How to evaluate $\int x^{5/2}\sqrt{1+x}dx$? $$\int x^{5/2}\sqrt{1+x}dx \,$$
I don't know how to start. I tried with a change of variable $t=\sqrt{x+1}$.
EDIT:
EDIT2: I don't know how to get to $\frac{1}{4}x^{5/2}(1+x)^{3/2} - \int \frac{5}{12}x^{3/2}\sqrt{1+x} dx$. 
Any hint will be greatly appreciated.
 A: This is equivalent to a so-called incomplete Beta integral. From Wolfram Integrator output, you can suspect that it is irreducible (cannot be expressed in elementary functions).
This answer is based on a former problem statement.

Hint on the current problem:
The integrand is of the form
$$x^p\sqrt{1+x}.$$
Integrating by parts on $\sqrt{1+x}$, you will get a term $x^p(1+x)^{3/2}$ and a new integrand $x^{p-1}(1+x)^{3/2}$ (coefficients ignored).
Then decomposing $(1+x)^{3/2}=(1+x)\sqrt{1+x}$, you will get a sum of integrands $x^{p-1}\sqrt{1+x}$ and $x^p\sqrt{1+x}$. Moving the latter to the LHS, you have achieved a reduction from $p$ to $p-1$.
$$\int x^p\sqrt{1+x}\,dx=\frac2{3+2p}x^p(1+x)^{3/2}-\frac{2p}{3+2p}\int x^{p-1}\sqrt{1+x}\,dx.$$
Repeating twice, you end up with
$$\int \sqrt x\sqrt{1+x}\,dx,$$ which you can solve using $2x+1=\cosh(t)$.
A: HINT:
$$\int x^{\frac{5}{2}}\sqrt{1+x}\space\space\text{d}x=$$

Substitute $u=\sqrt{x}$ and $\text{d}u=\frac{1}{2\sqrt{x}}\space\space\text{d}x$:

$$2\int u^6\sqrt{u^2+1}\space\space\text{d}u=$$

Substitute $u=\tan(s)$ and $\text{d}u=\sec^2(s)\space\space\text{d}s$:

$$2\int \tan^6(s)\sec^3(s)\space\space\text{d}s=$$
$$2\int \sec^3(s)(\sec^2(s)-1)^3\space\space\text{d}s=$$
$$2\int \left(\sec^9(s)-3\sec^7(s)+3\sec^5(s)-\sec^3(s)\right)\space\space\text{d}s=$$
$$2\left(\int\sec^9(s)\space\space\text{d}s-3\int\sec^7(s)\space\space\text{d}s+3\int\sec^5(s)\space\space\text{d}s-\int\sec^3(s)\space\space\text{d}s\right)$$
