$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arccos\bigl(-\frac{\beta+2\gamma x}{\sqrt{q}}\bigr)$? 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:
$$\frac{1}{r}=\frac{mk}{l^2}\left(1+\sqrt{1+\frac{2El^2}{mk^2}}\cdot \cos(\varphi-\varphi')\right)$$
 A: Assuming that $q$ is equal to $\beta ^2-4 \alpha \gamma$ otherwise the integral won't make any sense. In the integral $\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}$,  the coefficient of $x^2$ plays a major role.
Assumption: $q=\beta^2-4 \alpha \gamma>0$
When $\gamma <0$
$\begin{align} \int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}&=\int\frac{dx}{\sqrt{- \gamma (-x^2-\frac{\beta}{\gamma}x-\frac{\alpha}{\gamma})}} \\
&=\frac1{\sqrt{-\gamma}} \int \frac{dx}{\sqrt{\frac{\beta^2}{4 \gamma^2}-\frac{\alpha}{\gamma}-\left(x+\frac{\beta}{2\gamma}\right)^2}} 
\quad \;(\because -\gamma >0) \\
&=\frac1{\sqrt{-\gamma}} \int \frac{dx}{\sqrt{\frac{\beta^2-4 \alpha \gamma}{4 \gamma^2}-\left(x+\frac{\beta}{2\gamma}\right)^2}}\\ 
&=\frac1{\sqrt{-\gamma}} \int \frac{dx}{\sqrt{\left(\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 \gamma}\right)^2-\left(x+\frac{\beta}{2\gamma}\right)^2}} \text{ $\quad \;$ since ${\beta^2-4 \alpha \gamma}>0$} \\ 
&=\frac1{\sqrt{-\gamma}} \arcsin{\left( \frac{ \frac{2 \gamma x+\beta}{2\gamma}}{\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 |\gamma|}}\right)}+c \;\quad \left(\because \int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin\left(\frac{x}{\left|a\right|}\right)+C\right)\\
&=\frac1{\sqrt{-\gamma}} \arcsin{\left( -\frac{2 \gamma x+\beta}{\sqrt{\beta^2-4 \alpha \gamma}}\right)}+c \; \quad\; (\because |\gamma|= -\gamma)\\ 
&=-\frac1{\sqrt{-\gamma}} \arcsin{\left( \frac{2 \gamma x+\beta}{\sqrt{\beta^2-4 \alpha \gamma}}\right)}+c  \;\quad(\because \arcsin({-x}) = -\arcsin(x)) \\
&=-\frac{1}{\sqrt{-\gamma}}\arcsin\left(\frac{\beta+2\gamma x}{\sqrt{q}}\right)+c \tag{1}\end{align}$
When $\gamma>0$
$\begin{align} \int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}&=\int\frac{dx}{\sqrt{\gamma (x^2+\frac{\beta}{\gamma}x+\frac{\alpha}{\gamma})}} \\
&=\frac1{\sqrt{\gamma}} \int \frac{dx}{\sqrt{\left(x+\frac{\beta}{2\gamma}\right)^2-\left(\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 \gamma}\right)^2}} \text{ $\;\quad$ since ${\beta^2-4 \alpha \gamma}>0$} \\ 
&= \frac1{\sqrt{\gamma}} \cosh^{-1}{\left( \frac{ \frac{2 \gamma x+\beta}{2\gamma}}{\frac{\sqrt{\beta^2-4 \alpha \gamma}}{2 \gamma}}\right)}+c \; \quad \left(\because \int \frac{dx}{\sqrt{x^2-a^2}}=\cosh^{-1}\left(\frac{x}{a}\right)+C\right)\\ 
&= \frac1{\sqrt{\gamma}} \cosh^{-1}{\left( \frac{2 \gamma x+\beta}{\sqrt{\beta^2-4 \alpha \gamma}}\right)}+c\\
&=\frac1{\sqrt{\gamma}} \cosh^{-1}{\left( \frac{2 \gamma x+\beta}{\sqrt{ q}}\right)} \tag{2}\end{align}$

Since, in your case, $\gamma<0, \;$ $$\int\frac{dx}{\sqrt{\alpha+\beta x+\gamma x^2}}=\frac{1}{\sqrt{-\gamma}}\arcsin\left(-\frac{\beta+2\gamma x}{\sqrt{q}}\right)+c$$ And this result can be rewritten in terms of $\arccos()$ as $$-\frac{1}{\sqrt{-\gamma}}\arccos\left(-\frac{\beta+2\gamma x}{\sqrt{q}}\right)+c'\tag{3}$$ where $c'=c+{ \pi\over 2 \sqrt{-\gamma}}$
Clearly, $(3)$ and $$\frac{1}{\sqrt{-\gamma}}\arccos\left(-\frac{\beta+2\gamma x}{\sqrt{q}}\right) \tag4$$ don't just differ by a constant.
So, you are right, that is a misprint.
Edit: Proceeding with $(3)$ we get,
$\begin{align} &\int\frac{du}{ \sqrt{\frac{2mE}{l^2}+\frac{2mku}{l^2}-u^2}}=-\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)\\
\therefore  & \; \:\varphi=\varphi'+\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)\\
\implies & \cos(\varphi-\varphi')=\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\\
\implies &\frac{1}{r}=u=\frac{mk}{l^2}\left(1+\sqrt{1+\frac{2El^2}{mk^2}}\cdot \cos(\varphi-\varphi')\right)\end{align}$
which is same as the final result.
It can be noticed that it doesn't matter if $\varphi=\varphi'+\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)$ or $\varphi=\varphi'-\arccos\left(\frac{\frac{l^2u}{mk}-1}{\sqrt{1+\frac{2El^2}{mk^2}}}\right)$, the final result would be same since $\cos(\varphi-\varphi')=\cos(\varphi'-\varphi)$
A: If $\gamma<0$, the integral is equal to:
$-\frac{tan^{-1}\Big(\frac{2\gamma.x+\beta}{2\sqrt{-\gamma}.\sqrt{\gamma.x^{2}+\beta.x+\alpha}}\Big)}{\sqrt{-\gamma}}$
