A minimal solution is:
* 1 and 6 cross the bridge; 1 comes back (7)
* 1 and 3 cross the bridge, 1 comes back (4)
* 8 and 9 cross the bridge; 3 comes back (12)
* 1 and 3 cross the bridge, 1 comes back (4)
* 1 and 4 cross the bridge (4)
Total is $7+4+12+4+4 = 31$. There are minor variations that are still minimal. For example, 3 can return in step 2 instead of in step 3.
I have nothing useful to say about how to find the solution; I programmed the computer to do an exhaustive search. I only posted this because the other guy was getting upvotes for his wrong answer.
By request, [my search program is available ]