Since you notation is not quite concise I will do some guesswork in order to understand the underlying problem. The basic setting is that of a test particle that moves in a central potential $V(r)$. It holds that the change in the polar angle $\phi$ is given by $$ \Delta \phi = 2 \int_{r_\text{min}}^{r_\text{max}} \frac{\left(\frac{l}{r^2}\right)\text{d}r }{\sqrt{2m(E-V(r)) - \frac{l^2}{r^2}}},\qquad(1) $$ where $E,l$ is the initial energy and angular momentum of the system and $r_\text{min}, r_\text{max}$ are the minimal and maximal radius reached of the test mass $m$.
If I understand correctly you want to study the potential $$ V(r) = -\frac{\alpha}{r} + \beta r^2 $$ in the regimes where the first term dominates the second and vice versa (the situation roughly translates into (for all times) small and (for all times) large distances from the origin). It seems that OP is already aware of the method for example explained in "Mechanics. Vol. 1" by L.Landau and E. Lifshitz (chapter 15 exercise 3). I think the misunderstand comes from the fact that it makes no sense to include the "centrifugal term" into the $1/r$-potential. The bookkeeping should be: $$ V(r) = U(r) + \delta U (r),\qquad(2) $$ where in case one $$ U(r)=-\frac{\alpha}{r},\quad\delta U (r) = \beta r^2,\qquad(2.1) $$ and in case two $$ U(r)=-\beta r^2,\quad\delta U (r) = -\frac{\alpha}{r}.\qquad(2.2) $$$$ U(r)=\beta r^2,\quad\delta U (r) = -\frac{\alpha}{r}.\qquad(2.2) $$ Using this bookkeeping the insertion of (2.1), (2.2) in (2) then in (1) and then expanding in orders of $\delta U$ should result in the desired corrections for the two cases.