Suppose $$ \nabla_a X_b+ \nabla_b X_a = \Omega g_{ab} $$ for some smooth, positive function $\Omega$ on a (pseudo-)Riemannian manifold $(M, g)$. Can one find a nice expression, independent of $X$, for $\nabla_X \text{Rc}$ where $\text{Rc}$ is the Ricci tensor? I thought of trying the following, first recalling that
\begin{align} \mathcal{L}_X \text{Rc} (Y, Z) &= X (\text{Rc}(Y, Z)) - \text{Rc}([X, Y], Z) - \text{Rc}(Y, [X, Z]) \\ &= X (\text{Rc}(Y, Z)) - \text{Rc}(\nabla_X Y - \nabla_Y X, Z) - \text{Rc}(Y, \nabla_X Z - \nabla_Z X) \\ &= X (\text{Rc}(Y, Z)) - \text{Rc}(\nabla_X Y, Z) + \text{Rc}(\nabla_Y X, Z) - \text{Rc}(Y, \nabla_X Z) +\text{Rc}(Y, \nabla_Z X) \\ &= \nabla_X \text{Rc}(Y, Z) + \text{Rc}(\nabla_Y X, Z) + \text{Rc}(Y, \nabla_Z X) \end{align} so it suffices to compute $\mathcal L_X \text{Rc}$ and $\text{Rc}(\nabla_Y X, Z) + \text{Rc}(Y, \nabla_Z X)$. It seems like computing the Lie derivative should be doable by looking at formulas for conformal transformations of metrics, but the $\text{Rc}(\nabla_Y X, Z) + \text{Rc}(Y, \nabla_Z X)$ term eludes me. Is such a simplification actually possible?