TITLE: Integral $-\int_{0}^{a}{at^{3}\,{\rm d}t\over\,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,}$
QUESTION [8 upvotes]: This problem's taking me a lot longer than I like (probably doing things the hard way...).  I have 9 different terms to integrate, some of which are messier than others.  This one, though, is messier than most.
$$-{\Large\int}_{\!0}^{a}
\!\raise 0.8ex {at^{3}\,{\rm d}t
\over
\,\sqrt{\,\left(a^{2} -t^{2}\right)\left(b^{2}t^{2} +a^{4}-a^{2}t^{2}\right)\,}\,}
$$
I tried a whole series of substitutions on this.  First $t=a\sin\left(u\right)$, then $v=\cos\left(u\right)$ and split the resulting integral into two, only to find (after substitution number three) that the first half diverged (haven't tried the second half).  In any case, I'm assuming there has to be a better way to attempt this.  Any suggestions appreciated.

REPLY [8 votes]: $\newcommand{\+}{^{\dagger}}%
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$\ds{\fermi\pars{a,b}
\equiv
-\int_{0}^{a}{at^{3}\,{\rm d}t
\over
\,\sqrt{\,\left(a^{2} - t^{2}\right)\left(b^{2}t^{2} +a^{4}- a^{2}t^{2}\right)\,}\,}
\,,\quad \fermi\pars{0,b} = 0}$
When $a \not= 0$:
\begin{align}
\\[3mm]
\fermi\pars{a,b}
&=
-\sgn\pars{a}\int_{0}^{\verts{a}}{\verts{a}t^{3}\,{\rm d}t
\over
\,\sqrt{\,\left(a^{2} - t^{2}\right)\left(b^{2}t^{2} +a^{4}- a^{2}t^{2}\right)\,}\,}
\\[3mm]&=
-a\verts{a}{\rm F}\pars{\mu}\quad\mbox{where}\quad{\rm F}\pars{\mu} \equiv\int_{0}^{1}{t^{3}\,{\rm d}t
\over
\,\sqrt{\,\left(1 - t^{2}\right)\left(\mu t^{2} + 1\right)\,}\,}\,,\quad
\mu \equiv \pars{b \over a}^{2} - 1
\end{align}

\begin{align}
{\rm F}\pars{\mu} &=
\half\int_{0}^{1}{t\,{\rm d}t \over \root{\pars{1 - t}\pars{\mu t + 1}}}
\end{align}

It's clear that $\color{#00f}{{\rm F}\pars{0} = 2/3}$. When
$\color{#00f}{\mu \not= 0}$ we perform the change of variables
$t = 1 - x^{2}$ $\quad\iff\quad$ $x = \pars{1 - t}^{1/2}$:
\begin{align}
{\rm F}\pars{\mu}
&=
\half\int_{1}^{0}{1 - x^{2} \over x\bracks{\mu\pars{1 - x^{2}} + 1}}\,
\pars{-2x\,\dd x}
=
\int^{1}_{0}{1 - x^{2} \over \mu + 1 - \mu x^{2}}\,\dd x
\\[3mm]&=
{1 \over \mu}\int^{1}_{0}{\pars{\mu + 1- \mu x^{2}} - 1
                          \over \mu + 1 - \mu x^{2}}\,\dd x
={1 \over \mu} - {1 \over \mu^{2}}\int_{0}^{1}{\dd x \over \pars{1 + 1/\mu} - x^{2}}
\end{align}
$\color{#00f}{\large \mu < 0}$:
\begin{align}
{\cal F}\pars{\mu}
&=\color{#00f}{\large{1 \over \mu} + {1 \over \mu^{2}}\int_{0}^{1}
{\dd x \over x^{2} + \pars{\root{-1 - 1/\mu}}^{2}}}
\\[3mm]&={1 \over \mu} + {1 \over \mu^{2}}\,\root{-\mu \over \mu + 1}
\arctan\pars{\root{-\mu \over \mu + 1}}
\end{align}

$\color{#00f}{\large \mu > 0}$:
\begin{align}
{\cal F}\pars{\mu}
&={1 \over \mu} + {1 \over \mu^{2}}\int_{0}^{1}
{\dd x \over x^{2} - \pars{\root{1 + 1/\mu}}^{2}}
\\[3mm]&={1 \over \mu} + {1 \over \mu^{2}}
\int_{0}^{1}\pars{{1 \over x - \root{1 + 1/\mu}} - {1 \over x + \root{1 + 1/\mu}}}
{1 \over 2\root{1 + 1/\mu}}
\\[3mm]&=\color{#00f}{\large{1 \over \mu} + {1 \over 2\mu^{2}}\,
\root{\mu \over 1 + \mu}
\ln\pars{\verts{\root{\mu} - \root{1 + \mu} \over \root{\mu} + \root{1 + \mu}}}}
\end{align}

$\color{#00f}{\large \mu = 0}$:
    $${\cal F}\pars{\mu} = \color{#00f}{\large{\cal F}\pars{0} = {2 \over 3}} $$