There's a couple of conceptual problems at the base of your question that need to be addressed.
1) Free translation of particles in space is not discrete - the distribution of quantum states is continuous, and therefore so is the distribution of energies or of velocities - but rotation or vibration are quantised, so they have discrete distributions, so the formula definitely can't be applied to them. You could apply it to molecules as classical objects, ignoring quantum behaviour, but what you are describing then is not a molecule; you won't get realistic results that you could compare to experimental results.
2) The equipartition theorem states that rotation or vibration degrees of freedom will have the same energy as translation degrees of freedom, but the same energy will not translate into the same velocity.
For instance, in a classical treatment, the energy distribution function per degree of freedom in equilibrium that we obtain from Maxwell-Boltzmann is
$$ f_\epsilon = \sqrt{ \cfrac{1}{\pi \epsilon_i k_B T} } \exp \left[ \cfrac{-\epsilon_i}{k_B T} \right] $$
Taking into account that classical rotational energy is
$$ E_{rot} = \frac{1}{2} I \omega^2 $$
we can derive the classical angular velocity distribution:
$$ f_\omega = \sqrt{\cfrac{I}{2 \pi k_B T}} \exp \left[ \cfrac{-I \omega_i^2}{k_B T} \right] $$
with the moment of inertia $I$ being dependent on the geometry of the molecule. For instance, in the simplest case of a diatomic molecule approximated as a rigid rotor, it is
$$ I = \mu d^2 $$
where $\mu$ is the reduced mass and $d$ is the interatomic distance.
However, as we said, this is not a correct description of an actual molecule, in which vibration and rotation movements are quantised. In a quantum treatment, the allowed values for rotational energy in a linear molecule are
$$ E_{rot} = J \left( J + 1 \right) \cfrac{\hbar^2}{2I} ~ \mathrm{with} ~ J=0,1,2... $$
note that rotational levels are degenerate (with unequal degeneracy $g_j = 2J+1$) and discrete, so we need to use a discrete partition function $Z_{rot}$:
$$ Z_{rot} = \sum_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] $$
which in principle we can't calculate analytically. For molecules with a very large moment of inertia $I$ and/or at high enough temperatures $T$, the gap in energy between levels becomes very small and we can approximate this sum as a continuous integral over $J$ (the so-called high temperature limit):
$$ Z_{rot} \approx \int\limits_{J=0}^{\infty} \left( 2 J + 1 \right) \exp \left[ \cfrac{-J \left( J + 1 \right) \hbar^2 }{ 2 I k_B T} \right] dJ = \cfrac{ 2 I k_B T}{\hbar^2} $$
Under that assumption, the (discrete) angular momentum distribution of a rotational degree of freedom can be derived as
$$ F_J = \left( 2 J + 1 \right) \cfrac{ \pi \hbar^3}{ I k_B T} \exp \left[ \cfrac{-2 \pi \hbar^3 \left( J + 1 \right) }{ I k_B T} \right] $$
Note that we're using angular momentum and not angular velocity here because, while the former is a well-defined magnitude in quantum mechanics, it's not clear the later makes physical sense in a quantum treatment of molecules. We could naively calculate a semi-classical "angular velocity" distribution from that angular momentum distribution, but, again, we'd not be actually describing a molecule.
And a final conceptual problem:
3) Rotation and vibration (and electronic) motion is coupled; to give a simple example, in a diatomic molecule, as rotation angular momentum increases, interatomic distance (and therefore moment of inertia) changes; a simple approximation is to introduce a rotational term representing centrifugal distortion, but for higher precision both movements have to be considered at the same time, what makes for much more complex equations.
Keep also in mind that we've restricted ourselves to the simplest cases in rigid diatomic molecules; more complex systems have much more complex mathematics involved and often have to be treated empirically.
