3.47.16 \(\int \frac {e^{-x} (e^{e^{e^{-x} x}} (-12 e^x x^2+e^{e^{-x} x} (-4 x^3+4 x^4))+e^x (16 x^3+4 \log (9)))}{\log (9)} \, dx\)

Optimal. Leaf size=28 \[ 4 \left (x-\frac {\left (e^{e^{e^{-x} x}}-x\right ) x^3}{\log (9)}\right ) \]

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Rubi [F]  time = 1.77, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

4*x + (4*x^4)/Log[9] - (12*Defer[Int][E^E^(x/E^x)*x^2, x])/Log[9] - (4*Defer[Int][E^(E^(x/E^x) - x + x/E^x)*x^
3, x])/Log[9] + (4*Defer[Int][E^(E^(x/E^x) - x + x/E^x)*x^4, x])/Log[9]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int 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 ) \, dx}{\log (9)}\\ &=\frac {\int \left (4 e^{e^{e^{-x} x}-x} x^2 \left (-3 e^x-e^{e^{-x} x} x+e^{e^{-x} x} x^2\right )+4 \left (4 x^3+\log (9)\right )\right ) \, dx}{\log (9)}\\ &=\frac {4 \int e^{e^{e^{-x} x}-x} x^2 \left (-3 e^x-e^{e^{-x} x} x+e^{e^{-x} x} x^2\right ) \, dx}{\log (9)}+\frac {4 \int \left (4 x^3+\log (9)\right ) \, dx}{\log (9)}\\ &=4 x+\frac {4 x^4}{\log (9)}+\frac {4 \int e^{e^{e^{-x} x}-x} x^2 \left (-3 e^x+e^{e^{-x} x} (-1+x) x\right ) \, dx}{\log (9)}\\ &=4 x+\frac {4 x^4}{\log (9)}+\frac {4 \int \left (-3 e^{e^{e^{-x} x}} x^2+e^{e^{e^{-x} x}-x+e^{-x} x} (-1+x) x^3\right ) \, dx}{\log (9)}\\ &=4 x+\frac {4 x^4}{\log (9)}+\frac {4 \int e^{e^{e^{-x} x}-x+e^{-x} x} (-1+x) x^3 \, dx}{\log (9)}-\frac {12 \int e^{e^{e^{-x} x}} x^2 \, dx}{\log (9)}\\ &=4 x+\frac {4 x^4}{\log (9)}+\frac {4 \int \left (-e^{e^{e^{-x} x}-x+e^{-x} x} x^3+e^{e^{e^{-x} x}-x+e^{-x} x} x^4\right ) \, dx}{\log (9)}-\frac {12 \int e^{e^{e^{-x} x}} x^2 \, dx}{\log (9)}\\ &=4 x+\frac {4 x^4}{\log (9)}-\frac {4 \int e^{e^{e^{-x} x}-x+e^{-x} x} x^3 \, dx}{\log (9)}+\frac {4 \int e^{e^{e^{-x} x}-x+e^{-x} x} x^4 \, dx}{\log (9)}-\frac {12 \int e^{e^{e^{-x} x}} x^2 \, dx}{\log (9)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.80, size = 32, normalized size = 1.14 \begin {gather*} \frac {-4 e^{e^{e^{-x} x}} x^3+4 x^4+4 x \log (9)}{\log (9)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(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]

[Out]

(-4*E^E^(x/E^x)*x^3 + 4*x^4 + 4*x*Log[9])/Log[9]

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fricas [A]  time = 0.79, size = 28, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (x^{4} - x^{3} e^{\left (e^{\left (x e^{\left (-x\right )}\right )}\right )} + 2 \, x \log \relax (3)\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

2*(x^4 - x^3*e^(e^(x*e^(-x))) + 2*x*log(3))/log(3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {2 \, {\left (2 \, {\left (2 \, x^{3} + \log \relax (3)\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 \relax (3)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

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maple [A]  time = 0.05, size = 31, normalized size = 1.11

method	result	size
risch	\(\frac {2 x^{4}}{\ln \relax (3)}-\frac {2 x^{3} {\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{-x}}}}{\ln \relax (3)}+4 x\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

2/ln(3)*x^4-2/ln(3)*x^3*exp(exp(x*exp(-x)))+4*x

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maxima [A]  time = 0.41, size = 28, normalized size = 1.00 \begin {gather*} \frac {2 \, {\left (x^{4} - x^{3} e^{\left (e^{\left (x e^{\left (-x\right )}\right )}\right )} + 2 \, x \log \relax (3)\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

2*(x^4 - x^3*e^(e^(x*e^(-x))) + 2*x*log(3))/log(3)

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mupad [B]  time = 3.41, size = 28, normalized size = 1.00 \begin {gather*} \frac {2\,x\,\left (2\,\ln \relax (3)-x^2\,{\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}+x^3\right )}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

(2*x*(2*log(3) - x^2*exp(exp(x*exp(-x))) + x^3))/log(3)

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sympy [A]  time = 3.32, size = 27, normalized size = 0.96 \begin {gather*} \frac {2 x^{4}}{\log {\relax (3 )}} - \frac {2 x^{3} e^{e^{x e^{- x}}}}{\log {\relax (3 )}} + 4 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)

[Out]

2*x**4/log(3) - 2*x**3*exp(exp(x*exp(-x)))/log(3) + 4*x

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