Here's a simple Haskell program to do it:
(EDIT: this code is completely wrongheaded and does not work; it assumes that when sequencing regexes, you can take the product of their sizes to find the overall size. This is just not true. See reply, below, for an example.)
-- https://gist.github.com/rntz/03604e36888a8c6f08bb5e8c665ba9d0
import qualified Data.List as List
data Regex = Class [Char] -- character class
| Seq [Regex] -- sequence, ABC
| Choice [Regex] -- choice, A|B|C
| Star Regex -- zero or more, A*
deriving (Show)
data Size = Finite Int | Infinite deriving (Show, Eq)
instance Num Size where
abs = undefined; signum = undefined; negate = undefined -- unnecessary
fromInteger = Finite . fromInteger
Finite x + Finite y = Finite (x + y)
_ + _ = Infinite
Finite x * Finite y = Finite (x * y)
x * y = if x == 0 || y == 0 then 0 else Infinite
-- computes size & language (list of matching strings, if regex is finite)
eval :: Regex -> (Size, [String])
eval (Class chars) = (Finite (length cset), [[c] | c sequence langs)
where (sizes, langs) = unzip $ map eval regexes
eval (Choice regexes) = (size, lang)
where (sizes, langs) = unzip $ map eval regexes
lang = concat langs
size = if elem Infinite sizes then Infinite
-- finite, so just count 'em. inefficient but works.
else Finite (length (List.nub lang))
eval (Star r) = (size, lang)
where (rsize, rlang) = eval r
size | rsize == 0 = 1
| rsize == 1 && List.nub rlang == [""] = 1
| otherwise = Infinite
lang = [""] ++ ((++) [x | x lang)
size :: Regex -> Size
size = fst . eval
NB. Besides the utter wrong-headedness of the `product` call, the generated string-sets may not be exhaustive for infinite languages, and the original version (I have since edited it) was wrong in several cases for Star (if the argument was nullable or empty).