Entropy of a probability distribution (symbolically)

Timing can be improved with simplification with assumptions:

exp = Expand[
   FullSimplify[-Log[PDF[NormalDistribution[m, s], x]], 
    Assumptions -> {{m, x, s} ∈ Reals, s > 0}]];
Integrate[
  exp PDF[NormalDistribution[m, s], x], {x, -∞, ∞}, 
  Assumptions -> {{m, x, s} ∈ Reals, s > 0}] // Timing

yields:

{2.09375, 1/2 (1 + Log[2 π s^2])}

See comment by @gwr below.

Expectation 

yields better performance on simplified expression (as 'expected')...too bad I didn't think to use it :(

Let $X \sim N(\mu, \sigma^2)$ with pdf $f(x)$:

Then, the way we solve this in Chapter 1 of Mathematical Statistics with Mathematica (free download of book available here) is to find $E[-log(f)]$:

... which uses the mathStatica Expect function, and takes about 6 seconds to evaluate on my Mac in 11.2.

I was surprised at the OP's suggestion that:

Expectation[Log@PDF[NormalDistribution[m, s], x], 
Distributed[x, NormalDistribution[m, s]]] // AbsoluteTiming

... does not evaluate. When I tried it, it does evaluate, but it takes about 160 seconds to return the equivalent:

{161.93, 1/2 (-1 - Log[2 [Pi] s^2])}

... on the same computer in v11.1.1 and about 80 seconds under 11.2.