a integration with two constants I am trying to solve below integration
$$\int\frac{dx}{(x^2+a^2)(x^2+a^2+b^2)^{\frac12}}$$
I tried substituting $x=a \,tan\,u$. Then I ended up with $$\int\frac{du}{(a)(b^2+a^2sec^2\,u)^{\frac12}}$$. I am not able to continue after this. Appreciate a small help on this
 A: Since you ask for small help, here is a hint, that I normally wouldn't consider being an answer:
Try
$$
u=\frac{x}{\sqrt{a^2+b^2+x^2}}.
$$
If you can take it from here, fine. Otherwise, ask for more steps...
Update with further details:
You will get (at least, I got)
$$
x^2=\frac{(a^2+b^2)u^2}{1-u^2},\quad\text{and}\quad du=\frac{a^2+b^2}{(a^2+b^2+x^2)^{3/2}}\,dx
$$
Now insert this and simplify as much as you can. You will get something really simple.
A: Using rogerl's suggestion, letting $x=c\tan\theta$ where $c=\sqrt{a^2+b^2}$ gives
$\displaystyle\int\frac{1}{(x^2+a^2)\sqrt{x^2+a^2+b^2}}dx=\int\frac{1}{(c^2\tan^2\theta+a^2)(c\sec\theta)}\cdot c\sec^2\theta d\theta$
$\displaystyle=\int\frac{\sec\theta}{a^2\sec^2\theta+b^2\tan^2\theta} d\theta$ $\;$[since $c^2\tan^2\theta+a^2=(a^2+b^2)\tan^2\theta+a^2=a^2\sec^2\theta+b^2\tan^2\theta$]
$=\displaystyle\int\frac{\cos\theta}{a^2+b^2\sin^2\theta} d\theta=\frac{1}{b}\int\frac{1}{a^2+u^2} du$  $\;\;\;$ (letting $u=b\sin\theta$)
$\displaystyle=\frac{1}{ab}\tan^{-1}\frac{u}{a}+C=\frac{1}{ab}\tan^{-1}\frac{b\sin\theta}{a}+C=\frac{1}{ab}\tan^{-1}\bigg(\frac{bx}{a\sqrt{x^2+a^2+b^2}}\bigg)+C$
