3.22 \(\int \frac {1}{x (1-\log (x))} \, dx\)

Optimal. Leaf size=9 \[ -\log (1-\log (x)) \]

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Rubi [A]  time = 0.02, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2302, 29} \[ -\log (1-\log (x)) \]

Antiderivative was successfully verified.

[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 2302

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*} \int \frac {1}{x (1-\log (x))} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,1-\log (x)\right )\\ &=-\log (1-\log (x))\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 7, normalized size = 0.78 \[ -\log (\log (x)-1) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(1 - Log[x])),x]

[Out]

-Log[-1 + Log[x]]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x (1-\log (x))} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[1/(x*(1 - Log[x])),x]

[Out]

Could not integrate

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fricas [A]  time = 1.03, size = 7, normalized size = 0.78 \[ -\log \left (\log \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(x) - 1)

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giac [B]  time = 1.00, size = 22, normalized size = 2.44 \[ -\frac {1}{2} \, \log \left (\frac {1}{4} \, \pi ^{2} {\left (\mathrm {sgn}\relax (x) - 1\right )}^{2} + {\left (\log \left ({\left | x \right |}\right ) - 1\right )}^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.02, size = 8, normalized size = 0.89

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(-1+ln(x))

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maxima [A]  time = 0.44, size = 7, normalized size = 0.78 \[ -\log \left (\log \relax (x) - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(x) - 1)

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mupad [B]  time = 0.21, size = 7, normalized size = 0.78 \[ -\ln \left (\ln \relax (x)-1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x*(log(x) - 1)),x)

[Out]

-log(log(x) - 1)

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sympy [A]  time = 0.10, size = 7, normalized size = 0.78 \[ - \log {\left (\log {\relax (x )} - 1 \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(log(x) - 1)

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