Integral of (x-7)(3-2x)^(3/2)

Hey guys, I have this problem I have been trying to solve for the longest time.

Not looking for a complete solution, maybe part of one would be nice.

I'm trying to integrate $\displaystyle \int(x-7)\sqrt{3-2x}^3$

So far I have tried 2 ways of going at it.

First I completely expanded the integral

$\displaystyle \int(x-7)\sqrt{3-2x}^3=\int(-2x^2)(\sqrt{3-2x})+17x\sqrt{3-2x}-21\sqrt{3-2x}$

I have no problem integrating

$\displaystyle -21\sqrt{3-2x}$

I just simply use the substitution technique.

$\displaystyle u=3-2x$

The next part gets tricky, I'm stumped. Any of the functions with x variables I simply can't seem to do.

I tried substituting by parts by either making $\displaystyle u=x$ $\displaystyle dv=3-2x$ or $\displaystyle dv=\sqrt{3-2x}$ so on and so forth

I went to wolframalpha.com and tried to understand what they did step by step

For example, I tried to integrate $\displaystyle \int -21x^2\sqrt{3-2x}$ just like the site did. They say to substitute $\displaystyle u=\sqrt{3-2x}$ and $\displaystyle du=1/\sqrt{3-2x}$ to get

$\displaystyle 2\int(1/4)u^2(3-2u^2)^2du$

Either I am feeling really dumb right now and the answer is right in from of me or the answer is completely wrong. How did wolframalpha manage to get

$\displaystyle \int -21x^2\sqrt{3-2x}$=$\displaystyle 2\int(1/4)u^2(3-2u^2)^2du$

All help is greatly appreciated