The problem is complicated, because in order to integrate $\boldsymbol{e}{z}^{\prime}$ we have to invert the relationships between $\omega{i}$ and $\theta_{j}$. Let us assume that the angular velocities are constant and take the small angle approximation: \begin{align} \dot{\theta}{i}=\text{const.}\ \ \Longrightarrow\ \ \theta{i}(t)=\dot{\theta}_{i}t \end{align} the small angle approximation implies that the angular velocity components in the body frame become: \begin{align} \Omega_{x} & \approx\dot{\theta}{2}\ \Omega{y} & \approx\dot{\theta}{1}\ \Omega{z} & \approx\dot{\theta}_{3} \end{align}
With these assumptions we obtain: \begin{align} \boldsymbol{e}{z}^{\prime}(t) & =\cos\theta{3}\sin\theta_{1}\boldsymbol{e}{x}+\sin\theta{1}\sin\theta_{3}\boldsymbol{e}{y}+\cos\theta{1}\boldsymbol{e}{z}\ & =\cos\left(\Omega{z}t\right)\sin\left(\Omega_{y}t\right)\boldsymbol{e}{x}+\sin\left(\Omega{y}t\right)\sin\left(\Omega_{z}t\right)\boldsymbol{e}{y}+\cos\left(\Omega{y}t\right)\boldsymbol{e}_{z} \end{align} which can be integrated in order to derive $\boldsymbol{r}{0}(t)$: \begin{align} \boldsymbol{r}{0}(t) & =U\intop{0}^{t}\boldsymbol{e}{z}^{\prime}\left(\tau\right)d\tau=\frac{U}{\Omega{y}-\Omega_{z}}\left[\Omega_{y}-\Omega_{y}\cos\left(\Omega_{y}t\right)\cos\left(\Omega_{z}t\right)-\Omega_{z}\sin\left(\Omega_{y}t\right)\sin\left(\Omega_{z}t\right)\right]\boldsymbol{e}{x}\ & \hspace{8em}+\frac{U}{\Omega{y}-\Omega_{z}}\left[-\Omega_{y}\cos\left(\Omega_{y}t\right)\sin\left(\Omega_{z}t\right)+\Omega_{z}\sin\left(\Omega_{y}t\right)\cos\left(\Omega_{z}t\right)\right]\boldsymbol{e}{y}\ & \hspace{8em}+\frac{U}{\Omega{y}}\sin\left(\Omega_{y}t\right)\boldsymbol{e}_{z} \end{align} This is in general a complex and very intriguing curve, but we will limit our study here to the case where $\Omega{y}\ll\Omega_{z}$. 804