Planetary Motion: Integral I am solving central force problem to deduce equations of orbit of planets. During the calculation, I am stuck over an integral which I am unable to solve. Can anyone help or guide me in this?

The integral is:
$$\int \dfrac{dr}{r \left(2\mu Er^2 + 2 \mu Cr - l^2 \right)^{1/2}} .$$

Here, $\mu$, $E$, $C$ and $l$ are constants.
Thanks in advance!
 A: Make the substitution $\frac1r = x$. Then,
$$
\begin{align*}
\int \frac{dr}{r \left(2\mu Er^2 + 2 \mu Cr - l^2 \right)^{1/2}} 
&= \int \frac{dr}{r^2 \left( 2\mu E + \frac{2 \mu C}{r} - \frac{l^2}{r^2} \right)^{1/2}} 
\\ &= - \int \frac{dx}{\left( 2\mu E + 2 \mu C x - l^2x^2 \right)^{1/2}} 
\\ &= - \int \frac{dx}{\left( 2\mu E+ \frac{\mu^2 C^2}{l^2}  - \left( lx - \frac{\mu C}{l} \right)^2 \right)^{1/2}} 
\end{align*}
$$
Substitute $lx - \frac{\mu C}{l} = \sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \sin \theta$. Then $l dx = \sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta d \theta$. Therefore, the integral simplifies to
$$
\begin{align*}
- \frac{1}{l} \int \frac{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta d \theta}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta}
&= - \frac{1}{l} \int d \theta = - \frac{\theta}{l}
\\ &= - \frac{1}{l} \arcsin \left( \frac{lx - \frac{\mu C}{l}}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}}} \right)
\\ &= - \frac{1}{l} \arcsin \left( \frac{\frac{l}{r} - \frac{\mu C}{l}}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}}} \right)
\end{align*}
$$

Intuition for some of the steps. The above answer might appear a bit too slick, but it really is natural. In the step where we got
$$
- \int \frac{dx}{\left( 2\mu E+ \frac{\mu^2 C^2}{l^2}  - \left( lx - \frac{\mu C}{l} \right)^2 \right)^{1/2}} ,
$$
the natural thing to do is to substitute $lx - \frac{\mu C}{l} = y$. Moreover, we will denote the constant $2\mu E+ \frac{\mu^2 C^2}{l^2}$ by $A^2$. Then the integral becomes
$$
- \frac{1}{l} \int \frac{dx}{\sqrt{A^2 - y^2}} .
$$
The reason for calling that quantity $A^2$ instead of $A$ is to make the expression inside the square root look homogeneous (in physical terms, now $A$ has the same dimension as $y$). The last integral is a standard integral. It is fairly common to try substitutions involving trigonometric or hyperbolic functions. After some trial and error, we find that $y = A \sin \theta$ and $y = A \cos \theta$ will both work. [Of course, in this case, substituting $y$ as $A \tan \theta$, $A \sec \theta$, $A \cosh \theta$, $A \sinh \theta$ etc. do not seem to give anything.] My solution uses $y = A \sin \theta$.
