How do I solve $-\int \frac{dy}{y(ay^2+by+c)^{1/2}}$? I have the following integral which I am trying to solve, but I am unsure of how to proceed! Any help is much appreciated! 
\begin{equation}
-\int \frac{dy}{y(ay^2+by+c)^{1/2}}
\end{equation}
My strategy is to complete the square on the denominator and find a viable trig substitution. I am currently at stage of having completed the square, and performed a substitution. 
\begin{equation}
-\int \frac{du}{\bigg(u-b/(2a)\bigg)\bigg(u^2+ \frac c a -\frac 1 4 \bigg(\frac ba \bigg)^2\bigg)^{\frac 12}}
\end{equation}
where 
\begin{equation}
u=y + (b/2a)
\end{equation}
The result should take the form 
\begin{equation}
\frac{1}{\sqrt{-c}} \cos ^{-1}\frac{by +2c}{y\sqrt{b^2-4ac}}
\end{equation}
Where $c<0$.
 A: Assuming $a, c\gt0, 4ac-b^2>0$.
Switching to $x$ as a variable because I'm used to it and denoting the integral with $I$.
First complete the square:
$$ -I=\int\frac{dx}{x\sqrt{(\sqrt ax+\frac{b}{2\sqrt a})^2+c-\frac{b^2}{4a}}}=2\sqrt a\int\frac{dx}{x\sqrt{(2ax+b)^2+4ac-b^2}}$$
Now perform substitution $u=2ax+b$ and you get
$$2\sqrt a\int\frac{du}{(u-b)\sqrt{u^2+4ac-b^2}}$$
Now substitution $u=\sqrt{4ac-b^2}\sinh (v)$ and simplifying you get 
$$2\sqrt a\int\frac{dv}{\sqrt{4ac-b^2}\sinh(v)-b}$$
Now half-angle hyperbolic substitution $w=\tanh(\frac v2)$ yields
$$2\sqrt a\int\frac{2\,dw}{bw^2+2\sqrt{(4ac-b^2)}w-b}$$
Now factor the denominator (a bit tedious) and perform PFD, and you'll get a few easy integrals, undo the substitutions and you are done.
A: Hint on the general method:
This is an example of an abelian integral, i.e. an integral of the form
$$\int R(y,u)\,\mathrm dy,\qquad\text{where $u$ and $y$ are linked by a polynomial relation}\enspace p(y,u)=0 $$
Here, setting $u=\sqrt{ay^2+by+c}$, we obtain the quadratic relation $\;ay^2-u^2+by+c=0$.
If $a\ne 0$, this integral can be calculated by substitution with the following steps:


*

*First  write the quadratic polynomial $f(y)=ay^2+by+c\;$ in canonical form:
$$ay^2+by+c=a\biggl[\Bigl(y+\frac b{2a}\Bigr)^2+\frac{4ac-b^2}{4a^2}\biggr].$$
Set $\;t=y+\smash[t]{\dfrac b{2a}}$ and, as usual $\Delta=b^2-4ac$.

*Depending on the signs of $a$ and  $\Delta$, the square root takes one of the forms:
$$\sqrt{ay^2+by+c}=\begin{cases}
\lvert a\rvert\sqrt{t^2 -D^2}&\bigl(D=\sqrt{\lvert\Delta\rvert}\bigr)\\[1ex]
\lvert a\rvert\sqrt{t^2 +D^2}\\[1ex]
\lvert a\rvert\sqrt{D^2-t^2}
\end{cases}$$

*The trick is now to find a substitution that eliminates the square root:

*For $\sqrt{t^2 -D^2}$, one can set $\;t=D\cosh\theta$ $\;(\theta\ge 0)$.

*For $\sqrt{t^2 +D^2}$, one can set $\;t=D\sinh\theta$, or $\;t=D\tan\theta$ $\;(-\frac\pi2<\theta<\frac\pi2)$.

*For $\sqrt{D^2 -t^2}$, one can set $\;t=D\sin\theta$ $\;(-\frac\pi2<\theta<\frac\pi2)$, or $\;t=D\tanh\theta$.


These substitutions turn  the integral into the integral of a rational function of trigonometric or hyperbolic functions, which in turn can be changed into an integral of a rational function by a suitable second substitution.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With
  Euler-sub$\ds{\ldots\ \root{ay^{2} + by + c} = \root{a}y + t}$: 

\begin{align}
\int{\dd y \over y\root{ay^{2} + by + c}} & =
\int{2\,\dd t \over t^{2} - c} =
{1 \over \root{c}}\int\pars{%
{1 \over t - \root{c}} - {1 \over t + \root{c}}}\,\dd t
\\[5mm] & =
{1 \over \root{c}}\,\ln\pars{t - \root{c} \over t + \root{c}}
\\[5mm] = &\
{1 \over \root{c}}\,\ln\pars{\root{ay^{2} + by + c} - \root{a}y - \root{c} \over \root{ay^{2} + by + c} - \root{a}y + \root{c}} +
\pars{~\mbox{a}\ constant~}
\end{align}
