Integrand size = 67, antiderivative size = 28 \[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=4 \left (x-\frac {\left (e^{e^{e^{-x} x}}-x\right ) x^3}{\log (9)}\right ) \] Output:
4*x-2*(exp(exp(x/exp(x)))-x)/ln(3)*x^3
Time = 1.82 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14 \[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=\frac {-4 e^{e^{e^{-x} x}} x^3+4 x^4+4 x \log (9)}{\log (9)} \] Input:
Integrate[(E^E^(x/E^x)*(-12*E^x*x^2 + E^(x/E^x)*(-4*x^3 + 4*x^4)) + E^x*(1 6*x^3 + 4*Log[9]))/(E^x*Log[9]),x]
Output:
(-4*E^E^(x/E^x)*x^3 + 4*x^4 + 4*x*Log[9])/Log[9]
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.
\(\displaystyle \int \frac {e^{-x} \left (e^x \left (16 x^3+4 \log (9)\right )+e^{e^{e^{-x} x}} \left (e^{e^{-x} x} \left (4 x^4-4 x^3\right )-12 e^x x^2\right )\right )}{\log (9)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -4 e^{-x} \left (e^{e^{e^{-x} x}} \left (3 e^x x^2+e^{e^{-x} x} \left (x^3-x^4\right )\right )-e^x \left (4 x^3+\log (9)\right )\right )dx}{\log (9)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 \int e^{-x} \left (e^{e^{e^{-x} x}} \left (3 e^x x^2+e^{e^{-x} x} \left (x^3-x^4\right )\right )-e^x \left (4 x^3+\log (9)\right )\right )dx}{\log (9)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4 \int \left (-4 x^3-e^{e^{e^{-x} x}-x} \left (e^{e^{-x} x} x^2-e^{e^{-x} x} x-3 e^x\right ) x^2-\log (9)\right )dx}{\log (9)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \left (-\int e^{e^{-x} x-x+e^{e^{-x} x}} x^4dx+\int e^{e^{-x} x-x+e^{e^{-x} x}} x^3dx+3 \int e^{e^{e^{-x} x}} x^2dx-x^4-x \log (9)\right )}{\log (9)}\) |
Input:
Int[(E^E^(x/E^x)*(-12*E^x*x^2 + E^(x/E^x)*(-4*x^3 + 4*x^4)) + E^x*(16*x^3 + 4*Log[9]))/(E^x*Log[9]),x]
Output:
$Aborted
Time = 2.39 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11
method | result | size |
risch | \(\frac {2 x^{4}}{\ln \left (3\right )}-\frac {2 x^{3} {\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{-x}}}}{\ln \left (3\right )}+4 x\) | \(31\) |
parallelrisch | \(\frac {4 x^{4}-4 x^{3} {\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{-x}}}+8 x \ln \left (3\right )}{2 \ln \left (3\right )}\) | \(31\) |
Input:
int(1/2*(((4*x^4-4*x^3)*exp(x/exp(x))-12*exp(x)*x^2)*exp(exp(x/exp(x)))+(8 *ln(3)+16*x^3)*exp(x))/ln(3)/exp(x),x,method=_RETURNVERBOSE)
Output:
2/ln(3)*x^4-2/ln(3)*x^3*exp(exp(x*exp(-x)))+4*x
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=\frac {2 \, {\left (x^{4} - x^{3} e^{\left (e^{\left (x e^{\left (-x\right )}\right )}\right )} + 2 \, x \log \left (3\right )\right )}}{\log \left (3\right )} \] Input:
integrate(1/2*(((4*x^4-4*x^3)*exp(x/exp(x))-12*exp(x)*x^2)*exp(exp(x/exp(x )))+(8*log(3)+16*x^3)*exp(x))/log(3)/exp(x),x, algorithm="fricas")
Output:
2*(x^4 - x^3*e^(e^(x*e^(-x))) + 2*x*log(3))/log(3)
Time = 1.88 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=\frac {2 x^{4}}{\log {\left (3 \right )}} - \frac {2 x^{3} e^{e^{x e^{- x}}}}{\log {\left (3 \right )}} + 4 x \] Input:
integrate(1/2*(((4*x**4-4*x**3)*exp(x/exp(x))-12*exp(x)*x**2)*exp(exp(x/ex p(x)))+(8*ln(3)+16*x**3)*exp(x))/ln(3)/exp(x),x)
Output:
2*x**4/log(3) - 2*x**3*exp(exp(x*exp(-x)))/log(3) + 4*x
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=\frac {2 \, {\left (x^{4} - x^{3} e^{\left (e^{\left (x e^{\left (-x\right )}\right )}\right )} + 2 \, x \log \left (3\right )\right )}}{\log \left (3\right )} \] Input:
integrate(1/2*(((4*x^4-4*x^3)*exp(x/exp(x))-12*exp(x)*x^2)*exp(exp(x/exp(x )))+(8*log(3)+16*x^3)*exp(x))/log(3)/exp(x),x, algorithm="maxima")
Output:
2*(x^4 - x^3*e^(e^(x*e^(-x))) + 2*x*log(3))/log(3)
\[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=\int { \frac {2 \, {\left (2 \, {\left (2 \, x^{3} + \log \left (3\right )\right )} e^{x} - {\left (3 \, x^{2} e^{x} - {\left (x^{4} - x^{3}\right )} e^{\left (x e^{\left (-x\right )}\right )}\right )} e^{\left (e^{\left (x e^{\left (-x\right )}\right )}\right )}\right )} e^{\left (-x\right )}}{\log \left (3\right )} \,d x } \] Input:
integrate(1/2*(((4*x^4-4*x^3)*exp(x/exp(x))-12*exp(x)*x^2)*exp(exp(x/exp(x )))+(8*log(3)+16*x^3)*exp(x))/log(3)/exp(x),x, algorithm="giac")
Output:
integrate(2*(2*(2*x^3 + log(3))*e^x - (3*x^2*e^x - (x^4 - x^3)*e^(x*e^(-x) ))*e^(e^(x*e^(-x))))*e^(-x)/log(3), x)
Time = 3.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=\frac {2\,x\,\left (2\,\ln \left (3\right )-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}+x^3\right )}{\ln \left (3\right )} \] Input:
int(-(exp(-x)*((exp(exp(x*exp(-x)))*(12*x^2*exp(x) + exp(x*exp(-x))*(4*x^3 - 4*x^4)))/2 - (exp(x)*(8*log(3) + 16*x^3))/2))/log(3),x)
Output:
(2*x*(2*log(3) - x^2*exp(exp(x*exp(-x))) + x^3))/log(3)
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{-x} \left (e^{e^{e^{-x} x}} \left (-12 e^x x^2+e^{e^{-x} x} \left (-4 x^3+4 x^4\right )\right )+e^x \left (16 x^3+4 \log (9)\right )\right )}{\log (9)} \, dx=\frac {2 x \left (-e^{e^{\frac {x}{e^{x}}}} x^{2}+2 \,\mathrm {log}\left (3\right )+x^{3}\right )}{\mathrm {log}\left (3\right )} \] Input:
int(1/2*(((4*x^4-4*x^3)*exp(x/exp(x))-12*exp(x)*x^2)*exp(exp(x/exp(x)))+(8 *log(3)+16*x^3)*exp(x))/log(3)/exp(x),x)
Output:
(2*x*( - e**(e**(x/e**x))*x**2 + 2*log(3) + x**3))/log(3)