Integration in the van der Waals attractive potential

Question

The van der Waals attractive potential between 2 point particles is, $$ V = -kr^{-6} $$ The case of circular orbits was easy, the radius being $r_0 = (6k/l^2)^{1/4}$ and time period $\tau \propto r_0^4$, I didn't need integration at all. But for the general bounded motion, the $u(=r^{-1})$ versus $\theta$ curve is given by : $$ \theta = \theta' -\int \frac {du}{\left( \frac {2mE}{l^2} -u^2 + \frac {2mk}{l^2}u^6 \right)^{1/2}} $$ Had $E$ been equal to $k$ I could use the substitution $u + \frac {1}{u} = v$ to reduce it to elliptic integrals. But this is not generally true. Yet Wolfy does return a very messy result (that includes the roots of the sextic in the denominator) involving elliptic integrals which went up my head. Can we really use such a substitution, like $Au + \frac {B}{u} = v$ or there is some other trick? Thanks!

Answer 1

Change notations and write $$I=\int \frac {du}{\left( \frac {2mE}{l^2} -u^2 + \frac {2mk}{l^2}u^6 \right)^{1/2}}=\int \frac {du}{\left(a u^6-u^2+b \right)^{1/2}}$$ Now $$a u^6-u^2+b=a (u^2-r)(u^2-s)(u^2-t)$$ where $(r,s,t)$ are the roots of the depressed cubic equation in $u^2$. So

$$I=\frac 1 {\sqrt a}\int \frac {du} {\sqrt{(u^2-r)(u^2-s)(u^2-t) }}$$ $$I=\frac 1 {\sqrt{as(t-r)}}\,F\left(\sin ^{-1}\left(\sqrt{\frac{u^2 (t-r)}{t \left(u^2-r\right)}}\right)|\frac{t (r-s)}{s (r-t)}\right)$$