beta and gamma helping solve an integral I want to   compute the following integral:
$$I= \int_0 ^1 e^x\frac{x^2}{\sqrt[4]{1-x^4}} dx$$
I am looking for a clever trick that will give a concrete result for this integral, without much fuss or complicated calculations.
How may I compute that definite integral in an elementary way?
Find a closed formula containing just elementary functions for this integral or if it does not exist I would be curious about a proof that shows it cannot be expressed in this way.
 A: Trying with $x=\sqrt{\sin (t)}$
$$I= \int_0 ^1 e^x\frac{x^2}{\sqrt[4]{1-x^4}}\, dx=\frac 12 \int_0 ^{\frac \pi 2} e^{\sqrt{\sin (t)}} \sqrt{\sin (t) \cos (t)}\, dt$$
$$I=\frac 12\sum_{n=0}^\infty \frac 1 {n!}\,\int_0 ^{\frac \pi 2}  \sin ^{\frac{n+1}{2}}(t)\,\cos^{\frac 12} (t)\,dt$$ The antiderivative is known in terms of the Gaussian hypergeometric function :
$$\int \sin ^{\frac{n+1}{2}}(t)\,\cos^{\frac 12} (t)\,dt=-\frac{2}{3} \cos ^{\frac{3}{2}}(t) \,
   _2F_1\left(\frac{3}{4},\frac{1-n}{4};\frac{7}{4};\cos ^2(t)\right)$$
$$J_n=\int_0 ^{\frac \pi 2}  \sin ^{\frac{n+1}{2}}(t)\,\cos^{\frac 12} (t)\,dt=\frac{2 }{3}\Gamma \left(\frac{7}{4}\right)\frac{\Gamma \left(\frac{n+3}{4}\right)}{\Gamma \left(\frac{n+6}{4}\right)}$$  So
$$I=\frac{1 }{3}\Gamma \left(\frac{7}{4}\right)\sum_{n=0}^\infty \frac{\Gamma \left(\frac{n+3}{4}\right)}{\Gamma \left(\frac{n+6}{4}\right)\,\Gamma(n+1)}$$
If
$$a_n=\frac{\Gamma \left(\frac{n+3}{4}\right)}{\Gamma \left(\frac{n+6}{4}\right)\,\Gamma(n+1)}\quad\implies\quad \frac {a_{n+1}}{a_n}=\frac 1 n \Bigg[1-\frac{7}{4 n}+O\left(\frac{1}{n^2}\right) \Bigg]$$ and the summation should converge quite fast.
Wolfram Alpha gives the awful result in terms of hypergeometric functions as @LCandleEater already commented. But we do not need many terms; computing the partial sums
$$S_p= \sum_{n=0}^p \frac{\Gamma \left(\frac{n+3}{4}\right)}{\Gamma \left(\frac{n+6}{4}\right)\,\Gamma(n+1)}$$
$$S_{12}=\color{red}{3.089258529}253$$ while the infinite summation gives
$\color{red}{3.089258529315}$ (difference $=6.23 \times 10^{-11}$)
