If the integral is
[tex]\displaystyle \int (ax^2+2bx+c)^{3/2}(2ax+2b)\,\mathrm dx[/tex]
substitute [tex]u = ax^2+2bx+c[/tex] and [tex]\mathrm du=(2ax+2b)\,\mathrm dx[/tex]. Then you end up with
[tex]\displaystyle \int u^{3/2}\,\mathrm du = \frac25u^{5/2} + C[/tex]
which in terms of x is
[tex]\boxed{\dfrac25(ax^2+2bx+c)^{5/2} + C}[/tex]
