\(\int \frac {1}{x (1-\log (x))} \, dx\) [22]

Optimal result

Integrand size = 12, antiderivative size = 9 \[ \int \frac {1}{x (1-\log (x))} \, dx=-\log (1-\log (x)) \]

[Out]

-ln(1-ln(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2339, 29} \[ \int \frac {1}{x (1-\log (x))} \, dx=-\log (1-\log (x)) \]

[In]

Int[1/(x*(1 - Log[x])),x]

[Out]

-Log[1 - Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{x} \, dx,x,1-\log (x)\right ) \\ & = -\log (1-\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1-\log (x))} \, dx=-\log (-1+\log (x)) \]

[In]

Integrate[1/(x*(1 - Log[x])),x]

[Out]

-Log[-1 + Log[x]]

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89

method	result	size
norman	\(-\ln \left (-1+\ln \left (x \right )\right )\)	\(8\)
risch	\(-\ln \left (-1+\ln \left (x \right )\right )\)	\(8\)
parallelrisch	\(-\ln \left (-1+\ln \left (x \right )\right )\)	\(8\)
derivativedivides	\(-\ln \left (1-\ln \left (x \right )\right )\)	\(10\)
default	\(-\ln \left (1-\ln \left (x \right )\right )\)	\(10\)

[In]

int(1/x/(1-ln(x)),x,method=_RETURNVERBOSE)

[Out]

-ln(-1+ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1-\log (x))} \, dx=-\log \left (\log \left (x\right ) - 1\right ) \]

[In]

integrate(1/x/(1-log(x)),x, algorithm="fricas")

[Out]

-log(log(x) - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1-\log (x))} \, dx=- \log {\left (\log {\left (x \right )} - 1 \right )} \]

[In]

integrate(1/x/(1-ln(x)),x)

[Out]

-log(log(x) - 1)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1-\log (x))} \, dx=-\log \left (\log \left (x\right ) - 1\right ) \]

[In]

integrate(1/x/(1-log(x)),x, algorithm="maxima")

[Out]

-log(log(x) - 1)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 22 vs. \(2 (9) = 18\).

Time = 0.33 (sec) , antiderivative size = 22, normalized size of antiderivative = 2.44 \[ \int \frac {1}{x (1-\log (x))} \, dx=-\frac {1}{2} \, \log \left (\frac {1}{4} \, \pi ^{2} {\left (\mathrm {sgn}\left (x\right ) - 1\right )}^{2} + {\left (\log \left ({\left | x \right |}\right ) - 1\right )}^{2}\right ) \]

[In]

integrate(1/x/(1-log(x)),x, algorithm="giac")

[Out]

-1/2*log(1/4*pi^2*(sgn(x) - 1)^2 + (log(abs(x)) - 1)^2)

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.78 \[ \int \frac {1}{x (1-\log (x))} \, dx=-\ln \left (\ln \left (x\right )-1\right ) \]

[In]

int(-1/(x*(log(x) - 1)),x)

[Out]

-log(log(x) - 1)