Form of radial velocity along null geodesic under the Kerr metric

[Bertin]

Homework Statement

Given the affine parameter $\lambda$ of a null geodesic on the equator ($\theta = \pi/2$), prove that that the $r$ coordinate satisfies the following equation:
$$\left(\frac{dr}{d\lambda}\right)^2 = \frac{\Sigma^2}{\rho^4}(E - L W_-(r))(E - L W_+(r))$$
for some $W(r)$ that might depend on $E,L$ and $r$, and for $E, L$ constants of motion.

Relevant Equations

The Kerr metric, in the Boyer-Lindquist coordinates and on the equator, reads
$$ds^2 = -(1 - \frac{R}{r})dt^2 - \frac{R}{r}a (dtd\phi + d\phi dt) + \frac{r^2}{r^2 + a^2 - R r} dr^2 + \frac{\Sigma^2}{r^2} d\phi^2$$
for $\Sigma^2 = r^4 + a^2 r^2 + R r a$.

By the symmetries of the metric, k = \partial_t and l = \partial_\phi are Killing vectors. Since they are Killing vectors, they satisfy k_\mu \dot{x}^\mu = E and l_\mu \dot{x}^\mu = L, for the same constants appearing in the expression we must prove, and where the dot means the derivative w.r.t. to the affine parameter. Hence it follows that
$$E = -(1 - \frac{R}{r})\dot{t} - \frac{R}{r}a\dot{\phi}$$
$$L = -\frac{R}{r}a\dot{t} + \frac{\Sigma^2}{r^2}\dot{\phi}$$.
Moreover, since x(\lambda) is a null geodesic, we have that \dot{x}_\mu\dot{x}^\mu = 0, whence
$$ 0 = \frac{r^2}{r^2 + a^2 - R r}\dot{r}^2 - (1 - \frac{R}{r})\dot{t}^2 - 2\frac{R}{r}a\dot{t}\dot{\phi} + \frac{\Sigma^2}{r^2}\dot{\phi}^2$$

We can then solve the equations of E and L for \dot{t} and \dot{\phi} to later replace those values inside last equation. Nevertheless, this leads to a very messy expression for \dot{r}^2 that does not look that the one we must prove, first and foremost because the resulting expression doesn't seem to include any \frac{\Sigma^2}{r^4}E^2 (unless both Mathematica and I are missing a possible simplification, which could be the case), so I probably have done some mistake (not calculatory, though, because my results agree with Mathematica) or I am missing something.

I would appreciate if someone could show me how do we derive above expression. Thank you in advance.