How does one compute the ripple currents seen by an active PFC output capacitor?
The ripple current seen by the capacitor has two components: the fundamental, and the high-frequency switching. First, the fundamental:

Assuming the PFC to be perfect, the current output through the diode is a rectified sine wave. Over any given 180-degree period:

$$
1: I_d = I_{pk}sin \Theta
$$

The average load current is known and fixed. The instantaneous current seen by the capacitor is the difference between these currents:

$$
2: I_{cap} = I_d - I_{load}
$$

We can find the load current in terms of the peak current, by taking the average of a sine wave over a half-period:

$$
3: I_{load} = \frac{2I_{pk}}{\pi}
$$

Substituting (1) and (3) into (2), then factoring:

$$
4: I_{cap} = I_{pk}sin \Theta - \frac{2I_{pk}}{\pi}\\
5: I_{cap} = I_{pk}(sin \Theta - \frac{2}{\pi})
$$

The RMS current seen by the capacitor:

$$
6: I_{RMS}=\sqrt{\frac{\int_{0}^{\pi}{I_{cap}^2}d\Theta}{\pi}}\\
7: I_{RMS}=\sqrt{\frac{\int_{0}^{\pi}{I_{pk}^2(sin \Theta - \frac{2}{\pi})^2}d\Theta}{\pi}}\\
7: I_{RMS}=I_{pk}\sqrt{\frac{\int_{0}^{\pi}{(sin \Theta - \frac{2}{\pi})^2}d\Theta}{\pi}}\\
8: I_{RMS}=I_{pk}\sqrt{\frac{\int_{0}^{\pi}{sin^2 \Theta}d\Theta - \int_{0}^{\pi}{\frac{4sin \Theta}{\pi}}d\Theta + \int_{0}^{\pi}{\frac{4}{\pi^2}}d\Theta}{\pi}}\\
9: I_{RMS}=I_{pk}\sqrt{\frac{\frac{\pi}{2} - \frac{sin{2\pi}}{4} - \frac{0}{2} + \frac{sin{0}}{4}+ \frac{4cos \pi}{\pi} - \frac{4cos 0}{\pi}+ \frac{4}{\pi}}{\pi}}\\
10: I_{RMS}=I_{pk}\sqrt{\frac{\frac{\pi}{2} - \frac{8}{\pi}+ \frac{4}{\pi}}{\pi}}\\
11: I_{RMS}=I_{pk}\sqrt{\frac{1}{2} - \frac{4}{\pi^2}}\\
$$

Solving (3) for peak current and substituting into (11):
$$
12: I_{RMS}=I_{load}\frac{\pi}{2}\sqrt{\frac{1}{2} - \frac{4}{\pi^2}}=I_{load}\sqrt{\frac{\pi^2}{8}-1}\\
13: I_{RMS}\approx.4834 I_{load}
$$

The high-frequency switching component is more complex. Starting with this question, we can see that the RMS current seen by the capacitor will vary as the duty and input voltage change across the sine period. Unfortunately, this function becomes exceptionally complex, making an exact integral impractical. Numerical approximations give a ripple of .96 the load current, with voltage, frequency, and inductance having little effect at all practical values.