Simplifying $b(abb + abbbb + aba^*bbb)^*$
Note that $abbbb+aba^*bbb = aba^*bbb$. Also, note that you can &quot;factorize&quot; common expressions in the parentheses.
Your regular expression is equivalent to $b(a(\varepsilon + ba^*b)bb)^*$ (I don't know if it can be simplified again).