Question on simplification of terms in a Hopf algebra

Question

Let $H=(H,\mu,\eta,\Delta,\epsilon,S)$ be a Hopf algebra and $S$ is the antipode in $H$. By the definition of antipode $S*id_H=id_H*S=\eta \circ\epsilon $. If I denote the Sweedler's notation by $ \Delta(x)=x_1 \otimes x_2$, then the antipode relation implies that $Sx_1x_2=x_1Sx_2=\epsilon(x)1_H$. Using this how can we simplify the expressions involving tensors? That is like $x_1 \otimes x_2 \otimes Sx_3 \otimes x_4 \otimes x_5$ into the form $x_1 \otimes x_2 \otimes x_3$? By the convolution $S*id_H= \mu \circ (S \otimes id_H) \circ \Delta$ and $id_H*S= \mu \circ (id_H \otimes S)\circ \Delta$. How can we simplify the expression $x_1 \otimes x_2 \otimes Sx_3 \otimes x_4 \otimes x_5$ into $x_1 \otimes \epsilon(x_2) \otimes x_3 \otimes x_4 $ ? I agree that we can combine $x_2Sx_3=\epsilon(x_2)1_H$. I dnt how to use it here?