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

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

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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} $$\begin{array}{c}{e}^{-e^2}\log (x)-e^{-e^2}x\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =e^{-e^2}\int \frac{1-x}{x}dx\\ & =e^{-e^2}\int (-1+\frac{1}{x})dx\\ & =-e^{-e^2}x+e^{-e^2}\log (x)\end{array}\end{array}$$

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

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 $$\begin{array}{c}-(x-\log (x))e^{(-e^2)}\end{array}$$

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 $$\begin{array}{c}-(x-\log \left(\left|x\right|\right))e^{(-e^2)}\end{array}$$

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	${\mathrm{e}}^{-{\mathrm{e}}^2}(\ln (x)-x)$	$13$
norman	$-x{\mathrm{e}}^{-{\mathrm{e}}^2}+{\mathrm{e}}^{-{\mathrm{e}}^2}\ln (x)$	$18$
risch	$-x{\mathrm{e}}^{-{\mathrm{e}}^2}+{\mathrm{e}}^{-{\mathrm{e}}^2}\ln (x)$	$18$

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 $$\begin{array}{c}-(x-\log (x))e^{(-e^2)}\end{array}$$

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 $$\begin{array}{c}-{\mathrm{e}}^{-{\mathrm{e}}^2}(x-\ln (x))\end{array}$$

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 $$\begin{array}{c}\frac{-x+\log (x)}{e^{e^2}}\end{array}$$

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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