How to integrate $\int_{l1}^{l2}\frac{e^{\pm i a x}}{\sqrt{bx^2+cx+d}}dx$ I have the above mentioned integral
$$
\int_{l_1}^{l_2}\frac{e^{\pm i a x}}{\sqrt{bx^2+cx+d}}dx
$$
which I want to solve. I expect some special functions in its solution, but so far I am out of ideas now.
I have tried the following:
I substitute $t=\sqrt{bx^2+cx+d}$ and get an expression for $x$ that I can insert in the exponential:
$$
t=\sqrt{bx^2+cx+d}
$$
$$
\frac{t^2}{b}=x^2+\frac{c}{b}x+\frac{d}{b}=\left (x+\frac{c}{2b}\right )^2 + \frac{d}{b} - \left (\frac{c}{2b}\right )^2
$$
$$
\sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b}=x
$$
The derivative of the substitution t is:
$$
\frac{dt}{dx}=\frac{2bx+c}{2\sqrt{bx^2+cx+d}}=\frac{2bx+c}{2t}
$$
inserting everything in the integral gives
$$\int_{t(l_1)}^{t(l_2)}\frac{e^{\pm i a \left( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b} \right )}}{t}\frac{2t}{2bx+c}dt=2
\int_{t(l_1)}^{t(l_2)}\frac{e^{\pm i a \left( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b} \right )}}{2b\left ( \sqrt{\frac{t^2}{b}-\frac{d}{b} + \left (\frac{c}{2b}\right )^2}-\frac{c}{2b}\right ) + c}dt
$$ 
This looks nearly like an exponential integral. I would expect a solution involving this kind of integral, but this is where I am stuck right now. Is it possible to give a closed form solution involving special functions?
 A: Too long for a comment (a work in progress)
$$
\int \frac{\mathrm{e}^{\pm iax}}{\sqrt{bx^2+cx+d}}dx
$$
Focusing on the denominator
$$
b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) =b\left[\left(x+\frac{c}{2b}\right)^2+\frac{d}{b}-\left(\frac{c}{2b}\right)^2\right]
$$
lets change variables
$$
t = \frac{x+\frac{c}{2b}}{\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}}\implies x = -\frac{c}{2b} + \left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right) t
$$
leads to
$$
b\left(x^2+\frac{c}{b}x+\frac{d}{b}\right) = b\left[\left(\frac{c}{2b}\right)^2-\frac{d}{b}\right]\left(t^2-1\right)
$$
therefore we have
$$
\left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right)\int \frac{\mathrm{e}^{\pm ia\left(-\frac{c}{2b} + \left(\sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right) t\right)}}{\sqrt{b\left[\left(\frac{c}{2b}\right)^2-\frac{d}{b}\right]\left(t^2-1\right)}}dt
$$
let $t=\cosh \theta$ we have
$$
t^2-1 = \sinh^2 \theta\\
dt = \sinh \theta d\theta
$$
thus we get
$$
\frac{\mathrm{e}^{\mp i\frac{ac}{2b}}}{\sqrt{b}}\int_{\bar{l_1}}^{\bar{l_2}} \mathrm{e}^{\pm i\lambda_1\cosh \theta}d\theta
$$
where
$$
\lambda_1 = a\left[-\frac{c}{2b} + \sqrt{\left(\frac{c}{2b}\right)^2-\frac{d}{b}}\right]
$$
I am still working on the last integral.
