I wasn't able to integrate this function using standard techniques of integration. Using the computer algebra systems of both MATLAB and WolframAlpha I was able to have the computer generate an answer.
F(x)=1/6 (-(3 sqrt(π) "erfi"(1 - x))/e + 2 x^3 + 3 e^((x - 2) x))+C
Where the term "erfi"(1-x) refers to the imaginary error function defined as
"erfi"(1-x)=2/sqrt(π) int_0^(1-x) e^(t^2) dt
At F(0)=1, we have
1/6 (-(3 sqrt(π) "erfi"(1 - 0))/e + 2 (0)^3 + 3 e^((0 - 2) 0))+C=1
1/6 (-(3 sqrt(π) "erfi"(1))/e + 3 e^(0)+C)=1
1/6 (-(3 sqrt(π) "erfi"(1))/e + 3+C)=1
-(3 sqrt(π) "erfi"(1))/e+3+C=6
-(3 sqrt(π) "erfi"(1))/e+C=3
C=3+(3 sqrt(π) "erfi"(1))/e
Plugging the right hand side into MATLAB, gives
C~~6.2285
So the anti-derivative is
F(x)~=1/6 (-(3 sqrt(π) "erfi"(1 - x))/e + 2 x^3 + 3 e^((x - 2) x))+6.2285