smc

Since we want non-congruent triangles that are neither isosceles nor equilateral, we can just list side lengths $(a,b,c)$ with $a<b<c$. Furthermore, "positive area" tells us that $c<a+b$ and the perimeter constraints means $a+b+c<15$. There are no triangles when $a=1$ because then $c$ must be less than $b+1$, implying that $b\ge c$, contrary to $b<c$. When $a=2$, similar to above, $c$ must be less than $b+2$, so this leaves the only possibility $c=b+1$. This gives 3 triangles $(2,3,4),(2,4,5),(2,5,6)$ within our perimeter constraint. When $a=3$, $c$ can be $b+1$ or $b+2$, which gives triangles $(3,4,5),(3,4,6),(3,5,6)$. Note that $(3,4,5)$ is a right triangle, so we get rid of it and we get only 2 triangles. All in all, this gives us $3+2=\boxed{5}$ triangles.