Deriving Supergravity Equations of Motion The other day I decided to quickly make sure I can derive the supergravity equations of motion for the NS/NS sector using the following action:
$$
S=\int_{M_{10}}d^{10}x \space \sqrt{-g} e^{-2\phi}\left(R+4(\partial \phi)^2-\frac{1}{12}H_{abc}H^{abc}\right)
$$
Varying the metric, the Ricci-scalar term (R) will just yield an Einstein's equation term $R_{ab}-\frac{1}{2}Rg_{ab}$. The scalar field term should be simple enough since it is really just 4$g^{ab}\partial_a\phi \partial_b \phi$. This would just give a term in the field equations that looks like $4\partial_a\phi \partial_b \phi$. The flux term, noting that $H_{abc}=\partial_a b_{bc}-\partial_b b_{ac}+\partial_c b_{ab}$, will give a term $-\frac{1}{4}(H^2)_{ab}$.
According to various papers (specifically https://arxiv.org/abs/1205.2274) one should get the following equation of motion for the metric variation:
$$
R_{ab}+2D_a D_b \phi-\frac{1}{4}(H^2)_{ab}=0
$$
As you can see, I have no problem getting the last term, but I'm failing to see how the Einstein and scalar field term combine to get $R_{ab}+2D_a D_b \phi$.
Am I just missing some simplification here or is does it look like I actually made an error in my calculation? 
 A: I think this is not a big deal. Taking it into consideration the variation of your action with respect to $\phi$, these two terms $g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi$ and $g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\phi$ are actually the same.
Let us figure out
$$
\delta\int_{\Omega}e^{-2\phi}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi\,{\rm d}V=0,
$$
where I kept ${\rm d}V=\sqrt{-g}\,{\rm d}^{10}x$ as an invariant volume element. In this way, apply Euler-Lagrange equation (but in its covariant form for simplicity, i.e.,
$$
\frac{\partial\mathcal{L}}{\partial\phi}=\nabla_{\sigma}\left(\frac{\partial\mathcal{L}}{\partial\left(\nabla_{\sigma}\phi\right)}\right),
$$
where $\nabla_{\sigma}$ denotes the covariant derivative; and of course, for the scalar $\phi$, $\nabla_{\sigma}\phi=\partial_{\sigma}\phi$, while for the metric $g^{\mu\nu}$, $\nabla_{\sigma}g^{\mu\nu}=0$) to
$$
\mathcal{L}=e^{-2\phi}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi,
$$
and we have
$$
\frac{\partial\mathcal{L}}{\partial\phi}=-2e^{-2\phi}g^{\mu\nu}\partial_{\mu}\phi\partial_{\nu}\phi
$$
as well as
$$
\frac{\partial\mathcal{L}}{\partial\left(\nabla_{\sigma}\phi\right)}=2e^{-2\phi}g^{\mu\sigma}\partial_{\mu}\phi.
$$
Therefore
$$
\nabla_{\sigma}\left(\frac{\partial\mathcal{L}}{\partial\left(\nabla_{\sigma}\phi\right)}\right)=2e^{-2\phi}\left(-2g^{\mu\sigma}\partial_{\mu}\phi\partial_{\sigma}\phi+g^{\mu\sigma}\nabla_{\mu}\nabla_{\sigma}\phi\right).
$$
As a consequence, $\phi$ must satisfy
$$
g^{\mu\sigma}\partial_{\mu}\phi\partial_{\sigma}\phi=g^{\mu\sigma}\nabla_{\mu}\nabla_{\sigma}\phi.
$$
So you can see, if you replace the $g^{\mu\sigma}\partial_{\mu}\phi\partial_{\sigma}\phi$ term in the original action by $g^{\mu\sigma}\nabla_{\mu}\nabla_{\sigma}\phi$, and take the variational derivative of $g^{\mu\sigma}$, you will obtain the $\nabla_{\mu}\nabla_{\sigma}\phi$ term as is expected.
Hope this could be helpful for you :-)
