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How to find indefinite integrals for the rational trigonometric functions of the form

$\qquad\Lambda^{pq}_m = \int \frac{\cos^p x\sin^q x}{(a+b\sin x +c\cos x)^m} dx$

where $(p,q,m)>0$ using Mathematica?

Any suggestions for preventing the results from being given in terms of AppellF1, but rather to obtain them as simpler summation functions for the ranges

$\qquad 0 \leq p \leq 5$, $0 \leq q \leq 1$ and $3 \leq m \leq 9$.

For example: ClearAll[x, a, b, c, p, q, m];Integrate[1/(a + b Cos[x] + c Sin[x])^m, x]

yields an output in terms of AppellF1

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  • $\begingroup$ No chance under the above assumptions. Big elementary results are produced for integer exponents $p,q,m$. $\endgroup$
    – user64494
    Dec 27 '17 at 13:04
  • $\begingroup$ Any suggestions for preventing the results from being given in terms of AppellF1 it always helps to show the Mathematica code you used to obtain this result you claim to contain AppellF1 as I can't get Mathematica to give any result using the image you posted. So may be you typed something else, but without seeing what you typed, one can only guess. This is what I typed ClearAll[x,a,b,c,p,q]; int=(Cos[x]^p Sin[x]^q)/(a+b Sin[x]+c Cos[x])^m; Assuming[p>0&&q>0&&m>0,Integrate[int,x]] $\endgroup$
    – Nasser
    Dec 27 '17 at 16:56
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    $\begingroup$ People here generally like users to post code as Mathematica code instead of images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. You may find this this meta Q&A helpful $\endgroup$
    – Michael E2
    Dec 27 '17 at 22:31
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If I do it this way

Lambda[p_, q_, m_] := 
 Integrate[(Cos[x]^p Sin[x]^q)/(a + b Cos[x] + c Sin[x])^m, x]

Table[Lambda[p, q, m], {p, 0, 5}, {q, 0, 1}, {m, 3, 9}]

That will take awhile and will produce very large output. I did not wait for it to finish, but spot checking a few, such as

Lambda[0, 0, 3]
Lambda[2, 1, 5]
Lambda[5, 1, 9]

all produced standard Trig functions in output. If you want to use values for p,q,m other than integers, you are probably out of luck.

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