I want to know if it is possible to express the operation

$$
\nabla \phi \times (\nabla \times \mathbf A) 
$$

as the divergence of second order tensor field $T$. Here $ \phi$ is a scalar field and $\mathbf A$ is a solenoidal vector field ($\nabla \cdot \mathbf A=0$)

I have used all possible identities and finally I can only get 

$$ 
\nabla \phi \times (\nabla \times \mathbf A) = \nabla \phi \cdot(\nabla \mathbf A - \nabla \mathbf A^T) 
$$

Is it possible to find something like $ \nabla \phi \times (\nabla \times \mathbf A) = \nabla \cdot(\text{tensor}) $  ?

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EDIT: I could derive an expression that may be equivalent to Carlo's answer. 

$$
\nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A-\nabla \mathbf A^T) \cdot \nabla \phi
$$
Now, from the identity $\nabla \cdot (\phi  \mathbf T)=\phi\nabla\cdot  \mathbf T+ \mathbf T^T \nabla \phi$ we can write

$$
(\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi = -\nabla \cdot(\phi(\nabla \mathbf A - \nabla \mathbf A^T))+\phi\nabla\cdot(\nabla \mathbf A - \nabla \mathbf A^T)
$$
Considering that $\nabla \cdot(\nabla \mathbf A)=\nabla^2  \mathbf A$ and $\nabla \cdot(\nabla \mathbf A^T)=\nabla (\nabla \cdot \mathbf A)=0   $:

$$
\nabla \phi \times (\nabla \times \mathbf A) = (\nabla \mathbf A^T-\nabla \mathbf A) \cdot \nabla \phi = -\nabla\cdot(\phi(\nabla \mathbf A-\nabla \mathbf A^T)+\phi\nabla^2 \mathbf A
$$ 
This form is convenient since the LHS of the equation already contains $\nabla^2 \mathbf A$.