Is there a general formula for the antiderivative of x times a function of a quadratic? like $$\int a f(\alpha x^2+\beta x+ \gamma)xdx$$ and the antiderivative of f is F.

I tried this but I could only get it equal to $$\frac {aF(u)}{2\alpha] \pm \frac{a\beta}{8\alpha ^2} \int f(-\frac{1}{4\alpha}v+(\frac{\beta ^2}{4\alpha}-\gamma))v^(-1/2)dv$$
where $u=\alpha x^2+\beta x+ \gamma$ and $v=\beta^2-4\alpha (\gamma + u)$. There's no +C because there's still an indefinite integral in the expression.

I'm betting it isn't possible to find a formula simpler than the original integral