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I'll assume the answer to the second question in my comment is 'no'. My answer is perhaps a bit too long to stick in a comment.

It's certainly not free for every group homomorphism $\phi: \mathbb{Z} \to Aut(F_\infty)$; for example, if $\phi$ is trivial, then the semidirect product would be the cartesian product $F_\infty \times \mathbb{Z}$. This cannot be a free group, because every subgroup of a free group is free, but $F_\infty \times \mathbb{Z}$ contains a subgroup of the form $\mathbb{Z} \times \mathbb{Z}$ which is not free.

On the other hand, there are plenty of examples of $\phi$ where it is free. Take for example any surjective group homomorphism $F_\infty \to \mathbb{Z}$, and let $K$ be the kernel. Then $K$, being a subgroup of a free group, is free. Moreover, the exact sequence

$$1 \to K \to F_\infty \to \mathbb{Z} \to 1$$

splits, and this implies $F_\infty$ is a semidirect product of $\mathbb{Z}$ with $K$ in some way (see for example Wikipedia). Notice also $K$ cannot be finitely generated (if it were, then so would be the semidirect product), so $K$ must be a countably generated free group, and we are done.

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I'll assume the answer to the second question in my comment is 'no'. My answer is perhaps a bit too long to stick in a comment.

It's certainly not free for every group homomorphism $\phi: \mathbb{Z} \to Aut(F_\infty)$; for example, if $\phi$ is trivial, then the semidirect product would be the cartesian product $F_\infty \times \mathbb{Z}$. This cannot be a free group, because every subgroup of a free group is free, but $F_\infty \times \mathbb{Z}$ contains a subgroup of the form $\mathbb{Z} \times \mathbb{Z}$ which is not free.

On the other hand, there are plenty of examples of $\phi$ where it is free. Take for example any surjective group homomorphism $F_\infty \to \mathbb{Z}$, and let $K$ be the kernel. Then $K$, being a subgroup of a free group, is free. Moreover, the exact sequence

$$1 \to K \to F_\infty \to \mathbb{Z} \to 1$$

splits, and this implies $F_\infty$ is a semidirect product of $\mathbb{Z}$ with $K$ in some way (see for example Wikipedia). Notice also $K$ cannot be finitely generated (if it were, then so would be the semidirect product), so $K$ must be a countably generated free group, and we are done.