How do I integrate this (or is there a solution in a table)? 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.
 A: This may be the hard way, but  


*

*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.  

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

*Use partial fractions to do the resulting problem. 
A: 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.
A: 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) ) )
