The Apéry's constant and the Airy function Several weeks ago, while I was playing with a CAS, Wolfram Alpha online calculator I found the closed-form that provided this calculator for $$\int_0^\infty\operatorname{Ai}(x)\log^3(x)dx,\tag{1}$$
involving the Airy function $\operatorname{Ai}(x)$, see this special function from the Wikipedia Airy function.
But I don't know nor if it is in the literature neither nor any hint to get the corresponding indefinite integral (if it is feasible). 

Question. Do you know if the closed-form for $$\int_0^\infty\operatorname{Ai}(x)(\log(x))^3dx,\tag{1}$$ in terms of well-known constants and $\zeta(3)$ is in the literature? Provide the reference, and if I can I search it. In other case, can you provide an hint to calculate this kind of definite integral? Many thanks.

 A: 
A little note. Another way to derive the first formula from Sangchul Lee's answer.

We use the fact that:
$$\mathrm{Ai}''(x)=x \mathrm{Ai}(x)$$
Let's consider the following integral:
$$I_n(y)=\int_0^\infty \mathrm{Ai}(y x) x^n dx$$
$$I'_n(y)=\int_0^\infty \mathrm{Ai}'(y x) x^{n+1} dx$$
$$I''_n(y)=\int_0^\infty \mathrm{Ai}(y x) x^{n+3} dx=I_{n+3}(y)$$
Now we can rewrite it:
$$I_n(y)= \frac{1}{y^{n+1}}\int_0^\infty \mathrm{Ai}(x) x^n dx=\frac{I_n(1)}{y^{n+1}}$$
$$I''_n(y)= \frac{(n+1)(n+2) I_n(1)}{y^{n+3}}=I_{n+3}(y)$$
So we have obtained an explicit recurrence relation for $y=1$:

$$I_{n+3}=(n+1)(n+2) I_n \tag{1}$$

We know that:
$$I_0=\frac{1}{3}$$ (a well known integral)
$$I_1=-\mathrm{Ai}'(0)=\frac{1}{3^{1/3} \Gamma(1/3)}$$
$$I_2=\int_0^\infty \mathrm{Ai}(x) x^2 dx= \int_0^\infty \mathrm{Ai}''(x) x dx=\mathrm{Ai}'(x) x \big|_0^\infty-\int_0^\infty \mathrm{Ai}'(x) dx=\mathrm{Ai}(0)=\frac{1}{3^{2/3} \Gamma(2/3)}$$
So, we now know the general recurrence relation for $I_n$ and three initial conditions needed to find any term.
After we derive the general expression for $I_n$, it's easy enough to use continuity of all the functions involved to generalize for non-integer parameters and obtain the desired Mellin transform.
A: Using the various provided answers, I present to you the craziest integral I've ever seen. Behold: 

$$\int_0^\infty \mathrm{Ai}(x)\ln^3x\ dx=-\frac1{81}\left[(4\pi^2+\beta^2)\beta+52\zeta(3)\right]$$
  where $\beta=2\gamma+\ln3$.

A: You may refer to the entry 9.10.17 of DLMF, which describes the Mellin transform of the Airy function $\operatorname{Ai}$:
$$ \int_{0}^{\infty} t^{\alpha-1}\operatorname{Ai}(t)\,\mathrm{d}t = \frac{\Gamma(\alpha)}{3^{(\alpha+2)/3}\Gamma\big(\frac{\alpha+2}{3}\big)}. \tag{9.10.17}$$
Differentiating both sides with respect to $\alpha$ 3-times and plugging $\alpha = 1$ gives the answer to your integral in terms of polygamma functions with further simplifications available.

Sketch of proof of $\text{(9.10.17)}$. We begin with the integral representation
$$ \operatorname{Ai}(x) = \frac{1}{\pi} \int_{0}^{\infty} \cos\left(\frac{t^3}{3} + xt \right) \, \mathrm{d}t. $$
Taking Mellin transform and switching the order of integration,
\begin{align*}
\int_{0}^{\infty} x^{\alpha-1}\operatorname{Ai}(x)\,\mathrm{d}x
&= \frac{1}{\pi} \operatorname{Re}\bigg[ \int_{0}^{\infty}\int_{0}^{\infty} x^{\alpha-1} e^{\frac{1}{3}it^3 + ixt} \, \mathrm{d}x\mathrm{d}t \bigg] \\
&= \frac{1}{\pi} \operatorname{Re}\bigg[ \int_{0}^{\infty} \frac{\Gamma(\alpha)}{(-it)^{\alpha}} e^{\frac{1}{3}it^3} \, \mathrm{d}t \bigg] \\
&= \frac{1}{\pi} \operatorname{Re}\bigg[ \Gamma(\alpha) i^{\alpha} \frac{\Gamma\big(\frac{1-\alpha}{3}\big)}{3\cdot(-i/3)^{\frac{1-\alpha}{3}}} \bigg] \\
&= \frac{1}{\pi} \operatorname{Re}\bigg[ \Gamma(\alpha) i^{\frac{2\alpha+1}{3}} \frac{\Gamma\big(\frac{1-\alpha}{3}\big)}{3^{\frac{\alpha+2}{3}}} \bigg] \\
&= \frac{\Gamma(\alpha)\Gamma\big(\frac{1-\alpha}{3}\big)}{\pi 3^{\frac{\alpha+2}{3}}}\cos\big( \tfrac{1}{6}\pi + \tfrac{1}{3}\alpha\pi \big).
\end{align*}
Now applying the Euler's reflection formula $\Gamma(s)\Gamma(1-s) = \pi \csc(\pi s)$ with $s = \frac{1-\alpha}{3}$ yields the right-hand side of $\text{(9.10.17)}$.
