How do I solve $-\int \frac{dy}{y(ay^2+by+c)^{1/2}}$?

Question

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-\frac{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 + \left(\dfrac{b}{2a}\right) \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 $c<0$.

Answer 1

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.

Answer 2

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.

Answer 3

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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}