I noticed while studying integration that $\int \sqrt{1-x^2} \mathrm dx $ has a relatively simple antiderivative found by doing a trigonometric substitution.
On the other hand, $\int \sqrt{1-x^3} \mathrm dx $ can only be expressed with elliptic integrals (according to WA). The same thing occured when $a=4, 5, 6, 7...$ etc. I was wondering if there exists a proof that $\sqrt{1-x^a} $ cannot be integrated without hypergeometric or elliptic integrals when $a>2$.
Using Newton's generalised binomial theorem: $$\sqrt{1-x^a}=\sum_{k=0}^{\infty}{\frac{1}{2}\choose k}(-x)^{ak}$$ The anti derivative is: $$\int\sqrt{1-x^a}dx=C+\sum_{k=1}^{\infty}{\frac{1}{2}\choose k}\frac{(-x)^{ak+1}}{ak+1}$$
Since you asked for finding the antiderivative in another form besides a series. Let’s introduce a notation:
$$\text {Arc Length function of} f(x)=\text{ArcLength}(f(x))=\int_a^b\sqrt{(f’(x))^2+1}\,dx=\int_a^b\sqrt{f’^2+1}dx$$
Since your function is:
$$\int\sqrt{1-x^a}dx\mathop=^\text {set}\int \sqrt{f’^2+1}dx\implies f’^2+1=1-x^a\implies f(x)=c\pm\frac{2 i x^{\frac a2+1}}{a+2}$$
Therefore: $$\boxed{\int\sqrt{1-x^a}dx=\text{ArcLength}\left(c\pm\frac{2 i x^{\frac a2+1}}{a+2}\right)}$$
It is known that finding the arc length of a cubic function, and higher real powers, is usually nonelementary. Note that the arc length of the imaginary function is needed to have the $\sqrt{1-x^a}$ otherwise, the arc length of the boxed result without the $i$ would have $\sqrt{1+x^a}$ and also note the constant of integration. To find the arclength from $x=a$ to $x=b$, use the Fundamental theorem of Calculus and evaluate $$\text {Arclength}(f(x))\big|_b-\text {Arclength}(f(x))\big|_a$$
See these results.
Please correct me and give me feedback!
Also please see this $\int \sqrt{1+x^n}dx$ as an arclength post
