$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arccos\bigl(-\frac{\beta+2\gamma x}{\sqrt{q}}\bigr)$? I'm studying Kepler's laws from Classical Mechanics, 2nd ed. Goldstein. In page 95 there is given an indefinite integral
$$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arccos\biggl(-\frac{\beta+2\gamma x}{\sqrt{q}}\biggr).$$
However, when I took a look the source given in a book (A Short Table of Integrals), there is the result
$$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=-\frac{1}{\sqrt{-\gamma}}\arcsin\biggl(\frac{\beta+2\gamma x}{\sqrt{q}}\biggr).$$
Then, I tried relations of $\arcsin$ and $\arccos$, so $\arcsin(x)=\frac{\pi}{2}-\arccos(x)$ and also negative argument $-\arcsin(x)=\arcsin(-x)$, but just ended up to result
$$-\frac{1}{\sqrt{-\gamma}}\arcsin\biggl(\frac{\beta+2\gamma x}{\sqrt{q}}\biggr)=-\frac{1}{\sqrt{-\gamma}}\arccos\biggl(-\frac{\beta+2\gamma x}{\sqrt{q}}\biggr)+\frac{\pi}{2\sqrt{-\gamma}}.$$
So is there something I don't see or understand, or is there just a misprint in the book?
For the clarification, the result is used to solve this equation:
$$\varphi=\varphi_{0}-\int\frac{du}{ \sqrt{\frac{2mE}{l^2}+\frac{2mku}{l^2}-u^2}}$$
and the book ends up to result
$$\varphi=\varphi'-\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)$$
and by solving the $u=1/r$ we got final result:
$$\frac{1}{r}=\frac{mk}{l^2}\left(1+\sqrt{1+\frac{2El^2}{mk^2}}\cdot \cos(\varphi-\varphi')\right)$$

If $\gamma<0$, the integral is equal to:
$-\frac{tan^{-1}\Big(\frac{2\gamma.x+\beta}{2\sqrt{-\gamma}.\sqrt{\gamma.x^{2}+\beta.x+\alpha}}\Big)}{\sqrt{-\gamma}}$
