Let $_RM_R:={}_R(L^2R\otimes L^2R)_R$.
Its algebra of endomorphisms is $R^{\mathrm{op}}\,\bar\otimes\, R$.
Using (misleading!) intuition from the representation theory of separable $C^*$-algebras, one might guess that every MASA in $R^{\mathrm{op}}\bar\otimes R$ gives rise to a direct integral decomposition of $_RM_R$ into irreducible $R$-$R$-bimodules. But that is not true! There is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.
Indeed, suppose that ${}_R(M_x)_R$, $x\in X$, are irreducible bimodules, and that $\int^\oplus_{x\in X} {}_R(M_x)_R dx$ is an $R$-$R$-bimodule that is isomorphic to $_RM_R$. Then for every $M_x$, the left and right actions of $R$ induce an action of $R\,\bar\otimes\, R^{\mathrm{op}}$ on $M_x$. But $R\,\bar\otimes\, R^{\mathrm{op}}$ is type $II$ and does not admit irreducible representations. Contradiction.
My argument above is not valid. Nevertheless, I still maintain that there is no way of writing $_RM_R$ as a direct integral of irreducible bimodules.