Theorem: Primitive of x squared over Root of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x^2 \rd x} {\sqrt {a x^2 + b x + c} } = \frac {2 a x - 3 b} {4 a^2} \sqrt {a x^2 + b x + c} + \frac {3 b^2 - 4 a c} {8 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }$

Proof:
First:
{{begin-eqn}}
{{eqn | l = z
      | r = a x^2 + b x + c
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a x + b
      | c = Derivative of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {x^2 \rd x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {a x^2 \rd x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {\paren {a x^2 + b x + c - b x - c} \rd x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {\paren {a x^2 + b x + c} \rd x} {\sqrt {a x^2 + b x + c} } - \frac b a \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} } - \frac c a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 a \int \sqrt {a x^2 + b x + c} \rd x - \frac b a \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} } - \frac c a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{eqn | r = \frac 1 a \paren {\frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } }
      | c = Primitive of $\sqrt {a x^2 + b x + c}$
}}
{{eqn | o = 
      | ro= -
      | r = \frac b a \int \frac {x \rd x} {\sqrt {a x^2 + b x + c} } - \frac c a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac 1 a \paren {\frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a} + \frac {4 a c - b^2} {8 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac b a \paren {\frac {\sqrt {a x^2 + b x + c} } a - \frac b {2 a} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } } - \frac c a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = Primitive of $\dfrac x {\sqrt {a x^2 + b x + c} }$
}}
{{eqn | r = \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a^2} + \frac {4 a c - b^2} {8 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{eqn | o = 
      | ro= -
      | r = \frac {b \sqrt {a x^2 + b x + c} } {a^2} + \frac {b^2} {2 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac c a \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac {\paren {2 a x + b} \sqrt {a x^2 + b x + c} } {4 a^2} + \frac {4 a c - b^2} {8 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = common denominators
}}
{{eqn | o = 
      | ro= -
      | r = \frac {4 b \sqrt {a x^2 + b x + c} } {4 a^2} + \frac {4 b^2} {8 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} } - \frac {8 a c} {8 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac {\paren {2 a x + b - 4 b} \sqrt {a x^2 + b x + c} } {4 a^2} + \frac {4 a c - b^2 + 4 b^2 - 8 a c} {8 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = combining like terms
}}
{{eqn | r = \frac {2 a x - 3 b} {4 a^2} \sqrt {a x^2 + b x + c} + \frac {3 b^2 - 4 a c} {8 a^2} \int \frac {\d x} {\sqrt {a x^2 + b x + c} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}