3.54.57 \(\int \frac {e^x x \log (x)+e^x x^2 \log ^2(x)+e^{e^{3 e^{-x}}} (-3 e^{3 e^{-x}} x \log (x)+e^x x \log ^2(x))+(-e^{e^{3 e^{-x}}+x}-e^x x) \log (e^{e^{3 e^{-x}}}+x)}{e^{e^{3 e^{-x}}+x} x \log ^2(x)+e^x x^2 \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ x+\frac {\log \left (e^{e^{3 e^{-x}}}+x\right )}{\log (x)} \]

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Rubi [F]  time = 1.36, 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 x \log (x)+e^x x^2 \log ^2(x)+e^{e^{3 e^{-x}}} \left (-3 e^{3 e^{-x}} x \log (x)+e^x x \log ^2(x)\right )+\left (-e^{e^{3 e^{-x}}+x}-e^x x\right ) \log \left (e^{e^{3 e^{-x}}}+x\right )}{e^{e^{3 e^{-x}}+x} x \log ^2(x)+e^x x^2 \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

x + Defer[Int][1/((E^E^(3/E^x) + x)*Log[x]), x] - 3*Defer[Int][E^(E^(3/E^x) + 3/E^x - x)/((E^E^(3/E^x) + x)*Lo
g[x]), x] - Defer[Int][Log[E^E^(3/E^x) + x]/(x*Log[x]^2), x]

Rubi steps

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

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Mathematica [A]  time = 0.70, size = 21, normalized size = 1.00 \begin {gather*} x+\frac {\log \left (e^{e^{3 e^{-x}}}+x\right )}{\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + Log[E^E^(3/E^x) + x]/Log[x]

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fricas [A]  time = 0.77, size = 31, normalized size = 1.48 \begin {gather*} \frac {x \log \relax (x) + \log \left ({\left (x e^{x} + e^{\left (x + e^{\left (3 \, e^{\left (-x\right )}\right )}\right )}\right )} e^{\left (-x\right )}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*exp(exp(3/exp(x)))-exp(x)*x)*log(exp(exp(3/exp(x)))+x)+(-3*x*log(x)*exp(3/exp(x))+x*exp(x)
*log(x)^2)*exp(exp(3/exp(x)))+x^2*exp(x)*log(x)^2+x*exp(x)*log(x))/(x*exp(x)*log(x)^2*exp(exp(3/exp(x)))+x^2*e
xp(x)*log(x)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 31, normalized size = 1.48 \begin {gather*} \frac {x \log \relax (x) + \log \left ({\left (x e^{x} + e^{\left (x + e^{\left (3 \, e^{\left (-x\right )}\right )}\right )}\right )} e^{\left (-x\right )}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*exp(exp(3/exp(x)))-exp(x)*x)*log(exp(exp(3/exp(x)))+x)+(-3*x*log(x)*exp(3/exp(x))+x*exp(x)
*log(x)^2)*exp(exp(3/exp(x)))+x^2*exp(x)*log(x)^2+x*exp(x)*log(x))/(x*exp(x)*log(x)^2*exp(exp(3/exp(x)))+x^2*e
xp(x)*log(x)^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.08, size = 19, normalized size = 0.90

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{3 \,{\mathrm e}^{-x}}}+x \right )}{\ln \relax (x )}+x\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-exp(x)*exp(exp(3/exp(x)))-exp(x)*x)*ln(exp(exp(3/exp(x)))+x)+(-3*x*ln(x)*exp(3/exp(x))+x*exp(x)*ln(x)^2
)*exp(exp(3/exp(x)))+x^2*exp(x)*ln(x)^2+x*exp(x)*ln(x))/(x*exp(x)*ln(x)^2*exp(exp(3/exp(x)))+x^2*exp(x)*ln(x)^
2),x,method=_RETURNVERBOSE)

[Out]

1/ln(x)*ln(exp(exp(3*exp(-x)))+x)+x

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maxima [A]  time = 0.53, size = 21, normalized size = 1.00 \begin {gather*} \frac {x \log \relax (x) + \log \left (x + e^{\left (e^{\left (3 \, e^{\left (-x\right )}\right )}\right )}\right )}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*exp(exp(3/exp(x)))-exp(x)*x)*log(exp(exp(3/exp(x)))+x)+(-3*x*log(x)*exp(3/exp(x))+x*exp(x)
*log(x)^2)*exp(exp(3/exp(x)))+x^2*exp(x)*log(x)^2+x*exp(x)*log(x))/(x*exp(x)*log(x)^2*exp(exp(3/exp(x)))+x^2*e
xp(x)*log(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} -\int \frac {\ln \left (x+{\mathrm {e}}^{{\mathrm {e}}^{3\,{\mathrm {e}}^{-x}}}\right )\,\left ({\mathrm {e}}^{x+{\mathrm {e}}^{3\,{\mathrm {e}}^{-x}}}+x\,{\mathrm {e}}^x\right )+{\mathrm {e}}^{{\mathrm {e}}^{3\,{\mathrm {e}}^{-x}}}\,\left (3\,x\,{\mathrm {e}}^{3\,{\mathrm {e}}^{-x}}\,\ln \relax (x)-x\,{\mathrm {e}}^x\,{\ln \relax (x)}^2\right )-x^2\,{\mathrm {e}}^x\,{\ln \relax (x)}^2-x\,{\mathrm {e}}^x\,\ln \relax (x)}{x^2\,{\mathrm {e}}^x\,{\ln \relax (x)}^2+x\,{\mathrm {e}}^{x+{\mathrm {e}}^{3\,{\mathrm {e}}^{-x}}}\,{\ln \relax (x)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x + exp(exp(3*exp(-x))))*(exp(exp(3*exp(-x)))*exp(x) + x*exp(x)) + exp(exp(3*exp(-x)))*(3*x*exp(3*ex
p(-x))*log(x) - x*exp(x)*log(x)^2) - x^2*exp(x)*log(x)^2 - x*exp(x)*log(x))/(x^2*exp(x)*log(x)^2 + x*exp(exp(3
*exp(-x)))*exp(x)*log(x)^2),x)

[Out]

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

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sympy [A]  time = 12.75, size = 15, normalized size = 0.71 \begin {gather*} x + \frac {\log {\left (x + e^{e^{3 e^{- x}}} \right )}}{\log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-exp(x)*exp(exp(3/exp(x)))-exp(x)*x)*ln(exp(exp(3/exp(x)))+x)+(-3*x*ln(x)*exp(3/exp(x))+x*exp(x)*l
n(x)**2)*exp(exp(3/exp(x)))+x**2*exp(x)*ln(x)**2+x*exp(x)*ln(x))/(x*exp(x)*ln(x)**2*exp(exp(3/exp(x)))+x**2*ex
p(x)*ln(x)**2),x)

[Out]

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

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