3.19.94 \(\int \frac {e^{-x} (e^x (1+x)+e^{5 e^{-x} x} (-5 x+5 x^2))}{x} \, dx\)

Optimal. Leaf size=19 \[ -e^{5 e^{-x} x}+\log \left (e^x x\right ) \]

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Rubi [F]  time = 0.39, 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^x (1+x)+e^{5 e^{-x} x} \left (-5 x+5 x^2\right )\right )}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(1 + x) + E^((5*x)/E^x)*(-5*x + 5*x^2))/(E^x*x),x]

[Out]

x + Log[x] - 5*Defer[Int][E^(-(((-5 + E^x)*x)/E^x)), x] + 5*Defer[Int][x/E^(((-5 + E^x)*x)/E^x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1+5 e^{-e^{-x} \left (-5+e^x\right ) x} (-1+x)+\frac {1}{x}\right ) \, dx\\ &=x+\log (x)+5 \int e^{-e^{-x} \left (-5+e^x\right ) x} (-1+x) \, dx\\ &=x+\log (x)+5 \int \left (-e^{-e^{-x} \left (-5+e^x\right ) x}+e^{-e^{-x} \left (-5+e^x\right ) x} x\right ) \, dx\\ &=x+\log (x)-5 \int e^{-e^{-x} \left (-5+e^x\right ) x} \, dx+5 \int e^{-e^{-x} \left (-5+e^x\right ) x} x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.09, size = 16, normalized size = 0.84 \begin {gather*} -e^{5 e^{-x} x}+x+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(1 + x) + E^((5*x)/E^x)*(-5*x + 5*x^2))/(E^x*x),x]

[Out]

-E^((5*x)/E^x) + x + Log[x]

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fricas [A]  time = 0.95, size = 14, normalized size = 0.74 \begin {gather*} x - e^{\left (5 \, x e^{\left (-x\right )}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-5*x)*exp(5*x/exp(x))+(x+1)*exp(x))/exp(x)/x,x, algorithm="fricas")

[Out]

x - e^(5*x*e^(-x)) + log(x)

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giac [A]  time = 0.23, size = 31, normalized size = 1.63 \begin {gather*} {\left (x e^{\left (-x\right )} + e^{\left (-x\right )} \log \relax (x) - e^{\left (5 \, x e^{\left (-x\right )} - x\right )}\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-5*x)*exp(5*x/exp(x))+(x+1)*exp(x))/exp(x)/x,x, algorithm="giac")

[Out]

(x*e^(-x) + e^(-x)*log(x) - e^(5*x*e^(-x) - x))*e^x

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maple [A]  time = 0.04, size = 15, normalized size = 0.79

method	result	size
risch	\(x +\ln \relax (x )-{\mathrm e}^{5 x \,{\mathrm e}^{-x}}\)	\(15\)
norman	\(\left ({\mathrm e}^{x} x -{\mathrm e}^{x} {\mathrm e}^{5 x \,{\mathrm e}^{-x}}\right ) {\mathrm e}^{-x}+\ln \relax (x )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x^2-5*x)*exp(5*x/exp(x))+(x+1)*exp(x))/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

x+ln(x)-exp(5*x*exp(-x))

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maxima [A]  time = 0.64, size = 14, normalized size = 0.74 \begin {gather*} x - e^{\left (5 \, x e^{\left (-x\right )}\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x^2-5*x)*exp(5*x/exp(x))+(x+1)*exp(x))/exp(x)/x,x, algorithm="maxima")

[Out]

x - e^(5*x*e^(-x)) + log(x)

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mupad [B]  time = 1.16, size = 14, normalized size = 0.74 \begin {gather*} x-{\mathrm {e}}^{5\,x\,{\mathrm {e}}^{-x}}+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*(exp(x)*(x + 1) - exp(5*x*exp(-x))*(5*x - 5*x^2)))/x,x)

[Out]

x - exp(5*x*exp(-x)) + log(x)

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sympy [A]  time = 0.17, size = 12, normalized size = 0.63 \begin {gather*} x - e^{5 x e^{- x}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*x**2-5*x)*exp(5*x/exp(x))+(x+1)*exp(x))/exp(x)/x,x)

[Out]

x - exp(5*x*exp(-x)) + log(x)

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