Integral for a particle in a central field I am attempting to solve an integral for the path of a particle in a central field given by
$$U=-\frac{\alpha}{r}+\frac{\beta}{r^2}$$
The integral in question is
$$\theta-\theta_0=L\int\frac{dr}{r^2\sqrt{2m(E+\frac{\alpha}{r}-\frac{\beta}{r^2}-\frac{L^2}{2mr^2})}}$$
Attempt
Making the substitution $r=\frac{1}{s-x}$, where $x$ is some constant to be determined we are left with
$$-L\int\frac{ds}{\sqrt{2mE+2m\alpha(s-x)-2m\beta(s-x)^2-L^2(s-x)^2}}$$
I then chose $x=-\frac{m\alpha}{L^2}$ so that the term linear in $s$ would drop out leaving
$$-L\int\frac{ds}{\sqrt{2mE+\frac{m^2\alpha^2}{L^2}-2m\beta s^2-4\frac{m^2\beta\alpha}{L^2}s-2\frac{m^3\alpha^2\beta}{L^4}-L^2s^2}}$$
I had in mind from here a conversion into the standard form for the $\arcsin$ however the $\beta$ terms will not allow this. I do not know how to proceed from here, if that is even possible.
 A: Following the tip that Sal posted, I begin by changing variables before integration
$$\frac{d\theta}{dr}=\frac{L}{mr^2}\sqrt{\frac{m}{2(E-U_{eff})}}$$
Substituting $s=\frac{1}{r}$
$$\left(\frac{ds}{d\theta}\right)^2=-\frac{m}{L^2}\left(2E+2\alpha s-2\beta s^2-\frac{L^2}{m}s^2\right)$$
From here, I set $$A=-\frac{2mE}{L^2},\ \ \ B=\frac{2m\alpha}{L^2},\ \ \ C=\left(\frac{2\beta m}{L^2}+1\right)$$
Completing the square gives
$$\left(\frac{ds}{d\theta}\right)^2=C\left(s+\frac{B}{2A}\right)^2+D, \ \ \ \text{where} \ D=A-\frac{B^2}{4C}$$
A substitution of $s+\frac{B}{2A}=u$ gives
$$\left(\frac{du}{d\theta}\right)^2=Cu^2+D$$
Now the integral is
$$\theta-\theta_0=\int\frac{du}{\sqrt{Cu^2+D}}$$
This is computed via a substitution of $u=\sqrt{\frac{D}{C}}v$
$$\theta-\theta_0=\frac{1}{\sqrt{C}}\int\frac{dv}{\sqrt{v^2+1}}$$
Which after undoing all substitutions leaves us with:
$$\theta-\theta_0= \frac{1}{\sqrt{C}}\ln\left|\sqrt{\frac{C}{D}}(s+\frac{B}{2A})+\sqrt{\frac{C}{D}(s+\frac{B}{2A})^2+1}\right|$$
Where the constants $A,B,C,D$ are given by
$$A=-\frac{2mE}{L^2},\ \ \ B=\frac{2m\alpha}{L^2},\ \ \ C=\left(\frac{2\beta m}{L^2}+1\right),\ \ \ D=A-\frac{B^2}{4C}$$
Needless to say, this to me does not look like the correct equation for the trajectory of a particle in a central field. If anyone could advise further on the matter It would be greatly appreciated.
