Integration of functions of the form $(mx+n)/\sqrt{ax^2+bx+c}$ and $(mx+n)/(ax^2+bx+c)$

Question

is there some general method for the evaluation of integrals such as: $$\int \frac{mx+n}{\sqrt{ax^2+bx+c}}\,dx$$ and $$\int \frac{mx+n}{{ax^2+bx+c}}\,dx$$

Answer 1

To integrate $\int \frac{mx+n}{\sqrt{ax^2+bx+c}}\,dx.$ or $\int \frac{mx+n}{ax^2+bx+c}\,dx.$ you need to complete the square of the denominator:

by so doing, we have: $ax^2+bx+c=a((x+\frac{b}{2a})^2-\frac{b^2-4ac}{4a^2})$

Therefore, $\int \frac{mx+n}{\sqrt{ax^2+bx+c}}\,dx.=\int \frac{mx+n}{\sqrt{a((x+\frac{b}{2a})^2-\sqrt{\frac{b^2-4ac}{4a^2}}^2)}}\,dx.$

by substituting $\cosh{\theta}=\frac{x+\frac{b}{2a}}{\frac{\sqrt{b^2-4ac}}{2a}}$ this gives:

$x=\frac{\sqrt{b^2-4ac}}{2a}\cosh{\theta}-\frac{b}{2a}$ and $dx=\frac{\sqrt{b^2-4ac}}{2a}\sinh{\theta}d\theta$

Substituting gives

$\int \frac{mx+n}{\sqrt{a((x+\frac{b}{2a})^2-\sqrt{\frac{b^2-4ac}{4a^2}}^2)}}\,dx.=\frac{1}{\sqrt{a}}\int \frac{m(A\cosh{\theta}-B)+n}{\sinh{\theta}}\,d\theta.$

where $A=\frac{\sqrt{b^2-4ac}}{2a}$ and $B=\frac{b}{2a}$

Answer 2

Hints: $~\big(ax^2+bx+c\big)'=2ax+b,\quad mx+n~=~\dfrac m{2a}\big(2ax+b\big)~+~\bigg(n-\dfrac{mb}{2a}\bigg).~$ Thus, the latter integral will become a linear combination of a logarithm and something else, whereas the former will yield a linear combination of a radical and something else. That something else may be either an arctangent, provided the quadratic is irreducible over the reals, or yet another logarithm $($which is basically either a hyperbolic arctangent, or a hyperbolic arcsine$)$ if not. Of course, either factoring the quadratic, or otherwise completing the square and then using appropriate hyperbolic or trigonometric substitutions is key.