Integration of 1 / [sqrt(f(x))+g(x)] ?

[lavoisier]

Hi everyone, today I came across a problem at work that requires integrating this function (indefinitely, with respect to c):

\frac {1} {\sqrt {(k+p-c)^2 + 4 k c}-k-p+c}

k, p and c are all real and positive.

I submitted it to Maxima, but it stayed implicit.

Can you please suggest any substitution or other technique to solve it?

Thanks
L

 

[micromass]

First rationalize the denominator: so try to bring the square root to the numerator by multiplying with $\sqrt{(k+p-c)^2 + 4kc} + k + p - c$. Then an appropriate trigonometric substitution (like $x = \tan(u)$ or similar) will help.

 

[lavoisier]

Thank you micromass, it worked!
I multiplied both numerator and denominator by the factor you said, expanded, and got this:

\frac {\sqrt {(k+p-c)^2 + 4 k c}+k+p-c} {4 k c}

which I submitted to Maxima's integrate function, and I got the result.
Apologies for not writing it down, it's a long sum of logarithms, a square root, a linear term in c and asinh functions.
I don't exactly see how integrating this must involve inverse hyperbolic functions, except perhaps for the known relationship:

asinh(x)=Ln(\sqrt{1+x^2}+x)

Anyway, it does the job, so...
Thanks!
L