I'm trying to resolve the elliptic integral that appears in the Kepler problem, i.e., \begin{align} I=\int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}. \end{align} Goldstein says the the integral is \begin{align} \int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arccos \left(-\frac{\beta+2\gamma x}{\sqrt{\beta^2-4\alpha\gamma}}\right). \end{align} I don't know how to resolve the integral.
My attempt: \begin{align} \int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}=\int \frac{dx}{\sqrt{\alpha-\frac{\beta^2}{4\gamma}+\gamma\left(x+\frac{\beta}{2\gamma}\right)^2}} \end{align} Changing variables like $x=\frac{\beta}{2\gamma}\tan^2 \theta\rightarrow dx=\frac{\beta}{2\gamma}2\tan\theta \sec\theta\; d\theta$, \begin{align} \therefore I=\int\frac{\frac{\beta}{2\gamma}2\tan\theta\sec\theta\;d\theta}{\sqrt{\alpha-\frac{\beta^2}{4\gamma^2}+\gamma\left[\frac{\beta}{2\gamma}(\tan^2\theta+1)\right]^2}}=\int\frac{\frac{\beta}{\gamma}\tan\theta\sec\theta}{\sqrt{\alpha+\sec^4\theta}}\;d\theta=\frac{\beta}{\gamma}\int\frac{\sin\theta}{\sqrt{\alpha \cos^4\theta+1}}\;d\theta. \end{align}
=================================================
EDIT AND SOLUTION
\begin{align} I&=\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\int\frac{dx}{\sqrt{\alpha+\gamma\left(\frac{\beta}{\gamma}x+x^2\right)}}=\int\frac{dx}{\sqrt{\alpha+\gamma\left(x+\frac{\beta}{2\gamma}\right)^2-\frac{\beta^2}{4\gamma}}} \end{align} If $y=x+\frac{\beta}{2\gamma}$: \begin{align} I&=\int\frac{dy}{\sqrt{\alpha+\gamma y^2-\frac{\beta^2}{4\gamma}}}=\int\frac{dy}{\sqrt{\frac{4\alpha\gamma-\beta^2}{4\gamma}+\gamma y^2}} \end{align} If $q=\beta^2-4\alpha\gamma$, \begin{align} I=\int\frac{dy}{\sqrt{\gamma y^2-\frac{q}{4\gamma}}} \end{align} Then $y=-\frac{\sqrt{q}}{2\gamma}\cos\theta$ and $dy=\frac{\sqrt{q}}{2\gamma}\sin\theta\;d\theta$, \begin{align} \therefore I=\frac{\sqrt{q}}{2(-\gamma)}\int\frac{-\sin\theta\;d\theta}{\sqrt{\frac{q}{4\gamma}(\cos^2\theta-1)}}=\frac{\sqrt{q}}{2(-\gamma)}\int\frac{-\sin\theta\;d\theta}{\sqrt{\frac{q}{4(-\gamma)}(1-\cos^2\theta)}}=-\frac{1}{\sqrt{-\gamma}}\;\theta\\ \boxed{\therefore \int \frac{dx}{\sqrt{\alpha +\beta x+\gamma x^2}}=-\frac{1}{\sqrt{-\gamma}}\arccos\left(-\frac{\beta+2\gamma x}{\sqrt{\beta^2-4\alpha\gamma}}\right)} \end{align}