Put it this way
$\int {\frac{x}{{\sqrt {a{x^2} + bx + c} }}dx} = \frac{1}{{2a}}\int {\frac{{2ax + b}}{{\sqrt {a{x^2} + bx + c} }}dx} - \frac{b}{{2a}}\int {\frac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} $
Then
$\int {\frac{x}{{\sqrt {a{x^2} + bx + c} }}dx} = \frac{1}{a}\sqrt {a{x^2} + bx + c} - \frac{b}{{2a}}\int {\frac{{dx}}{{\sqrt {a{x^2} + bx + c} }}} $
So let's focus on the last integral.
$a{x^2} + bx + c = a\left[ {{{\left( {x + \frac{b}{{2a}}} \right)}^2} + \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}}} \right]$
Thus by making ${x + \frac{b}{{2a}}} = u$
we have a new polynomial to integrate. Consider the cases:
$\displaystyle \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}} < 0 = - {k^2}$. So you have:
$\frac{1}{{\sqrt a }}\int {\frac{{du}}{{\sqrt {{u^2} - {k^2}} }}} = \frac{1}{{\sqrt a }}\cosh^{-1} \frac{u}{k}$
$\displaystyle \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}} = 0$. You get
$\frac{1}{{\sqrt a }}\int {\frac{{du}}{u}} = \frac{1}{{\sqrt a }}\log u$
- $\displaystyle \frac{c}{a} - \frac{{{b^2}}}{{4{a^2}}} > 0 = {k^2}$
$\frac{1}{{\sqrt a }}\int {\frac{{du}}{{\sqrt {{u^2} + {k^2}} }}} = \frac{1}{{\sqrt a }}{\sinh ^{ - 1}}\frac{u}{k}$
So it all depends on the polynomial.