Find the area between $f(x) = x^2+3x+7 $ and $g(x) = xe^{x^3+4}$ for $x in [3,5]$.

Calculate the area between the two functions, $f(x)$, $g(x)$, for $x in [3,5]$.
$$f(x)=x^2+3x+7$$
$$g(x)=xe^{x^3+4}$$

To determine the area between the functions I used the formula $A= int_a^b|f(x)-g(x)|dx$. Therefore, I have:

begin{align}
A&=int_3^5|x^2+3x+7-xe^{x^3+4}|dx \
&= int_3^5|x^2|dx+int_3^5|3x|dx+int_3^5|7|dx-int_3^5|e^{x^3+4}|dx\
&= left(frac{x^3}{3}+frac{3x^2}{2}+7xright)_3^5 – int_3^5|xe^{x^3+4}|dx
end{align}

Here I can use $u$-subsitution and the Incomplete Gamma Function to find the integral of $xe^{x^3+4}$. All in all I get:

begin{align}
A&=left(frac{x^3}{3}+frac{3x^2}{2}+7x-e^4frac{1}{2}left(-frac{x^2Γleft(frac{1}{frac{3}{2}},:-left(x^2right)^{frac{3}{2}}right)}{frac{3}{2}sqrt[frac{3}{2}]{-left(x^2right)^{frac{3}{2}}}}right)right)_3^5\
&=-frac{98}{3}-24-14+e^4frac{1}{2}left(-frac{25Γleft(frac{1}{frac{3}{2}},:-25^{frac{3}{2}}right)}{frac{3}{2}sqrt[frac{3}{2}]{-25^{frac{3}{2}}}}-left(-frac{9Γleft(frac{1}{frac{3}{2}},:-9^{frac{3}{2}}right)}{frac{3}{2}sqrt[frac{3}{2}]{-9^{frac{3}{2}}}}right)right)
end{align}

Can this be simplified further or is the given solution enough?