3.56.16 \(\int \frac {e^{-3+e^{\frac {-e^5 x+e^3 (9 x^2+9 x^3) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 (9 x^2+9 x^3) \log (2)}{9 e^3 \log (2)}} (-e^5 x+e^3 (9+18 x^2+27 x^3) \log (2))}{9 \log (2)} \, dx\)

Optimal. Leaf size=24 \[ e^{e^{x \left (x+x^2-\frac {e^2}{9 \log (2)}\right )} x} \]

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Rubi [F]  time = 4.15, 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 {\exp \left (-3+\exp \left (\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}\right ) x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}\right ) \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right )}{9 \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-3 + E^((-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2]))*x + (-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Lo
g[2])/(9*E^3*Log[2]))*(-(E^5*x) + E^3*(9 + 18*x^2 + 27*x^3)*Log[2]))/(9*Log[2]),x]

[Out]

(Log[512]*Defer[Int][E^(E^(x*(x + x^2 - E^2/Log[512]))*x + x^2 + x^3 - (E^2*x)/Log[512]), x])/(9*Log[2]) - Def
er[Int][E^(2 + E^(x*(x + x^2 - E^2/Log[512]))*x + x^2 + x^3 - (E^2*x)/Log[512])*x, x]/(9*Log[2]) + (Log[4]*Def
er[Int][E^(E^(x*(x + x^2 - E^2/Log[512]))*x + x^2 + x^3 - (E^2*x)/Log[512])*x^2, x])/Log[2] + (Log[8]*Defer[In
t][E^(E^(x*(x + x^2 - E^2/Log[512]))*x + x^2 + x^3 - (E^2*x)/Log[512])*x^3, x])/Log[2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \exp \left (-3+\exp \left (\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}\right ) x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}\right ) \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right ) \, dx}{9 \log (2)}\\ &=\frac {\int \exp \left (e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) \left (-e^2 x+9 x^2 \log (4)+9 x^3 \log (8)+\log (512)\right ) \, dx}{9 \log (2)}\\ &=\frac {\int \left (-\exp \left (2+e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) x+9 \exp \left (e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) x^2 \log (4)+9 \exp \left (e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) x^3 \log (8)+\exp \left (e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) \log (512)\right ) \, dx}{9 \log (2)}\\ &=-\frac {\int \exp \left (2+e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) x \, dx}{9 \log (2)}+\frac {\log (4) \int \exp \left (e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) x^2 \, dx}{\log (2)}+\frac {\log (8) \int \exp \left (e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) x^3 \, dx}{\log (2)}+\frac {\log (512) \int \exp \left (e^{x \left (x+x^2-\frac {e^2}{\log (512)}\right )} x+x^2+x^3-\frac {e^2 x}{\log (512)}\right ) \, dx}{9 \log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 4.95, size = 114, normalized size = 4.75 \begin {gather*} \frac {\int e^{-3+e^{\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} x+\frac {-e^5 x+e^3 \left (9 x^2+9 x^3\right ) \log (2)}{9 e^3 \log (2)}} \left (-e^5 x+e^3 \left (9+18 x^2+27 x^3\right ) \log (2)\right ) \, dx}{9 \log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 + E^((-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2]))*x + (-(E^5*x) + E^3*(9*x^2 + 9*x
^3)*Log[2])/(9*E^3*Log[2]))*(-(E^5*x) + E^3*(9 + 18*x^2 + 27*x^3)*Log[2]))/(9*Log[2]),x]

[Out]

Integrate[E^(-3 + E^((-(E^5*x) + E^3*(9*x^2 + 9*x^3)*Log[2])/(9*E^3*Log[2]))*x + (-(E^5*x) + E^3*(9*x^2 + 9*x^
3)*Log[2])/(9*E^3*Log[2]))*(-(E^5*x) + E^3*(9 + 18*x^2 + 27*x^3)*Log[2]), x]/(9*Log[2])

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fricas [B]  time = 1.06, size = 77, normalized size = 3.21 \begin {gather*} e^{\left (\frac {9 \, x e^{\left (-\frac {x e^{2} - 9 \, {\left (x^{3} + x^{2}\right )} \log \relax (2)}{9 \, \log \relax (2)}\right )} \log \relax (2) - x e^{2} + 9 \, {\left (x^{3} + x^{2} - 3\right )} \log \relax (2)}{9 \, \log \relax (2)} + \frac {x e^{2} - 9 \, {\left (x^{3} + x^{2}\right )} \log \relax (2)}{9 \, \log \relax (2)} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((27*x^3+18*x^2+9)*exp(3)*log(2)-x*exp(5))*exp(1/9*((9*x^3+9*x^2)*exp(3)*log(2)-x*exp(5))/exp(3)
/log(2))*exp(x*exp(1/9*((9*x^3+9*x^2)*exp(3)*log(2)-x*exp(5))/exp(3)/log(2)))/exp(3)/log(2),x, algorithm="fric
as")

[Out]

e^(1/9*(9*x*e^(-1/9*(x*e^2 - 9*(x^3 + x^2)*log(2))/log(2))*log(2) - x*e^2 + 9*(x^3 + x^2 - 3)*log(2))/log(2) +
1/9*(x*e^2 - 9*(x^3 + x^2)*log(2))/log(2) + 3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (9 \, {\left (3 \, x^{3} + 2 \, x^{2} + 1\right )} e^{3} \log \relax (2) - x e^{5}\right )} e^{\left (x e^{\left (\frac {{\left (9 \, {\left (x^{3} + x^{2}\right )} e^{3} \log \relax (2) - x e^{5}\right )} e^{\left (-3\right )}}{9 \, \log \relax (2)}\right )} + \frac {{\left (9 \, {\left (x^{3} + x^{2}\right )} e^{3} \log \relax (2) - x e^{5}\right )} e^{\left (-3\right )}}{9 \, \log \relax (2)} - 3\right )}}{9 \, \log \relax (2)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((27*x^3+18*x^2+9)*exp(3)*log(2)-x*exp(5))*exp(1/9*((9*x^3+9*x^2)*exp(3)*log(2)-x*exp(5))/exp(3)
/log(2))*exp(x*exp(1/9*((9*x^3+9*x^2)*exp(3)*log(2)-x*exp(5))/exp(3)/log(2)))/exp(3)/log(2),x, algorithm="giac
")

[Out]

integrate(1/9*(9*(3*x^3 + 2*x^2 + 1)*e^3*log(2) - x*e^5)*e^(x*e^(1/9*(9*(x^3 + x^2)*e^3*log(2) - x*e^5)*e^(-3)
/log(2)) + 1/9*(9*(x^3 + x^2)*e^3*log(2) - x*e^5)*e^(-3)/log(2) - 3)/log(2), x)

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maple [A]  time = 0.15, size = 35, normalized size = 1.46

method	result	size
risch	\({\mathrm e}^{x \,{\mathrm e}^{\frac {x \left (9 \,{\mathrm e}^{3} \ln \relax (2) x^{2}+9 \,{\mathrm e}^{3} \ln \relax (2) x -{\mathrm e}^{5}\right ) {\mathrm e}^{-3}}{9 \ln \relax (2)}}}\)	\(35\)
norman	\({\mathrm e}^{x \,{\mathrm e}^{\frac {\left (\left (9 x^{3}+9 x^{2}\right ) {\mathrm e}^{3} \ln \relax (2)-x \,{\mathrm e}^{5}\right ) {\mathrm e}^{-3}}{9 \ln \relax (2)}}}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((27*x^3+18*x^2+9)*exp(3)*ln(2)-x*exp(5))*exp(1/9*((9*x^3+9*x^2)*exp(3)*ln(2)-x*exp(5))/exp(3)/ln(2))*
exp(x*exp(1/9*((9*x^3+9*x^2)*exp(3)*ln(2)-x*exp(5))/exp(3)/ln(2)))/exp(3)/ln(2),x,method=_RETURNVERBOSE)

[Out]

exp(x*exp(1/9*x*(9*exp(3)*ln(2)*x^2+9*exp(3)*ln(2)*x-exp(5))*exp(-3)/ln(2)))

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maxima [A]  time = 0.68, size = 20, normalized size = 0.83 \begin {gather*} e^{\left (x e^{\left (x^{3} + x^{2} - \frac {x e^{2}}{9 \, \log \relax (2)}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((27*x^3+18*x^2+9)*exp(3)*log(2)-x*exp(5))*exp(1/9*((9*x^3+9*x^2)*exp(3)*log(2)-x*exp(5))/exp(3)
/log(2))*exp(x*exp(1/9*((9*x^3+9*x^2)*exp(3)*log(2)-x*exp(5))/exp(3)/log(2)))/exp(3)/log(2),x, algorithm="maxi
ma")

[Out]

e^(x*e^(x^3 + x^2 - 1/9*x*e^2/log(2)))

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mupad [B]  time = 3.90, size = 21, normalized size = 0.88 \begin {gather*} {\mathrm {e}}^{x\,{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^2}{9\,\ln \relax (2)}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-3)*exp(-(exp(-3)*((x*exp(5))/9 - (exp(3)*log(2)*(9*x^2 + 9*x^3))/9))/log(2))*exp(x*exp(-(exp(-3)*((
x*exp(5))/9 - (exp(3)*log(2)*(9*x^2 + 9*x^3))/9))/log(2)))*(x*exp(5) - exp(3)*log(2)*(18*x^2 + 27*x^3 + 9)))/(
9*log(2)),x)

[Out]

exp(x*exp(x^2)*exp(x^3)*exp(-(x*exp(2))/(9*log(2))))

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sympy [A]  time = 0.54, size = 36, normalized size = 1.50 \begin {gather*} e^{x e^{\frac {- \frac {x e^{5}}{9} + \frac {\left (9 x^{3} + 9 x^{2}\right ) e^{3} \log {\relax (2 )}}{9}}{e^{3} \log {\relax (2 )}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((27*x**3+18*x**2+9)*exp(3)*ln(2)-x*exp(5))*exp(1/9*((9*x**3+9*x**2)*exp(3)*ln(2)-x*exp(5))/exp(
3)/ln(2))*exp(x*exp(1/9*((9*x**3+9*x**2)*exp(3)*ln(2)-x*exp(5))/exp(3)/ln(2)))/exp(3)/ln(2),x)

[Out]

exp(x*exp((-x*exp(5)/9 + (9*x**3 + 9*x**2)*exp(3)*log(2)/9)*exp(-3)/log(2)))

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