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

Optimal. Leaf size=28 \[ e^{e^{-2+x}-e^x-x} \log \left (\frac {e^{-x} x}{2}\right ) \]

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Rubi [B]  time = 0.20, antiderivative size = 60, normalized size of antiderivative = 2.14, number of steps used = 1, number of rules used = 1, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {2288} \begin {gather*} \frac {e^{-x+e^{x-2}-e^x} \left (-e^{x-2} x+e^x x+x\right ) \log \left (\frac {e^{-x} x}{2}\right )}{\left (-e^{x-2}+e^x+1\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(E^(-2 + x) - E^x - x)*(x - E^(-2 + x)*x + E^x*x)*Log[x/(2*E^x)])/((1 - E^(-2 + x) + E^x)*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^{e^{-2+x}-e^x-x} \left (x-e^{-2+x} x+e^x x\right ) \log \left (\frac {e^{-x} x}{2}\right )}{\left (1-e^{-2+x}+e^x\right ) x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 28, normalized size = 1.00 \begin {gather*} e^{e^{-2+x}-e^x-x} \log \left (\frac {e^{-x} x}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 30, normalized size = 1.07 \begin {gather*} e^{\left (-{\left (x e^{2} + {\left (e^{2} - 1\right )} e^{x} - e^{2} \log \left (\log \left (\frac {1}{2} \, x e^{\left (-x\right )}\right )\right )\right )} e^{\left (-2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.43, size = 272, normalized size = 9.71

method	result	size
risch	\(\frac {2 i \left (-\ln \relax (2)+\ln \relax (x )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )}{2}+\frac {i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}{2}+\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}{2}-\frac {i \pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}}{2}\right ) \left (-\ln \relax (2)+\ln \relax (x )-\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x}\right )\right )}{2}\right ) {\mathrm e}^{-{\mathrm e}^{x}+{\mathrm e}^{x -2}-x}}{2 i \ln \relax (x )-2 i \ln \left ({\mathrm e}^{x}\right )+\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )+\pi \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{3}-2 i \ln \relax (2)-\pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}-\pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (i x \,{\mathrm e}^{-x}\right )^{2}}\)	\(272\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*(-ln(2)+ln(x)-ln(exp(x))-1/2*I*Pi*csgn(I*x)*csgn(I*exp(-x))*csgn(I*x*exp(-x))+1/2*I*Pi*csgn(I*x)*csgn(I*x*
exp(-x))^2+1/2*I*Pi*csgn(I*exp(-x))*csgn(I*x*exp(-x))^2-1/2*I*Pi*csgn(I*x*exp(-x))^3)/(2*I*ln(x)-2*I*ln(exp(x)
)+Pi*csgn(I*x)*csgn(I*exp(-x))*csgn(I*x*exp(-x))+Pi*csgn(I*x*exp(-x))^3-2*I*ln(2)-Pi*csgn(I*x)*csgn(I*x*exp(-x
))^2-Pi*csgn(I*exp(-x))*csgn(I*x*exp(-x))^2)*(-ln(2)+ln(x)-ln(exp(x))-1/2*I*Pi*csgn(I*x*exp(-x))*(-csgn(I*x*ex
p(-x))+csgn(I*x))*(-csgn(I*x*exp(-x))+csgn(I*exp(-x))))*exp(-exp(x)+exp(x-2)-x)

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maxima [A]  time = 0.70, size = 23, normalized size = 0.82 \begin {gather*} -{\left (x + \log \relax (2) - \log \relax (x)\right )} e^{\left (-x + e^{\left (x - 2\right )} - e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} -\int \frac {{\mathrm {e}}^{{\mathrm {e}}^{x-2}-x-{\mathrm {e}}^x+\ln \left (\ln \left (\frac {x\,{\mathrm {e}}^{-x}}{2}\right )\right )}\,\left (x+\ln \left (\frac {x\,{\mathrm {e}}^{-x}}{2}\right )\,\left (x-x\,{\mathrm {e}}^{x-2}+x\,{\mathrm {e}}^x\right )-1\right )}{x\,\ln \left (\frac {x\,{\mathrm {e}}^{-x}}{2}\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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