Is there an easier way to solve this complicated integral?

[gaganaut]

While I was working on my research problem, I came across this integral.

$$\int_0^{\pi}\;\frac{k_1}{-k_2\left(2R\;(k_3\cos{\alpha}+k_4\sin{\alpha})-k_5\right)^{1.5}}$$

All the $$k_i\;'s$$ are constants and so is R.

I tried writing out an infinite series for $$\left(2R\;(k_3\cos{\alpha}+k_4\sin{\alpha})-k_5\right)^{-1.5}$$ in Maple and then integrating. But the solution is nasty with the first 4 terms considered. It has got quite a few terms to deal with.

I then tried performing numerical integration by using the Simpson's rule, but the solution again has too many terms in it.

So is there any easier way to solve this or will I have to live with the multi-term solution that I get from both the methods I used?

Any sort of help will be highly appreciated.

Thanks

 

[CompuChip]

I did the integration in Mathematica and it gave me
$$-\frac{2 F_1\left(-\frac{1}{2};\frac{1}{2},\frac{1}{2};\frac{1}{2};\left\{\frac{2 R c_3+c_5}{2 R \sqrt{\frac{c_3^2}{c_4^2}+1} c_4+c_5},\frac{c_5-2 R c_3}{2 R<br /> \sqrt{\frac{c_3^2}{c_4^2}+1} c_4+c_5}\right\},\left\{-\frac{2 R c_3+c_5}{2 R \sqrt{\frac{c_3^2}{c_4^2}+1} c_4-c_5},\frac{2 R c_3-c_5}{2 R<br /> \sqrt{\frac{c_3^2}{c_4^2}+1} c_4-c_5}\right\}\right) c_1 \left(\sqrt{2 R c_3-c_5} \sqrt{\frac{R \left(c_3^2+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_3+c_4^2\right)}{2<br /> R \left(c_3^2+c_4^2\right)-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}} \sqrt{\frac{R \left(c_3^2-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_3+c_4^2\right)}{2 R<br /> \left(c_3^2+c_4^2\right)+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}}+\sqrt{-2 R c_3-c_5} \sqrt{\frac{R \left(c_3^2-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4<br /> c_3+c_4^2\right)}{2 R \left(c_3^2+c_4^2\right)-\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}} \sqrt{\frac{R \left(c_3^2+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4<br /> c_3+c_4^2\right)}{2 R \left(c_3^2+c_4^2\right)+\sqrt{\frac{c_3^2}{c_4^2}+1} c_4 c_5}}\right)}{R c_2 c_4 \sqrt{-2 R c_3-c_5} \sqrt{2 R c_3-c_5}}$$

where F₁ is some Appell hypergeometric function...
so somehow I'm not convinced that there is a "nice" solution.

 

[jacophile]

Well its straight forward to get it into the form

$$\int_{0}^{\pi}\frac{Adx}{(sin(x+\alpha)-B)^\frac{3}{2}}$$

where A, B and alpha are functions of R and the K's

you could then try a substitution like

$$y=sin(x+\alpha)+A\;\;\;\;;dx=\frac{dy}{cos(arcsin(y-A)))}$$

to give

$$\int_{sin(\alpha)+A}^{sin(\alpha+\pi)+A}\frac{Ady}{sin(arcsin(y-A)-\frac{\pi}{2})x^\frac{3}{2}}$$

maybe then try integration by parts...