Integrand size = 33, antiderivative size = 183 \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx=\frac {2 a^2 e^{(a+b x)^3}}{b^5}-\frac {a^4 (a+b x) \Gamma \left (\frac {1}{3},-(a+b x)^3\right )}{3 b^5 \sqrt [3]{-(a+b x)^3}}+\frac {4 a^3 (a+b x)^2 \Gamma \left (\frac {2}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{2/3}}+\frac {4 a (a+b x)^4 \Gamma \left (\frac {4}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{4/3}}-\frac {(a+b x)^5 \Gamma \left (\frac {5}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{5/3}} \]
2*a^2*exp((b*x+a)^3)/b^5-1/3*a^4*(b*x+a)*GAMMA(1/3,-(b*x+a)^3)/b^5/(-(b*x+ a)^3)^(1/3)+4/3*a^3*(b*x+a)^2*GAMMA(2/3,-(b*x+a)^3)/b^5/(-(b*x+a)^3)^(2/3) +4/3*a*(b*x+a)^4*GAMMA(4/3,-(b*x+a)^3)/b^5/(-(b*x+a)^3)^(4/3)-1/3*(b*x+a)^ 5*GAMMA(5/3,-(b*x+a)^3)/b^5/(-(b*x+a)^3)^(5/3)
Time = 0.40 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.90 \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx=\frac {6 a^2 e^{(a+b x)^3} \left (-(a+b x)^3\right )^{2/3}-a^4 (a+b x) \sqrt [3]{-(a+b x)^3} \Gamma \left (\frac {1}{3},-(a+b x)^3\right )+4 a^3 (a+b x)^2 \Gamma \left (\frac {2}{3},-(a+b x)^3\right )-4 a (a+b x) \sqrt [3]{-(a+b x)^3} \Gamma \left (\frac {4}{3},-(a+b x)^3\right )+(a+b x)^2 \Gamma \left (\frac {5}{3},-(a+b x)^3\right )}{3 b^5 \left (-(a+b x)^3\right )^{2/3}} \]
(6*a^2*E^(a + b*x)^3*(-(a + b*x)^3)^(2/3) - a^4*(a + b*x)*(-(a + b*x)^3)^( 1/3)*Gamma[1/3, -(a + b*x)^3] + 4*a^3*(a + b*x)^2*Gamma[2/3, -(a + b*x)^3] - 4*a*(a + b*x)*(-(a + b*x)^3)^(1/3)*Gamma[4/3, -(a + b*x)^3] + (a + b*x) ^2*Gamma[5/3, -(a + b*x)^3])/(3*b^5*(-(a + b*x)^3)^(2/3))
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^4 e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} \, dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int x^4 e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3}dx\) |
3.3.8.3.1 Defintions of rubi rules used
\[\int {\mathrm e}^{b^{3} x^{3}+3 a \,b^{2} x^{2}+3 a^{2} b x +a^{3}} x^{4}d x\]
Time = 0.08 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.86 \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx=-\frac {2 \, {\left (6 \, a^{3} + 1\right )} \left (-b^{3}\right )^{\frac {1}{3}} b \Gamma \left (\frac {2}{3}, -b^{3} x^{3} - 3 \, a b^{2} x^{2} - 3 \, a^{2} b x - a^{3}\right ) - {\left (3 \, a^{4} + 4 \, a\right )} \left (-b^{3}\right )^{\frac {2}{3}} \Gamma \left (\frac {1}{3}, -b^{3} x^{3} - 3 \, a b^{2} x^{2} - 3 \, a^{2} b x - a^{3}\right ) - 3 \, {\left (b^{4} x^{2} - 2 \, a b^{3} x + 3 \, a^{2} b^{2}\right )} e^{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )}}{9 \, b^{7}} \]
-1/9*(2*(6*a^3 + 1)*(-b^3)^(1/3)*b*gamma(2/3, -b^3*x^3 - 3*a*b^2*x^2 - 3*a ^2*b*x - a^3) - (3*a^4 + 4*a)*(-b^3)^(2/3)*gamma(1/3, -b^3*x^3 - 3*a*b^2*x ^2 - 3*a^2*b*x - a^3) - 3*(b^4*x^2 - 2*a*b^3*x + 3*a^2*b^2)*e^(b^3*x^3 + 3 *a*b^2*x^2 + 3*a^2*b*x + a^3))/b^7
\[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx=e^{a^{3}} \int x^{4} e^{b^{3} x^{3}} e^{3 a b^{2} x^{2}} e^{3 a^{2} b x}\, dx \]
\[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx=\int { x^{4} e^{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \,d x } \]
\[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx=\int { x^{4} e^{\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right )} \,d x } \]
Timed out. \[ \int e^{a^3+3 a^2 b x+3 a b^2 x^2+b^3 x^3} x^4 \, dx=\int x^4\,{\mathrm {e}}^{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3} \,d x \]