4},\frac{1}{2};\frac{3}{2};-\zeta ^2,-\frac{\zeta ^2}{\alpha ^2}\right)=\frac{\left(\zeta ^2+1\right)^{\frac{1}{2}-\frac{n}{4}}}{\sqrt{\frac{\zeta ^2}{\alpha ^2}+1}}F_1\left(1;\frac{n-2}{4},\frac{1}{2};\frac{3}{2};\frac{\zeta ^2}{\zeta ^2+1},\frac{\zeta ^2}{\alpha ^2+\zeta ^2}\right)$$
The limit now becomes
$$\lim_{\zeta\to\infty}\left[\zeta \left(\frac{\left(\zeta ^2+1\right)^{\frac{1}{2}-\frac{n}{4}} F_1\left(1;\frac{n-2}{4},\frac{1}{2};\frac{3}{2};\frac{\zeta ^2}{\zeta ^2+1},\frac{\zeta ^2}{\alpha ^2+\zeta ^2}\right)}{\sqrt{\alpha ^2+\zeta ^2}}+\frac{2 \zeta ^{-\frac{n}{2}}}{n-2}\right)\right]$$
Now mathematica evaluates the limit to be $0$ at both end points..... I feel like there has to be something flawed with these transformations but all the given criteria are met. What's going on here?