EoM for scalar field in Brans-Dicke Theory The action is given by 
$$ S^{(BD)} = \int d^4 x \sqrt{|g|} \left[ \phi R - \frac{\omega}{\phi} g^{\mu \nu} \, \nabla_\mu \phi \nabla_\nu \phi  - V(\phi) \right]$$
I am trying to vary with respect to $\phi$ using Euler - Lagrange equations in curved spacetime, i.e. 
$$ \underbrace{\frac{\mathcal{\partial L}}{\partial \phi}}_{I} - \underbrace{\nabla_\mu \left( \frac{\mathcal{\partial L}}{\partial (\nabla_\mu \phi)} \right)}_{II} = 0 $$
I obtained the following
$$ I: R + \frac{\omega}{\phi^2}g^{\mu \nu} \, \nabla_\mu \phi \nabla_\nu \phi - \frac{\partial V}{\partial \phi} $$
\begin{align} 
\begin{aligned} 
II&: \nabla_\mu \left[ - \frac{\omega}{\phi} g^{\mu \nu} \left( \nabla_\nu \phi + \underbrace{\nabla_\mu \phi \, \delta^\mu{}_\nu}_{\nabla_\nu \phi}\right)\right] \\[3 ex]
&= - \nabla_\mu \left( \frac{2\omega}{\phi} \underbrace{g^{\mu \nu} \, \nabla_\nu \phi}_{\nabla^\mu \phi}\right) \\[3 ex]
&= -2 \omega \, \nabla_\mu \left( \frac{1}{\phi} \nabla^{\mu} \phi\right) \\[3 ex]
&= -2 \omega \left( \nabla_\mu \frac{1}{\phi} \cdot \nabla^\mu \phi + \frac{1}{\phi} \nabla_\mu \nabla^\mu \phi\right)\\[3 ex]
& = -2 \omega \left( \nabla_\mu \frac{1}{\phi} \cdot \nabla^\mu \phi + \frac{1}{\phi} \Box \phi\right)
\end{aligned}
\end{align}
According to literature, EoM should have been
\begin{align}
I - II =  R - \frac{\omega}{\phi^2}g^{\mu \nu} \, \nabla_\mu \phi \nabla_\nu \phi - \frac{\partial V}{\partial \phi} + \frac{2\omega}{\phi} \Box \phi 
\end{align}
So I have an extra term
$$ \frac{2\omega}{\phi} \nabla_\mu \frac{1}{\phi} \cdot \nabla^\mu \phi$$
I've tried to write step-by-step, any help would be appreciated
 A: Seems like there is no mistake, I should have evaluated my 'extra term'. For the ones who are looking for EoM of BD field here is the full calculation
EoM for scalar field
Euler - Lagrange Equations in curved spacetime
\begin{align}
\frac{\partial\mathcal{L}}{\partial\phi}=\frac{1}{\sqrt{-g}}\partial_{\mu}\left[\sqrt{-g}\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right]
\end{align}
OR 
\begin{align}
\frac{\partial\mathcal{L}}{\partial\phi}=\nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right]
\end{align}
\begin{align}
 \frac{\partial\mathcal{L}}{\partial\phi} = \phi R + \frac{\omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \nabla_\nu \phi - \frac{\partial V}{\partial \phi}
 \end{align}
\begin{align}
 \begin{aligned}
 \nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right] &= \nabla_{\mu} \left[ - \frac{\omega}{\phi} g^{\mu \nu} \left(\nabla_\nu \phi + \underbrace{\nabla_{\mu} \phi \, \delta^\mu{}_\nu}_{\nabla_\nu \phi} \right)\right] \\[1ex]
 &= \nabla_{\mu} \left[ - \frac{\omega}{\phi} g^{\mu \nu} \left(2\nabla_\nu \phi  \right)\right]\\[1ex]
 &= -2 \omega \nabla_{\mu} \left( \frac{1}{\phi} g^{\mu \nu} \nabla_{\nu} \phi \right) \\[1ex]
 &= -2 \omega \nabla_{\mu} \left( \frac{1}{\phi} \nabla^{\mu} \phi \right) \\[1ex]
 &= -2 \omega \left[ \left(\nabla_{\mu} \frac{1}{\phi}\right) \nabla^\mu \phi + \frac{1}{\phi} \underbrace{\nabla_\mu \nabla^\mu \phi}_{\Box \phi} \right] \\[1ex]
 &= -2 \omega \left[ \left(-\frac{1}{\phi^2} \nabla_{\mu} \phi \right) \nabla^\mu \phi + \frac{1}{\phi} \Box \phi \right]\\[1ex]
 &= \frac{2 \omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \, \nabla_{\nu} \phi - \frac{2\omega}{\phi} \Box \phi 
 \end{aligned}
 \end{align}
Therefore, 
 \begin{align}
\begin{aligned}
 \frac{\partial\mathcal{L}}{\partial\phi} - \nabla_{\mu}\left[\frac{\partial\mathcal{L}}{\partial\left(\partial_{\mu}\phi\right)}\right] = R + \frac{\omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \, \nabla_{\nu} \phi - \frac{\partial V}{\partial \phi} + \frac{2 \omega}{\phi} \Box \phi
\end{aligned}
 \end{align}
\begin{align}
 \boxed{R - \frac{\omega}{\phi^2} g^{\mu \nu} \nabla_{\mu} \phi \, \nabla_{\nu} \phi -  \frac{\partial V}{\partial \phi} + \frac{2 \omega}{\phi} \Box \phi = 0}
 \end{align}
