= Solution
Integrate the two velocity factors over the time square $0\le s,s'\le t$. The <white noise> contribution is $4Dt$. The orientation contribution depends only on the time separation, giving
$$
M(t)\equiv\langle|\mathbf r(t)-\mathbf r(0)|^2\rangle
=4Dt+2v^2\int_0^t(t-s)e^{-D_rs}\cos(\omega s)ds.
$$
This double-integral derivation fixes the delta contribution without ambiguity about a half-weight at an endpoint. Put $z=D_r-i\omega$. Elementary integration gives
$$
\boxed{M(t)=4Dt+2v^2\operatorname{Re}\left[\frac t z-\frac{1-e^{-zt}}{z^2}\right].}
$$
Using $\operatorname{Re}(1/z)=D_r/(D_r^2+\omega^2)$ and multiplying out the second term yields
$$
\boxed{\begin{aligned}
M(t)={}&4Dt+\frac{2v^2D_rt}{D_r^2+\omega^2}+\frac{2v^2(\omega^2-D_r^2)}{(D_r^2+\omega^2)^2}\\
&+\frac{2v^2e^{-D_rt}}{(D_r^2+\omega^2)^2}
\left[(D_r^2-\omega^2)\cos\omega t-2D_r\omega\sin\omega t\right].
\end{aligned}}
$$
The constant and decaying terms cancel at $t=0$, as required. The apparently singular case $D_r=\omega=0$ is evaluated from the integral, giving $M=4Dt+v^2t^2$.
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