Edit:
There seems to be a much easier way I overlooked, which I'll explain. My first answer is kept below for reference.
Your assembly consists of a small sector subtracted from a larger sector as shown below:
You calculate the moment of inertia of the sector about the horizontal axis as follows:
$$I=\frac{R^4}{24}(3\phi-3sin(\phi)-2sin(\phi)sin^2(\frac{\phi}{2}))$$
So you will have:
$$I_{xx}=2(I_{out}-I_{in})$$
Old Answer:
It will take some effort, but it can be done analytically:
You first calculate the moment of inertia of the complete annulus with:
$$I_{ann}=\frac{\pi}{4}(R^4-r^4)$$
Then you subract twice the moment of inertia, $I_{rec}$, of the lxb rectangle calculated with $\frac{lb^3}{12}$ and the parallel axis theorem.
Now you have subtracted the red hatched part which shouldn't be, but not yet the green part, which should be, so lets fix this quickly by adding the red moment twice, and subtracting the green moment twice.
The image below explains how to calculate the moment of inertia of the sectors about the centreline:
$$I_{sect}=\frac{r^4}{8}(\phi_i-sin(\phi_i)+2sin(\phi_i)sin^2(\frac{\phi_i}{2}))$$
$\phi_i$ is measured in radians.
So in the end you have:
$$I_{xx}=I_{ann}-2(I_{rec}+I_{sect-green}-I_{sect-red})$$