Integrating $\int_0^\infty \sqrt x e^{-4/3x^{3/2}}\left(\int_0^x \operatorname{Ai}(t)dt\int_0^x \operatorname{Bi}(t)dt\right)dx$ 
Show that
$$I= \int_0^\infty \sqrt x e^{\large -4/3x^{3/2}}\left(\int_0^x \operatorname{Ai}(t)dt\int_0^x \operatorname{Bi}(t)dt\right)dx$$
$$=\frac{1}{3}-\frac{\sqrt[3]{2\sqrt 3+3}+\sqrt[3]{2\sqrt 3-3}}{\sqrt 3 \cdot 3^{5/3}}$$
Where $\operatorname{Ai}(t)$ and $\operatorname{Bi}(t)$ are the Airy functions.

This integral originates from an unsolved MSE post, found here. One thing that I tried and it might be worth mentioning is to use the connection formulas and rewrite the integral as:
$$I=\frac{1}{4i}\int_0^\infty \frac{\left(x^{3/4} e^{-2/3 x^{3/2}} \operatorname{AB}_{+}(x)\right)^2-\left(x^{3/4} e^{-2/3 x^{3/2}} \operatorname{AB}_{-}(x)\right)^2}{x}dx$$
$$\operatorname{AB}_{\pm}(x)=\int_0^x\left(\operatorname{Ai}(t)\pm i\operatorname{Bi}(t)\right)dt=-2e^{\mp 2i\pi/3}\int_0^x \operatorname{Ai}\left(e^{\mp 2i\pi/3}t\right)dt$$
Furthermore I attempted to use Frullani's integral or Plancherel's theorem, but without any luck.
 A: $
\DeclareMathOperator{\Ai}{\mathrm{Ai}}
\DeclareMathOperator{\Bi}{\mathrm{Bi}}
$This is not a full answer (yet). As commented by @Timothy, the problem can be rewritten using modified Bessel functions $I_\nu$ and their antiderivatives $F_\nu$. As
\begin{align}
\Ai(x)&=\frac{\sqrt x}{3}\big[
 I_{-1/3}\big(\tfrac 2 3 x^{3/2}\big)
-I_{ 1/3}\big(\tfrac 2 3 x^{3/2}\big)
\big],\tag{1a}\\
\Bi(x)&=\frac{\sqrt x}{\sqrt 3}\big[
 I_{-1/3}\big(\tfrac 2 3 x^{3/2}\big)
+I_{ 1/3}\big(\tfrac 2 3 x^{3/2}\big)
\big]\tag{1b},
\end{align}
we can substitute $2/3 x^{3/2}\mapsto y$ and get
\begin{align}
I &= 3^{-3/2} \int_0^\infty \mathrm dy\,e^{-2y}
\big[F_{-1/3}(y)-F_{1/3}(y)\big]
\big[F_{-1/3}(y)+F_{1/3}(y)\big]\tag{2a}\\
&\stackrel{?}{=} \frac 1 3 
- \frac 1 9 \sqrt[3]{2+\sqrt 3}
- \frac 1 9 \sqrt[3]{2-\sqrt 3},\tag{2b}
\end{align}
with hypergeometric ${}_1\!F_2$
$$
F_\nu(y)=\int_0^y I_\nu(z)\,\mathrm dz =
\frac{2^{-\nu } y^{\nu+1}}{\Gamma(\nu +2)}
\,{}_1\!F_2\left(\tfrac{\nu+1}{2};\tfrac{\nu+3}{2},\nu+1;\tfrac{1}{4}y^2\right).
\tag{3}
$$
We notice that the linear expressions can easily be integrated,
\begin{align}
I_{\pm} &= 3^{-3/2} \int_0^\infty \mathrm dy\,e^{-2y}
\big[F_{-1/3}(y) \pm F_{1/3}(y)\big] \tag{4a}\\
&=  \frac 1 {18} \sqrt[3]{2+\sqrt 3}
\pm \frac 1 {18} \sqrt[3]{2-\sqrt 3},\tag{4b}
\end{align}
such that the OP's question is equivalent to
\begin{align}
\hat I &= 3I + 6I_{+} \tag{5a}\\
&= \frac 1{\sqrt 3} \int_0^\infty \mathrm dy\,e^{-2y}
\big[F_{-1/3}(y)-F_{1/3}(y)+2\big]
\big[F_{-1/3}(y)+F_{1/3}(y)  \big] \tag{5b}\\
&\stackrel{?}{=} 1.\tag{5c}
\end{align}
While the integral (5b) is not really simpler than (2a), the result is.
This simpler result might be related to the fact that
$$
\int_{-\infty}^0 \mathrm dt \Ai(t) = \frac 2 3,
\quad
\int_{-\infty}^0 \mathrm dt \Bi(t) = 0,\tag{6}
$$
which induces
$$
\hat I = 3\int_0^\infty \mathrm dx
\sqrt x e^{\large -\frac 4 3 x^{3/2}}
\left(\int_{-\infty}^x \mathrm dt \Ai(t)\right)
\left(\int_{-\infty}^x \mathrm dt \Bi(t)\right)
\stackrel{?}{=} 1.\tag{7}
$$
