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How do I integrate:

$$\int \frac{1}{p+q(x-r)^2}\frac{1}{\sqrt{s+t x^2}}\, dx$$

All variables other than $x$ can be assumed to be greater than $0$ and independent of $x$. Pointers to a formula from an integration table are also sufficient. This wikipedia article is almost what I need, except for the fact that I have an $x-r$ in just one of the terms.

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What is y? Is it a function of x? Or a constant? – Martin Leslie Sep 9 '12 at 1:39
@Martin Everything other than $x$ is independent of $x$. I've changed it to $e$ to avoid confusion :) – user39836 Sep 9 '12 at 1:40
Using $e$ will confuse people thinking it's the natural log base e. A variable like $m$ or $n$ would probably be better. – Envious Page Sep 9 '12 at 1:42
@EnviousPage Good point! Changed it and clarified in the text too. – user39836 Sep 9 '12 at 1:44
up vote 3 down vote accepted

This may be the hard way, but

  1. Substitute $x=\sqrt{s/t}\tan\theta$. That will get rid of the square root, and turn the integrand into a rational function of trig functions.

  2. Use the $\tan(t/2)$ substitution to turn the integrand into a rational function.

  3. Use partial fractions to do the resulting problem.

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As an alternative to Gerry's suggestion, you could Euler's substitution. Let's relabel $t$ to $t^2$, i.e. we are solving for $$ \int \frac{1}{p+q(x-r)^2} \frac{1}{\sqrt{s + t^2 x^2}} \mathrm{d} x $$

Specifically, make a change of variables $$ x = \frac{u^2-s}{2 t u}, \quad \mathrm{d}x = \frac{u^2+s}{2 t u^2} \mathrm{d} u, \quad \frac{1}{\sqrt{s + t^2 x^2}} = \frac{2 u}{s+u^2}, \quad \frac{\mathrm{d}x}{\sqrt{s + t^2 x^2}} =\frac{1}{t}\frac{\mathrm{d}u}{u} $$ Thus: $$ \int \frac{1}{p+q(x-r)^2} \frac{1}{\sqrt{s + t^2 x^2}} \mathrm{d} x = \int \frac{1}{p+ q \left(\frac{u^2-s}{2 t u}-r\right)^2} \frac{1}{t} \frac{\mathrm{d}u}{u} = \int \frac{4 t u \cdot \mathrm{d} u}{4 p t^2 u^2 + q \left(u^2 -s - 2 r t u\right)^2} $$ Now it is down to integration of the rational function.

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Wolfram alpha can solve this, but it is messy. Type

integral of ( (1 / (a + b*x + c*x^2) ) (1 / sqrt(d + e*x^2) ) )

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Yes, I have the result from W|A and Mathematica too, but I need either an outline to do it or an explicit reference (e.g. standard, published integral tables) so that I can cite it in a publication. "Integrated using W|A" won't fly in a review... – user39836 Sep 9 '12 at 1:49
@fsk: I have seen "integrated using mathematica" fly in a review. – binn Sep 9 '12 at 1:54
Yes, but it depends on the rigor of the journal... it works in a soft science/biology journal, but not so in an applied physics journal. – user39836 Sep 9 '12 at 1:56
I don't see how it's different than using an integral table. – binn Sep 9 '12 at 1:59
I don't either, but some traditions are hard to break... Abramowitz & Stegun, Gradshteyn & Ryzhik are classics and easily fly (although they do have errors occasionally). Besides, Mathematica won't necessarily give you the simplest form of the expression... it's just what it managed to find via its CAS routines – user39836 Sep 9 '12 at 2:06

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