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
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
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}}$