= Solution
Differentiate $B_{\alpha\beta}=\nabla_\beta X_\alpha$ along $X$ and commute the two <covariant derivative>[covariant derivatives]:
$$
X^\mu\nabla_\mu B_{\alpha\beta}
=X^\mu\nabla_\mu\nabla_\beta X_\alpha
=X^\mu\nabla_\beta\nabla_\mu X_\alpha
-X^\mu R^\nu{}_{\alpha\mu\beta}X_\nu,
$$
where the sign is the covector form of the <Ricci identity>. Apply the <product rule> to the first term:
$$
X^\mu\nabla_\beta\nabla_\mu X_\alpha
=\nabla_\beta(X^\mu\nabla_\mu X_\alpha)
-(\nabla_\beta X^\mu)(\nabla_\mu X_\alpha).
$$
The first term vanishes by the <geodesic equation>, and the second is
$$
-B^\mu{}_\beta B_{\alpha\mu}.
$$
Finally the pair symmetries and antisymmetries of the <Riemann curvature tensor> give
$$
-X^\mu X_\nu R^\nu{}_{\alpha\mu\beta}
=R_{\mu\beta\alpha}{}^\nu X^\mu X_\nu.
$$
Consequently
$$
\boxed{
X^\mu\nabla_\mu B_{\alpha\beta}
=-B^\mu{}_\beta B_{\alpha\mu}
+R_{\mu\beta\alpha}{}^\nu X^\mu X_\nu.
}
$$
This is the <Riccati equation for a timelike geodesic congruence>; taking its spatial trace is the starting point for the timelike <Raychaudhuri equation>.
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