An integral involving Airy functions $\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx$ I need your help with this integral:
$$\mathcal{K}(p)=\int_0^\infty\frac{x^p}{\operatorname{Ai}^2 x + \operatorname{Bi}^2 x}\mathrm dx,$$
where $\operatorname{Ai}$, $\operatorname{Bi}$ are Airy functions:
$$\operatorname{Ai}\,x=\frac{1}{\pi}\int_0^\infty\cos\left(x\,z+\frac{z^3}{3}\right)\,\mathrm dz,$$
$$\operatorname{Bi}\,x=\frac{1}{\pi}\int_0^\infty\left(\sin\left(x\,z+\frac{z^3}{3}\right)+\exp\left(x\,z-\frac{z^3}{3}\right)\right)\,\mathrm dz.$$
I am not sure that $\mathcal{K}(p)$ has a general closed form, but I hope so, because approximate numerical calculations suggest these conjectured values:
$$\mathcal{K}(3)\stackrel?=\frac{5\,\pi^2}{32},\ \ \mathcal{K}(6)\stackrel?=\frac{565\,\pi^2}{512}.$$
 A: Let us start with a warm-up exercise. Introduce the functions
$$g_{\pm}(x)=\operatorname{Ai}(x)\pm i\operatorname{Bi}(x).$$
Computing the Wronskian of these two solutions of the Airy equation, one can check that
$$\frac{1}{\operatorname{Ai}^2(x)+\operatorname{Bi}^2(x)}=\frac{\pi}{2i}\left[\frac{g_+'(x)}{g_+(x)}-\frac{g'_-(x)}{g_-(x)}\right]$$
This gives the integral $\mathcal{K}(0)$ as
$$\mathcal{K}(0)=\pi\left[\arg g_+(\infty)-\arg g_+(0)\right]=\pi\left[\pi-\frac{\pi}{3}\right]=\frac{\pi^2}{6}.$$

To compute the integral $\mathcal{K}(3n)$, we will need to develop a more sophisticated approach. First note that (see here)
$$g_{\pm}(x)=-2e^{\mp 2\pi i/3}\operatorname{Ai}\left(e^{\mp2\pi i/3}x\right).$$
Therefore 
\begin{align}\mathcal{K}(3n)&=\frac{\pi}{2i}\int_0^{\infty}x^{3n}\left[\frac{g_+'(x)}{g_+(x)}-\frac{g'_-(x)}{g_-(x)}\right]dx=\\
&=\frac{\pi}{2i}\lim_{R\rightarrow\infty}\int_{S_R}z^{3n}\frac{\operatorname{Ai}'(z)}{\operatorname{Ai}(z)}dz,
\end{align}
where the contour $S_R$  in the complex $z$-plane
is composed of two segments: one going from $Re^{2\pi i/3}$ to $ 0$ and another one going from $0$ to $ Re^{-2\pi i/3}$.
It is a well-known fact that the Airy function $\operatorname{Ai}(z)$ has zeros (i.e. our integrand has poles) on the negative real axis only. Therefore by residue theorem our integral is equal to
$$\mathcal{K}(3n)=-\frac{\pi}{2i}\lim_{R\rightarrow \infty}\int_{C_R}z^{3n}\left[\ln\operatorname{Ai}(z)\right]'dz,\tag{1}$$
where $C_R$ is the arc of the circle of radius $R$ centered at the origin going counterclockwise from $Re^{-2\pi i/3}$ to $Re^{2\pi i/3}$. 
The limit (1), on the other hand, can be computed using the asymptotics of the Airy function as $z\rightarrow\infty$ for $|\arg z|<\pi$:
\begin{align}
\ln\operatorname{Ai}(z)\sim -\frac23 z^{3/2}-\ln2\sqrt{\pi}-\frac14\ln z+
\ln\sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac16\right)_k\left(\frac56\right)_k}{k!}\left(\frac43 z^{3/2}\right)^{-k}.\tag{2}
\end{align}
Note that if we introduce instead of $z$ the variable $s=\frac43z^{3/2}$, then the integration will be done over the circle of radius $\Lambda=\frac43 R^{3/2}$, i.e. a closed contour in the complex $s$-plane. The corresponding integral can therefore be computed by residues by picking the necessary term in the large $s$ expansion of $\ln \operatorname{Ai}(z)$.
More precisely, we have the following formula:
\begin{align}
\mathcal{K}(3n)=-\frac{\pi}{2i}\lim_{\Lambda\rightarrow \infty}\oint_{|s|=\Lambda}
\left(\frac{3s}{4}\right)^{2n}d\left[-\frac16\ln s+\ln\sum_{k=0}^{\infty}\frac{(-1)^k\left(\frac16\right)_k\left(\frac56\right)_k}{k!}s^{-k}\right]\tag{3}
\end{align}
To compute the residue, it suffices to expand the logarithm-sum up to order $2n$ in $s^{-1}$. Note that the Pochhammer symbol coefficients are in fact some rational numbers.
In the simplest case $n=0$, the residue is determined by the term $-\frac16\ln s$ and we readily reproduce the previous result
$$\mathcal{K}(0)=-\frac{\pi}{2i}\cdot 2\pi i\cdot\left(-\frac16\right)=\frac{\pi^2}{6}.$$
The general formula for arbitrary $n$ would look a bit complicated (but straightforward to obtain) due to the need to expand the logarithm of the sum.
Example. The calculation of the corresponding values $M(n)=\mathcal{K}(3n)$ in Mathematica can be done using the command

\begin{align}\mathtt{\text{ M[n_] := -Pi^2 SeriesCoefficient[
 Series[(3 s/4)^(2 n) D[-Log[s]/6 +}} \\ \mathtt{\text{  Log[Sum[(-1)^k
 Pochhammer[1/6, k] Pochhammer[5/6, k]/(k! s^k), }} \\
 \mathtt{\text{{k, 0,  2 n}]], s], {s, Infinity, 1}], 1]}} \end{align}

This yields, for instance,
$$M(0)=\frac{\pi^2}{6},\quad M(1)=\frac{5\pi^2}{32},\quad M(2)=\frac{565\pi^2}{512},$$ $$\ldots, M(10)=\frac{2\,660\,774\,144\,147\,177\,521\,025\,228\,125\,\pi^2}{2\,199\,023\,255\,552},\ldots$$
and so on.

Added: The large $s$ expansion (2) can also be found directly by using Airy equation. Moreover, by transforming it into an equation for $\ln \operatorname{Ai}(z)$, one can avoid reexpanding the logarithm of the sum. The price to pay will be that the expansion coefficients will be determined by a nonlinear recurrence relation instead of explicit formulas.
A: I can give a partial answer. Using DLMF, Chapter 9 "Airy and Related Functions", Airy Functions:


*

*§9.2 (ii) Initial Values, formulae 9.2.3, 9.2.5, 

*§9.8 (i) Modulus and Phase, Definitions, formulae 9.8.3, 9.8.4, and

*§9.8 (ii) Modulus and Phase, Identities formula 9.8.14, 


we can see that 
$$\frac{d}{dx}\arctan\frac{\operatorname{Ai}x}{\operatorname{Bi}x}=-\frac1\pi\frac1{\operatorname{Ai}^2x+\operatorname{Bi}^2x}$$
and
$$\int_0^\infty\frac{dx}{\operatorname{Ai}^2x+\operatorname{Bi}^2x}=\frac{\pi^2}6.$$
This settles the question for the particular case $p=0$.
A: Here is an explicit formula based on O.L.'s answer:
$$\mathcal{K}(3n)=\frac{\pi^2}{6\cdot64^n}a_{2n},$$
where $a_n$ is the sequence defined recursively as follows:
$$a_0=1,\ \ a_{n+1}=(6\,n+4)\,a_n+\sum\limits_{i=0}^n a_i\,a_{n-i}$$
