Evaluate $\int \sqrt{1+x^{\frac{3}{2}}} \operatorname d x$ I can't figure out how to integrate
$$\int \sqrt{1+x^{\frac{3}{2}}} \operatorname d x$$
I've tried substitution by letting $u = x^3$, but it didn't go anywhere. I also tried to integrate using a trigonometric substitution, but that also got me nowhere. Then I tried Wolfram Alpha, and it got me even nowhere-er!
If you could give me a hint as to where to go, I'll try to answer this question at a later time. Thanks!
 A: Let $I = \int \sqrt{1 + x^{3/2}} \, \mathrm{d} x$. Integrating by parts:
$$
  \begin{eqnarray}
   I &=& x \sqrt{1 + x^{3/2}} - \int x \frac{3}{4} \, \frac{\sqrt{x}}{\sqrt{1+x^{3/2}}} \,  \mathrm{d} x = x \sqrt{1 + x^{3/2}} - \frac{3}{4} \int \sqrt{1+x^{3/2}} \, \mathrm{d} x + \frac{3}{4} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}} \\
  &=& -\frac{3}{4} I + x \sqrt{1 + x^{3/2}}  + \frac{3}{4} \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}}
  \end{eqnarray}
$$
Thus 
$$
  I = \frac{4}{7} x \sqrt{1 + x^{3/2}} + \frac{3}{7}  \int \frac{\mathrm{d} x}{\sqrt{1+x^{3/2}}}
$$
The latter integral is not elementary and can be evaluated in terms of Gauss hypergeometric function:
$$
  I = \frac{4}{7} x \sqrt{1 + x^{3/2}} + \frac{3 x}{7} \, {}_2F_1\left(\frac{1}{2}, \frac{2}{3} ; \frac{5}{3} ; -x^{3/2} \right)
$$
A: Let's rearrange the integral: 
$I=\int \sqrt{1^2+(x^\frac{3}{4})^2} dx$ , and make the substitution $x^\frac{3}{4}=t \Rightarrow \frac{3}{4}x^\frac{-1}{4} dx=dt$ , 
Since $x=t^\frac{4}{3} \Rightarrow dx=\frac{4}{3}t^\frac{1}{3} dt$; if we substitute this into integral we get the following:
$I=\frac{4}{3}\int \sqrt {(1^2+t^2)} t^\frac{1}{3} dt$,
This integral can be solved using integration by parts where
$u=t^\frac{1}{3}$ and $dv=\sqrt {1^2+t^2} dt$, and the integral $v=\int\sqrt {1^2+t^2} dt$ can be solved by applying the first formula from this list.
