Evaluation of $\int \frac{x\sin( \sqrt{ax^2+bx+c})}{ax^2+bx+c} \ dx\ $

Question

How do we find $$\int \frac{x\sin( \sqrt{ax^2+bx+c})}{ax^2+bx+c} \ dx\ $$

NB: It is not mandatory that $ax^2+bx+c$ has only a single root

Answer 1

Hint:

$\int\dfrac{x\sin\sqrt{ax^2+bx+c}}{ax^2+bx+c}dx$

$=\int\dfrac{x}{ax^2+bx+c}\sum\limits_{n=0}^\infty\dfrac{(-1)^n(ax^2+bx+c)^{n+\frac{1}{2}}}{(2n+1)!}dx$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nx(ax^2+bx+c)^{n-\frac{1}{2}}}{(2n+1)!}dx$

Answer 2

Too long for a comment:

If the polynomial has only a single root, i.e $ax^2+bx+c=A^2(x+B)^2$, the integral may be solved by substitution $y=A(x+B)$. The result is then a straightforward combination of the trigonometric integrals: $$\frac{B}{A}\left(\text{sinc}(y)-\text{Ci}(y)\right)+\frac{1}{A^2}\text{Si}(y)$$ I don't quite see how the non-degenerate case can be solved in closed form, though an approximation can be derived by expanding $\sin$ in a Taylor series, and then using Euler's substitution on each of the rational fractions in $\sqrt{ax^2+bx+c}$.