I have usually found in books/lectures that the Dirac theory, given by
$$S=\int d^4x\bar\psi(i\gamma^\mu\partial_\mu-m)\psi, $$ is invariant under $U(1)$ global transformations (which is evident) but not invariant under $U(1)$ local (gauge) transformations, due to an extra term appearing in the Lagrangian of the form
Nevertheless, if one integrates this Lagrangian to obtain the new action, you find when integrating by parts that this extra term vanishes.
$$\int d^4x (\partial_\mu\theta)\bar\psi\gamma^\mu\psi=-\int d^4x \theta \partial_\mu(\bar\psi\gamma^\mu\psi)=0,$$ since $$\partial_\mu(\bar\psi\gamma^\mu\psi)=\partial_\mu\bar\psi\gamma^\mu\psi-im\bar\psi\psi=i(\partial_\mu\bar\psi\gamma^\mu\psi+m\bar\psi\psi)=0,$$ where I used the Dirac equation both for $\psi$ and for $\bar\psi$. So one concludes $S=S’$. Is then the Dirac theory invariant to gauge transformation with no need of introducing the covariant derivative?