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Is there a solution for the integral $$ \int_0^b (a-x^m)^{1/n} dx? $$ WolframAlpha doesn't help a lot: $$ \frac1a x(1-x^m)^{\frac1n+1}{}_2F_1\left(1,\frac1n+1+\frac1m; 1+\frac1m;\frac{x^m}a\right)\Biggr|_0^b $$ Or can this be simpified?

Thanks a lot

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    What do you mean by a "solution" to the integral? If you mean: can it be expressed using elementary functions, then according to Wolfram Alpha, the answer is: no, you need hypergeometric functions. (And I would be inclined to trust Wolfram Alpha on this.) If that is not what you mean, perhaps you could let us know!2012-12-03
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    That's exactly what I mean. I hoped for something more elementary...2012-12-03
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    Then it looks like you're out of luck, I'm afraid.2012-12-03
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    Even $m=4,n=2$ gives you an elliptic integral, not elementary.2012-12-03

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As Tony K noted, there is no elementary antiderivative in general (or even for special cases such as $a=1,m=2,n=3$). However, you can get a series solution: assuming $a > b^m$, for $0 < x < b$ we have

$$(a - x^m)^{1/n} = a^{1/n} \sum_{k=0}^\infty {1/n \choose k} (-1/a)^k x^{mk} $$ and then

$$ \eqalign{\int_0^b (a-x^m)^{1/n}\ dx &= a^{1/n} \sum_{k=0}^\infty {1/n \choose k} (-1/a)^k \int_0^b x^{mk}\ dx\cr &= a^{1/n} \sum_{k=0}^\infty {1/n \choose k} (-1/a)^k \frac{b^{mk+1}}{mk+1}}$$