Does the integral $$\int (1-x^a)^\frac{1}{a} dx$$ have a closed form solution? The parameter $a$ is a constant and both $a$ and $x$ are positive real numbers less than or equal to $1$.
I tried plugging it in to Wolfram and got a result I don't understand. What e.g. does "$x_2 F_1(\ldots)$" mean? Or "$B_{x^a}(\ldots)$"?
On
Just rephrazing what was already in comments.
$$I=\int (1-x^a)^\frac{1}{a} dx=x \, _2F_1\left(-\frac{1}{a},\frac{1}{a};1+\frac{1}{a};x^a\right)$$
Now, as suggested, make $$x^a=z \implies x=z^\frac{1}{a} \implies dx=\frac{1}{a}{z^{\frac{1}{a}-1}}\,dz$$ to get $$I=\frac{1}{a} \int (1-z)^{\frac{1}{a}} z^{\frac{1}{a}-1}\,dz=\frac{1}{a}\,B_z\left(\frac{1}{a},\frac{1}{a}+1\right)$$
Yes, you can use binomial theorem then integrate.
$$ (1-x^a)^{1/a} = \sum_{k=0}^\infty \binom{1/a}{k}(-1)^k x^{ak} \\ \int(1-x^a)^{1/a} dx = \sum_{k=0}^\infty \binom{1/a}{k}(-1)^k \frac{x^{ak+1}}{ak+1} $$ Then do the usual manipulation to recognize this as a hypergeometric function \begin{align} \sum_{k=0}^\infty & \binom{1/a}{k}(-1)^k \frac{x^{ak+1}}{ak+1} = x \sum_{k=0}^\infty \frac{(1/a)(1/a-1)\cdots(1/a-k+1)}{k!}\;(-1)^k\frac{1}{ak+1}\;x^{ak} \\ &= x \sum_{k=0}^\infty \frac{(-1/a)(-1/a+1)\cdots(-1/a+k-1)}{k!}\frac{1}{a} \frac{1}{(1/a+k)}\;x^{ak} \\ &= x \sum_{k=0}^\infty \frac{(-1/a)(-1/a+1)\cdots(-1/a+k-1)}{k!}\; \frac{(1/a)(1/a+1)\cdots(1/a+k-1)}{(1/a+1)(1/a+2)\cdots(1/a+k)}\;x^{ak} \\&= x\;{}_2F_1\left(-\frac{1}{a},\frac{1}{a};\frac{1}{a}+1;x^a\right) \end{align} We used Gauss's definition $$ {}_2F_1(a,b;c;z) = \sum_{k=0}^\infty \frac{a(a+1)(a+2)\cdots (a+k-1)\;b(b+1)(b+2)\cdots(b+k-1)}{c(c+1)(c+2)\cdots(c+k-1)\; k!}\;z^k $$