3.73.39 \(\int \frac {e^{-e^2} (1-x)}{x} \, dx\)

Optimal. Leaf size=26 \[ -x+x \left (1-\frac {e^{-e^2} \left (x+\log \left (\frac {1}{x}\right )\right )}{x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 0.81, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 43} \begin {gather*} e^{-e^2} \log (x)-e^{-e^2} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x)/(E^E^2*x),x]

[Out]

-(x/E^E^2) + Log[x]/E^E^2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 15, normalized size = 0.58 \begin {gather*} -e^{-e^2} (x-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - x)/(E^E^2*x),x]

[Out]

-((x - Log[x])/E^E^2)

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fricas [A]  time = 0.69, size = 13, normalized size = 0.50 \begin {gather*} -{\left (x - \log \relax (x)\right )} e^{\left (-e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - log(x))*e^(-e^2)

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giac [A]  time = 0.15, size = 14, normalized size = 0.54 \begin {gather*} -{\left (x - \log \left ({\left | x \right |}\right )\right )} e^{\left (-e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - log(abs(x)))*e^(-e^2)

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maple [A]  time = 0.02, size = 13, normalized size = 0.50

method	result	size
default	\({\mathrm e}^{-{\mathrm e}^{2}} \left (\ln \relax (x )-x \right )\)	\(13\)
norman	\(-x \,{\mathrm e}^{-{\mathrm e}^{2}}+{\mathrm e}^{-{\mathrm e}^{2}} \ln \relax (x )\)	\(18\)
risch	\(-x \,{\mathrm e}^{-{\mathrm e}^{2}}+{\mathrm e}^{-{\mathrm e}^{2}} \ln \relax (x )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/exp(exp(2))*(ln(x)-x)

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maxima [A]  time = 0.37, size = 13, normalized size = 0.50 \begin {gather*} -{\left (x - \log \relax (x)\right )} e^{\left (-e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(x - log(x))*e^(-e^2)

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mupad [B]  time = 0.03, size = 13, normalized size = 0.50 \begin {gather*} -{\mathrm {e}}^{-{\mathrm {e}}^2}\,\left (x-\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-exp(2))*(x - 1))/x,x)

[Out]

-exp(-exp(2))*(x - log(x))

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sympy [A]  time = 0.08, size = 8, normalized size = 0.31 \begin {gather*} \frac {- x + \log {\relax (x )}}{e^{e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-x + log(x))*exp(-exp(2))

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