$\newcommand{\Kstar}[1]{#1^\ast}$ I would like to write the regular expression $$ a^\ast b^\ast + ( a + b )^\ast b a ( a + b )^\ast $$ as a Kleene closure of a union of catenations or in other words, a Kleene closure of something akin to a disjunctive normal form of regular expressions. I didn't immediately see what operations I would have to use to get there, so I started by applying the regular language function $K$ to the regex to see where it would take me: \begin{align*} &K(a^\ast b^\ast + ( a + b )^\ast b a ( a + b )^\ast ) \\ &= K(a^\ast b^\ast) \cup K( ( a + b )^\ast b a ( a + b )^\ast ) \\ &= K(a^\ast)K(b^\ast) \cup K(( a + b )^\ast) K( b a ) K( ( a + b )^\ast ) \\ &= K(a)^\ast K(b)^\ast \cup K( a + b )^\ast K( b a ) K( a + b )^\ast \\ &= K(a)^\ast K(b)^\ast \cup ( K(a) \cup K(b) )^\ast K( b a ) ( K(a) \cup K(b) )^\ast \\ &= K(a)^\ast K(b)^\ast \cup ( K(a)^\ast \cup K(b)^\ast )^\ast K( b a ) ( K(a)^\ast \cup K(b)^\ast )^\ast \,. \end{align*}
However, this does not look very promising, at least not at the moment. If I understood the question correctly, I would need to be able to somehow get a single Kleene closure out of the entire expression, but that does not seem possible.
Does anyone have any pointers towards possible calculation rules that might simplify this further and make the final form easier to spot?
Edit
It looks like there exists a rule that says the following: for any regular expressions $s$ and $r$, $(s + r)^\ast \equiv (s^\ast r^\ast)^\ast$, as in the two regular expressions are semantically equivalent. This already results in \begin{align} a^\ast b^\ast + ( a + b )^\ast b a ( a + b )^\ast\equiv a^\ast b^\ast + ( a^\ast b^\ast )^\ast b a ( a^\ast b^\ast )^\ast \end{align} and our regular expression is in disjunctive normal form. However, this is still ways away from only a single Kleene closure being present in the expression.