Cross section of elastic scattering in repulsive central potential field

today I was studying from Landau and Goldstein and I came with this problem

Cross section of elastic scattering in repulsive central potential field $U(r)=frac{alpha}{r}+frac{beta}{r^2}$, where $alpha$, $beta>0$

My approach: I know that cross section of elastic scattering is given by
$$frac{dsigma}{dtheta}=2pirho(theta)left|frac{drho}{dtheta}right|$$
where
$$hspace{2cm}theta=pi-2phi_0mbox{,}quadphi_0=int_{r_{min}}^{infty}frac{frac{M}{mr^2}}{sqrt{frac{2}{m}left(E-U(r)-frac{M^2}{2mr^2}right)}}dr$$
and
$$E=frac{mv_{infty}^2}{2}mbox{,}quad M=mv_{infty}rho.$$

After some calculations and a difficult algebra I obtained
$$theta=pi-frac{2}{sqrt{1+frac{2beta}{mv_{infty}^2rho^2}}} cos^{-1}left(frac{alpha}{sqrt{alpha^2+mv_{infty}^2(mv_{infty}^2rho^2+2beta)}}right).$$
At this part I do not know how to solve for $rho$, is there a way? or do I have to make an approximation? could Mathematica solve it? Thanks in advance.