Maxima of the function $\left \vert \int_{-1}^1 e^{i(ax+bx^2)}dx \right \vert$ 
I am looking for extrema of the function
$$g(a,b):=\left \vert \int_{-1}^1 e^{i(ax+bx^2)}dx \right \vert$$
where $a,b >0$  are real parameters. I already plotted this function and got the impression that there are periodically local extrema along some particular curve, but it was hard to be more precise. My question is: Can we say anything about extrema from the function itself analytically?
In particular, I am interested in the asymptotic distribution of the maxima (if we cannot catch them analytically) for $a,b>0$ that you can see in the plot above. Apparently, they start with "more or less" $a = b$, but how are they distributed for large $a,b$? They also seem to have some periodicity, can we say anything about this?
You might also want to look at the new representations given in the comments :-)
If anything is unclear about this, please let me know.
NOTICE: Anything( also periodicity etc.) you could say about the extrema is probably useful, so I deliberately want this question to be somewhat broad.
 A: Here it is a plot of $|f(a,b)|$ where $f(a,b)=\int_{-1}^{1}e^{i(ax+bx^2)}\,dx$:

Moreover, we have (we can assume $a,b\geq 0$ since $f$ is an even function with respect to both its arguments):
$$|f(a,b)|=2\left|\int_{0}^{1+a/(2b)}e^{ibx^2}\,dx\right|=\frac{1}{\sqrt{b}}\left|\int_{0}^{a+b+\frac{a^2}{4b}}\frac{1}{\sqrt{u}}e^{-iu}\,du\right|.$$
By assuming $c\triangleq a+b+\frac{a^2}{4b}\geq\pi$, we can exploit the periodicity of the complex exponential function, $e^{i(u+\pi)}=-e^{iu}$, in order to have:
$$\begin{eqnarray*}2\int_{0}^{c}\frac{1}{\sqrt{u}}e^{iu}\,du&=&\int_{0}^{\pi}\frac{1}{\sqrt{u}}e^{iu}\,du+\int_{\pi}^{c}\frac{1}{\sqrt{u}}e^{iu}\,du\\&+&\int_{0}^{c-\pi}\frac{1}{\sqrt{u}}e^{iu}\,du+\int_{c-\pi}^{\pi}\frac{1}{\sqrt{u}}e^{iu}\,du\end{eqnarray*}$$
where the two middle terms in the RHS almost cancel out, leaving:
$$\left|\int_{0}^{c}\frac{1}{\sqrt{u}}e^{iu}\,du\right|\leq K_1+\frac{K_2}{\sqrt{c-\pi}}.$$
Hence it is expected that $f(a,b)$, far enough from the origin, goes to zero like:

$$|f(a,b)|\ll \frac{1}{|b+a/2|}.$$

