Yes. ```heurisch(*)``` function does not take bounded variables. So this makes sense.
```
In[0]: heurisch(x**n,x)

Out[0]:
    n
 x⋅x 
─────
n + 1
In [1]: heurisch(x**n*y,x)

Out[1]:
   n  
x⋅x ⋅y
──────
n + 1 
In [2]: heurisch(x**n*y,x, y)

Out[2]:
   n  
x⋅x ⋅y
──────
n + 1 
In[3]: heurisch(x**n*y, y)

Out[3]:
 n  2
x ⋅y 
─────
  2
``