Simpifying boolean algebra

Question

I need some help simplifying a boolean algebra expression. $(~a*b*c*~d) + (a*~b*~c*~d) + (a*~b*c*~d) +(a*b*~c*d) + (a*b*c*~d)$ I have managed to simplify to $(~c*~d)*(~a+~b)+(a*b)*(~c*~d)+(a*~b*c*~d)$ but after this step, I manage to get a couple of xor gates which I don't understand how to use in order to simplify any further without getting several wrong answers.
I have tried the different results but they don't match the truth table. I would appreciate any help. Thank You

Answer 1

$$a'bcd+ab'c'd+ab'cd'+abcd+abcd'$$ $$=(a'+a)bcd+(c'+c)ab'd+abcd'$$ $$=bcd+ab'd+abcd'$$ $$=bc(d+ad')+ab'd'$$ $$=bc(a+d)+ab'd'$$ $$=abc+bcd+ab'd'$$