For rigid body rotation discrete time kinematics equation is as follows
$R_{k+1} = T_r \cdot R_k$
where:
$R$ is a rotation matrix
$T_r = e^{-[a\times]}$ is a transition matrix
$[a\times]$ means a skew symmetric matrix from vector $a$
$a=\omega \cdot \delta t$ is a rotation vector
$\omega$ is angular velocity vector in body frame
$\delta t$ is a time step.
Using Rodrigue's formula one can have a closed form solution for matrix exponent so no infinite series calculation required.
$T_r=I+sin(\theta)\left[e\times \right] + (1-cos(\theta))\left[e\times \right]^2$
where
$e = \frac{a}{||a||}$ is a unit vector of rotation
$\theta = ||a||$ is angle of rotation around that unit vector
When dealing with quaternions similar expression is
$q_{k+1} = T_q \cdot q_k$
where
$T_q = e^{\frac{1}{2}\begin{bmatrix}-\left[a\times\right] && a\\ -a^T&& 0 \end{bmatrix}}$
Hope I have put not many mistakes in the expressions above. But general idea should be clear... Is there a closed form solution for quaternion discrete kinematics transition matrix $T_q$?
Observe
$$ \begin{bmatrix}-[a\times] & a \\ -a^T & 0\end{bmatrix}^2=\|a\|^2\begin{bmatrix}-I_3 & 0 \\ 0 & -1\end{bmatrix} $$
implies
$$ \exp\Big(\frac{1}{2}\begin{bmatrix}-[a\times] & a \\ -a^T & 0\end{bmatrix}\Big)\,=\,\cos\!\Big(\frac{\theta}{2}\Big)I_4+\sin\!\Big(\frac{\theta}{2}\Big)\begin{bmatrix}-[e\times] & e \\ -e^T & 0\end{bmatrix}. $$