Let $AB$ and $CD$ meet at $P$ Then $PBC$ is an equilateral triangle

Possible values are 6,7 and 9.

Why is this so case bashy!! Anyways the only possible values for the sides are: $(EA,BC,CD,DE,BA)=(5,3,2,4,1),(6,2,4,5,3),(5,1,4,3,2),(4,3,1,5,2),(5,2,3,6,4),(3,1,2,5,4)$ And hence $AB+BC+CD \in \{6,9,7\}$ The sketch of my solution: Firstly let the sides be $a-2,a-1,a,a+1,a+2$ in some order, where $a \in N$. The key idea is to extend $EA, ED$, and $BC$. We then get some equilateral triangles (since $\angle E=60^{o}$). Using this, we get the 2 conditions: $EA=BC+CD$ and $DE=BA+BC$. Here, we mark that $EA>BC,CD$ and $DE>BA,BC$ to simplify our work. We now first assume that $EA>DE$ and make 8 (simple) cases. Out of these, only three work. (For instance, a case that would not work would be $EA>DE>BC>BA>CD$. This case implies $(EA,DE,BC,BA,CD)=(a+2,a+1,a,a-1,a-2)$. Now, $EA=BC+CD$ gives us $a=4$, but this does not satisfy $DE=BA+BC$. Hence this case is not valid.) Then for the next part, we assume that $DE>EA$ and simply swap the values: $DE \leftrightarrow EA$ and $BA \leftrightarrow CD$ in the three cases that worked before, to get the six possibilities as mentioned above. Done!