How to simplify the following vector cross products?

Question

Let the position vector be $\mathbf{x}$, normal vector $\mathbf{n}$, rotation angular vector $\mathcal{\alpha}$ and rotation matrix $\mathcal{M}$ which contains 2nd order of vector $\mathcal{\alpha}$. \begin{align} \hat{\mathbf{x}}=\mathbf{x}+\mathbf{\alpha}\times\mathbf{x}+\mathcal{M}\tag{1}\mathbf{x} \end{align} \begin{align} \hat{\mathbf{n}}=\mathbf{n}+\mathbf{\alpha}\times\mathbf{n}+\mathcal{M}\tag{2}\mathbf{n} \end{align} The cross product of (1) and (2) then takes a form \begin{align}\label{eq:xXn_expand} \hat{\mathbf{x}}\times\hat{\mathbf{n}}=&\mathbf{x}\times\mathbf{n}+\mathbf{x}\times(\mathcal{\alpha}\times\mathbf{n})+(\mathcal{\alpha}\times\mathbf{x})\times\mathbf{n}+(\mathcal{\alpha}\times\mathbf{x})\times(\mathcal{\alpha}\times\mathbf{n})+\\&(\mathcal{M}\mathbf{x})\times\mathbf{n}+\mathbf{x}\times(\mathcal{M}\mathbf{n})+(higherorder) \end{align} The 2nd and 3rd terms can be combined \begin{align} \mathbf{x}\times(\mathcal{\alpha}\times\mathbf{n})+(\mathcal{\alpha}\times\mathbf{x})\times\mathbf{n}&=\mathcal{\alpha}\times(\mathbf{x}\times\mathbf{n})\\ \end{align} and the 4th term expands \begin{align} (\mathcal{\alpha}\times\mathbf{x})\times(\mathcal{\alpha}\times\mathbf{n})&=[(\mathcal{\alpha}\times\mathbf{x})\cdot\mathbf{n}]\mathcal{\alpha}+\mathbf{n} [(\mathcal{\alpha}\times\mathbf{x})\cdot\mathcal{\alpha}]=[(\mathcal{\alpha}\times\mathbf{x})\cdot\mathbf{n}]\mathcal{\alpha} \end{align} However the matrix vector terms are left mysterious to me. I tried \begin{align} (\mathcal{M}\mathbf{x})\times\mathbf{n}+\mathbf{x}\times(\mathcal{M}\mathbf{n})&=\varepsilon_{ijk}(M_{jl}x_ln_k+x_jM_{kl}n_l)\\ &= \end{align} By brutal expansion, \begin{align} -\mathcal{M}_{11}x_1n_3 +\mathcal{M}_{11}x_3n_1 +\mathcal{M}_{11}x_1n_2 -\mathcal{M}_{11}x_2n_1 -\mathcal{M}_{12}x_2n_3 +\mathcal{M}_{12}x_3n_2 +\mathcal{M}_{12}x_2n_2 -\mathcal{M}_{12}x_2n_2 -\mathcal{M}_{13}x_3n_3 +\mathcal{M}_{13}x_3n_3 +\mathcal{M}_{13}x_3n_2 -\mathcal{M}_{13}x_2n_3 +\mathcal{M}_{21}x_1n_3 -\mathcal{M}_{21}x_3n_1 +\mathcal{M}_{21}x_1n_1 -\mathcal{M}_{21}x_1n_1 +\mathcal{M}_{22}x_2n_3 -\mathcal{M}_{22}x_3n_2 +\mathcal{M}_{22}x_1n_2 -\mathcal{M}_{22}x_2n_1 +\mathcal{M}_{23}x_3n_3 -\mathcal{M}_{23}x_3n_3 +\mathcal{M}_{23}x_1n_3 -\mathcal{M}_{23}x_3n_1 +\mathcal{M}_{31}x_2n_1 -\mathcal{M}_{31}x_1n_2 -\mathcal{M}_{31}x_1n_1 +\mathcal{M}_{31}x_1n_1 +\mathcal{M}_{32}x_2n_2 -\mathcal{M}_{32}x_2n_2 -\mathcal{M}_{32}x_1n_2 +\mathcal{M}_{32}x_2n_1 +\mathcal{M}_{33}x_2n_3 -\mathcal{M}_{33}x_3n_2 -\mathcal{M}_{33}x_1n_3 +\mathcal{M}_{33}x_3n_1 \end{align} The next simplification to be in terms of $\mathcal{M}(\mathbf{x}\times\mathbf{n})$?