approaching $\int \sqrt{x^5+2}\; dx$ I recently came across this integral $\int\sqrt{x^5+2}\; dx$. From Wolframalpha i can see that it has a closed form. how does one get to that closed form? what techniques should i approach?
 A: As said in comments, using Taylor or the binomial theorem, we have
$$\sqrt{x^5+2}=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{  \left(n-\frac{3}{2}\right)!}{
  2^n\, n!} x^{5 n}$$
$$\int \sqrt{x^5+2}\,dx=-\frac{1}{\sqrt{2 \pi }}\sum_{n=0}^\infty(-1)^{n}\frac{  \left(n-\frac{3}{2}\right)!}{
  2^n\,(5n+1)\, n!} x^{5 n+1}$$ Computing the infinite summation
$$S=x\sqrt{2}  \,\, _2F_1\left(-\frac{1}{2},\frac{1}{5};\frac{6}{5};-\frac{x^5}{2}\right)$$ which looks simpler that the result from Wolfram Alpha but which numerically does not agree with it. However, the formula given here matches the results obtained by numerical integration.
What happens ? That is the question !
Edit
If I use the result given by Wolfram Alpha,
$$\int \sqrt{x^5+2}\,dx=\frac{1}{7} x \left(5 \sqrt{2} \,
   _2F_1\left(\frac{1}{5},\frac{1}{2};\frac{6}{5};-\frac{x^5}{2}\right)+2   \sqrt{x^5+2}\right)$$ and differentiate is, hoping that I am not mistaken, the result is 
$$\frac{5 x^5}{2 \sqrt{x^5+2}}$$
On Wolfram Cloud, I obtained the result I gave.
