Calculating in closed form an integral in Airy function Can we hope for a nice closed form for the integral below?
$$\int_0^1 \frac{\displaystyle \text{Ai}\left(-\frac{t}{2^{2/3} \sqrt[3]{3-3 t}}\right)^2+\text{Bi}\left(-\frac{t}{2^{2/3} \sqrt[3]{3-3 t}}\right)^2}{ \sqrt[6]{1- t}} \, dt$$
What tools do you recommend me to use? 
Some information about Airy function here https://en.wikipedia.org/wiki/Airy_function.
EDIT: let me know if anything is wrong with the question, or if it happens that Airy function is annoying to some extent. I'm glad to improve the post. When I have ideas about the way to go,
I let you know. What else?

This is only a partial solution. I haven't obtained any closed form, but reduced to an integral of an elementary function.
One can simplify the above integral as follows
$$
I=\int_0^1 \frac{\displaystyle \text{Ai}\left(-\frac{t}{\sqrt[3]{12(1 -t)}}\right)^2+\text{Bi}\left(-\frac{t}{ \sqrt[3]{12(1- t)}}\right)^2}{ \sqrt[6]{1- t}} \, dt.
$$
Using the integral representation (see the book Vallee, Soares "Airy function and applications to physics")
$$
\operatorname{Ai}^2(t)+\operatorname{Bi}^2(t)=\frac{1}{\pi^{3/2}}\int_0^\infty\frac{1}{\sqrt{x}}e^{xt-\frac{x^3}{12}}dx
$$
one gets:
$$
I=\frac{1}{\pi^{3/2}}\int_0^1 \frac{dt}{ \sqrt[6]{1- t}} \int_0^\infty\frac{1}{\sqrt{x}}\exp\left\{ -\frac{xt}{ \sqrt[3]{12(1- t)}}-\frac{x^3}{12}\right\}dx=\\
\frac{12^{1/6}}{\pi^{3/2}}\int_0^1 dt \int_0^\infty\frac{1}{\sqrt{x}}e^{ -xt-x^3(1-t)}dx=\\
\frac{12^{1/6}}{\pi^{3/2}}\int_0^\infty\frac{dx}{\sqrt{x}}e^{-x^3}\int_0^1 e^{ (x^3-x)t}dt =\\
\frac{12^{1/6}}{\pi^{3/2}}\int_0^\infty\frac{e^{-x}-e^{-x^3}}{x^3-x}\frac{dx}{\sqrt{x}}
$$
