Integrating $\int \sqrt{2me-mkr^2-\frac{1}{2}m br^4 - \frac{a^2}{r^2}} \,dr$

Question

I was trying to find Hamiltons principle function, $S$, for the Hamiltonian:

$$ H = \frac{1}{2} m \left( P_{r}^2 + \frac{P_{\theta}^2}{r^2} \right) + \frac{1}{2}kr^2 + \frac{1}{4} b r^4$$

After considering the form $S = f(r, \theta) - Et = R(r) + \Theta(\theta) - Et$ and identifying the canonical momenta with the partial derivatives of $S$ I got to:

$$ \left(\frac{dR}{dr} \right)^2 r^2 + mkr^4 + \frac{1}{2}mbr^6 - 2mEr^2 = -\left(\frac{d \Theta}{d \theta} \right)^2 $$

Each side must be a constant, $-a^2$, and so:

$$ \frac{dR}{dr} = \sqrt{2mE-mkr^2-\frac{1}{2}m br^4 - \frac{a^2}{r^2}} $$

which leads to:

$$ R(r)= \int \sqrt{2mE-mkr^2-\frac{1}{2}m br^4 - \frac{a^2}{r^2}} \,dr$$

I've tried to find a substitution of variables to make it simpler but to no luck. How can I solve this?

Answer 1

$$\sqrt{2mE-mkr^2-\frac12mbr^4-\frac{a^2}{r^2}}\,dr=\frac1r\sqrt{2mEr^2-mkr^4-\frac12mbr^6-a^2}\,dr$$ Now with the substitution $u=r^2\Rightarrow dr=\frac{du}{2r}$ so: $$\frac{1}{2u}\sqrt{2mEu-mku^2-\frac12mbu^3-a^2}du$$ so we have reduced the order of the polynomial inside the square root. However, integrals of this form do not have "nice" closed-form solutions and so you are probably going to end up with a special function/series to represent this