Arora and Barak states (p. 230) the following:
What is the relation between $BQP$ and $NP$? It seems that quantum computers only offer a quadratic speedup (using Grover’s search) on $NP$-complete problems. There are also oracle results showing that $NP$ problems require exponential time on quantum computers [BBBV97]. So most researchers believe that $NP \subsetneq BPP$. On the other hand, there is a problem in $BQP$ (the Recursive Fourier Sampling or RFS problem [BV93]) that is not known to be in the polynomial-hierarchy, let alone in $NP$. Thus it seems that BQP and NP may be incomparable classes.
My question is: what is different between two classes are 'incomparable' or two classes are 'not equal'?
I knew that it is possible that $P \neq NP$, but $P \subsetneq NP$, so classes P and NP are comparable. Also, I am aware that when two classes have different type then they are incomparable, e.g. one is decision problems (NP) and another is counting problems (#P), then they are not comparable. But BQP and NP are both decision class of problems, so it is not clear to me why Arora and Barak said that both classes could be incomparable. For example, why not saying they are 'not equal' instead of 'incomparable'?