Solving the Rotation Differential Equation when the reference frame itself is rotating

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Consider the three-dimensional general rotation differential equation $\mathbf{\dot{R}}_{b}^{i} = \mathbf{{R}}_{b}^{i} [{\boldsymbol{\omega}_{ib}^{b}}]_\times$ where $i$ is an inertial frame of reference, $b$ is some body frame, a and ${\boldsymbol{\omega}_{ib}^{b}}$ is the angular rate of the body with respect to inertial space, resolved in the body frame (e.g. an ideal gyro measurement).

If the angular rate is constant, then this differential equation has the well-known solution $\mathbf{{R}}_{b(k+1)}^{i} = \mathbf{{R}}_{b(k)}^{i} \operatorname{expm}\left( [{\boldsymbol{\omega}_{ib}^{b}}]_\times \right)$. So far so good.

Consider now that the reference frame is NOT an inertial frame; instead, it is another body frame $v$ which rotates independently of body frame $b$. I am interested in the solution of the differential equation:

$\mathbf{\dot{R}}_{b}^{v} = \mathbf{{R}}_{b}^{v} [{\boldsymbol{\omega}_{vb}^{b}}]_\times$

Except, I don't have measurement ${\boldsymbol{\omega}_{vb}^{b}}$ but instead have each body frame with respect to the inertial frame ${\boldsymbol{\omega}_{iv}^{v}}$ and ${\boldsymbol{\omega}_{ib}^{b}}$. Making the appropriate substitutions:

$$ \mathbf{\dot{R}}_{b}^{v} = \mathbf{{R}}_{b}^{v} [{\boldsymbol{\omega}_{vb}^{b}}]_\times \\ = \mathbf{{R}}_{b}^{v} [{\boldsymbol{\omega}_{ib}^{b}} - {\boldsymbol{\omega}_{iv}^{b}}]_\times \\ = \mathbf{{R}}_{b}^{v} [{\boldsymbol{\omega}_{ib}^{b}} - \mathbf{{R}}_{v}^{b} {\boldsymbol{\omega}_{iv}^{v}}]_\times \\ $$

The angular rate in the square brackets is no longer constant over the interval of rotation because of the addition of the non-constant rotation term.

Is there a solution for this problem? Or indeed, have I formulated the problem correctly?