I have this following differential equation
$$ \left(\frac{dx}{d\gamma} \right)^{2} = \frac{M^2E^2}{L^2}-x^2 (1-2x) \,\, , $$
where the parameters $M,E,L \in \mathbb{R}^{+}$ are all constants, and the variable $x(\gamma), \gamma \in \mathbb{R}$. Performing a separation of variables I get,
$$ \int d\gamma = \int \frac{dx}{\sqrt{\frac{M^2E^2}{L^2}-x^2 (1-2x)}} \,\, , $$ where the left hand side of the equation is straightforward to get $\gamma + C$, but how could I solve the right hand side of this equation?
$$\int d\gamma = \int \frac{dx}{\sqrt{C^2-x^2 (1-2x)}} \quad;\quad C^2=\frac{M^2E^2}{L^2}$$
Well, I am not so smart than the Great Wolfram Mathematica, but I will do my best.
One know how to solve the cubic equation :$\quad C^2-r^2(1-2r)=0\quad$ for $r$.
I let you find the roots $r_1$ , $r_2$ , $r_3$. Now, consider that they are known functions of $C$. As a consequence, we obtain :
$$C^2-x^2 (1-2x)=2(x-r_1)(x-r_2)(x-r_3)$$ $$\int d\gamma = \int \frac{dx}{\sqrt{2(x-r_1)(x-r_2)(x-r_3) }} $$
$$\gamma = -\sqrt2\:\:F\left(X\:|\: m\right)+\text{constant, }$$ $F(X\:|\:m)$ is the elliptic integral of the first kind, with $\begin{cases} X=\sin^{-1}\left(\sqrt{\frac{r_2-r_1}{x-r_1}} \right) \\ m=\frac{r_1-r_3}{r_1-r_2} \end{cases}$
Thanks to WolframAlpha : http://www.wolframalpha.com/input/?i=integrate+1%2Fsqrt(2(x-a)(x-b)(x-c))&x=0&y=0
$$ F\left(X\:|\: m\right)= -\frac{\gamma}{\sqrt2}+c$$ The inverse function is the Jacobi amplitude function : $$X=\text{am}\left(-\frac{\gamma}{\sqrt2}+c\:\bigg|\: m\right)$$ $$x(\gamma)=r_1+\frac{r_2-r_1}{\sin^2\left(\text{am}\left(-\frac{\gamma}{\sqrt2}+c\:\bigg|\: \frac{r_1-r_3}{r_1-r_2}\right)\right)}$$