Calculate an integral of an irrational function How can I calculate this integral 
$$\int \frac{dx}{(x^2+b)\sqrt{x^2-a}} $$
Without using the substitution $$x=\sqrt{a}\sec {u} $$
I guess I'm having some problems while using secant substitution so I wanted to know if there is any other possible way to solve the above integral . 
Any hint will be appreciated.
Thank you in return!
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{\quad x \equiv {t^{2} + a \over 2t}\quad}$ and
  $\ds{\quad t = x - \root{x^{2} - a}}$:

\begin{align}
&\bbox[10px,#ffd]{\ds{\int{\dd x \over \pars{x^{2} + b}\root{x^{2} - a}}}} =
-\int{4t \over t^{4} + 2\pars{a + 2b}t^{2} + a^{2}}\,\dd t
\\[5mm] \stackrel{y\ =\ t^{2}}{=}\,\,\,&
-2\int{\dd y \over y^{2} + 2\pars{a + 2b}y + a^{2}}
\\[5mm] = &\
-2\int{\dd y \over \pars{y + a + 2b}^{2} + a^{2} - \pars{a + 2b}^{2}} =
-2\int{\dd y \over \pars{y + a + 2b}^{2} - 4b\pars{b - a}}
\end{align}

At this point, the integration is an elementary one. What is the relation between $\ds{a\ \mbox{and}\ b}$ ?.

