Is DCFL closed with regular always?

Suppose $$L=\{a^mb^n∣m≠n\}∪{(a+b)^∗b(b+a)^*a(a+b)^∗} =\{a^mb^n|mn\} \cup (a+b)^*b(a+b)^*a(a+b)^*$$

It is DCFL ∪ Regular, hence it should be DCFL, but not able to design DPDA, always it designed as NPDA.

Can anybody made DPDA for $$L$$?

New contributor
user19121278 is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.

$$L$$ is equivalent to $$(a+b) ^*-\{a^nb^n\}$$ which is well known DCFL, I mean $$L$$ is the complement of $$\{a^nb^n\}.$$