For the integral
$$ I=\int \sqrt{1+x^4}dx,$$
Mathematica can give a result like this:
$$I=\frac{x^5-2 \sqrt[4]{-1} \sqrt{x^4+1} F\left(\left.i \sinh ^{-1}\left(\sqrt[4]{-1} x\right)\right|-1\right)+x}{3 \sqrt{x^4+1}}$$
where $F(a|b)$ is the elliptic integral.
However, in a paper I find the result of the integral can be described by hypergeometric function. I cannot find the result in table of integrals. Anybody can help me?
The series expansion of the integrand is $$ \sqrt{1+x^4} = \sum_{k=0}^{\infty} \frac{\Gamma(k-1/2)}{\Gamma(-1/2)} \frac{(-x^4)^k}{k!} = {}_1 F_0(-\tfrac{1}{2};;-x^4). $$ Now, the indefinite integral is, suppressing arbitrary constants and so on, $$ \int \sqrt{1+x^4} \, dx = x\sum_{k=0}^{\infty} \frac{\Gamma(k-1/2)}{\Gamma(-1/2)} \frac{1}{4k+1} \frac{(-x^4)^{k}}{k!}, $$ by fiddling about with $ \int (-x)^k \, dx $. The hard bit now is $1/(4k+1)$, which can be expressed as $$ \frac{1}{4k+1} = \frac{1}{4} \frac{1}{k+1/4} = \frac{1}{4} \frac{\Gamma(k+1/4)}{\Gamma(k+5/4)} = \frac{\Gamma(5/4)}{\Gamma(1/4)} \frac{\Gamma(k+1/4)}{\Gamma(k+5/4)} $$ (this is a well-known trick for making series look hypergeometric).
We therefore end up with $$ \int \sqrt{1+x^4} \, dx = x\sum_{k=0}^{\infty} \frac{\Gamma(k-1/2)}{\Gamma(-1/2)} \frac{\Gamma(k+1/4)}{\Gamma(1/4)} \frac{\Gamma(5/4)}{\Gamma(k+5/4)} \frac{(-x^4)^{k}}{k!} = x \, {}_2 F_1 \left( -\tfrac{1}{2},\tfrac{1}{4} ; \tfrac{5}{4} ;-x^4 \right) $$
