3.65.71 \(\int \frac {e^{-3-\frac {x}{e^3}} (e^3 (1-x)-2 x+x^2-x \log (x))}{x} \, dx\)

Optimal. Leaf size=16 \[ e^{-\frac {x}{e^3}} (2-x+\log (x)) \]

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Rubi [A]  time = 0.07, antiderivative size = 25, normalized size of antiderivative = 1.56, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2288} \begin {gather*} \frac {e^{-\frac {x}{e^3}} \left (-x^2+2 x+x \log (x)\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x - x^2 + x*Log[x])/(E^(x/E^3)*x)

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.00 \begin {gather*} e^{-\frac {x}{e^3}} (2-x+\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 - x/E^3)*(E^3*(1 - x) - 2*x + x^2 - x*Log[x]))/x,x]

[Out]

(2 - x + Log[x])/E^(x/E^3)

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fricas [B]  time = 0.64, size = 35, normalized size = 2.19 \begin {gather*} -{\left (x - 2\right )} e^{\left (-{\left (x + 3 \, e^{3}\right )} e^{\left (-3\right )} + 3\right )} + e^{\left (-{\left (x + 3 \, e^{3}\right )} e^{\left (-3\right )} + 3\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+(-x+1)*exp(3)+x^2-2*x)/x/exp(3)/exp(x/exp(3)),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 18, normalized size = 1.12

method	result	size
norman	\(\left (2+\ln \relax (x )-x \right ) {\mathrm e}^{-{\mathrm e}^{-3} x}\)	\(18\)
risch	\({\mathrm e}^{-{\mathrm e}^{-3} x} \ln \relax (x )-\left (x -2\right ) {\mathrm e}^{-{\mathrm e}^{-3} x}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2+ln(x)-x)/exp(x/exp(3))

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maxima [B]  time = 0.40, size = 38, normalized size = 2.38 \begin {gather*} -{\left (x + e^{3}\right )} e^{\left (-x e^{\left (-3\right )}\right )} + e^{\left (-x e^{\left (-3\right )}\right )} \log \relax (x) + e^{\left (-x e^{\left (-3\right )} + 3\right )} + 2 \, e^{\left (-x e^{\left (-3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*log(x)+(-x+1)*exp(3)+x^2-2*x)/x/exp(3)/exp(x/exp(3)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.23, size = 14, normalized size = 0.88 \begin {gather*} {\mathrm {e}}^{-x\,{\mathrm {e}}^{-3}}\,\left (\ln \relax (x)-x+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-3)*exp(-x*exp(-3))*(2*x + exp(3)*(x - 1) + x*log(x) - x^2))/x,x)

[Out]

exp(-x*exp(-3))*(log(x) - x + 2)

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sympy [A]  time = 0.35, size = 12, normalized size = 0.75 \begin {gather*} \left (- x + \log {\relax (x )} + 2\right ) e^{- \frac {x}{e^{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*ln(x)+(-x+1)*exp(3)+x**2-2*x)/x/exp(3)/exp(x/exp(3)),x)

[Out]

(-x + log(x) + 2)*exp(-x*exp(-3))

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