 Eleven-dimensional three-form action variation

For a differential three-form $A$ with $F=dA$ on an oriented eleven-dimensional manifold, varying $\int (\frac{1}{2}F\wedge \star F+F\wedge F\wedge A)$ with fixed metric gives $\delta S=\int \delta A\wedge (d\star F+3F\wedge F)$ up to a boundary term. Indeed $d(\delta A\wedge F\wedge A)=d\delta A\wedge F\wedge A-\delta A\wedge F\wedge F$, so the two differentiated $F$ factors and the differentiated $A$ factor contribute equally. The shift $A\mapsto A+d\Lambda$ changes the density by $d(F\wedge F\wedge \Lambda )$, preserving the field equation.