3.17.99 \(\int \frac {1}{x \log (\frac {1}{5 x})} \, dx\) [1699]

Optimal. Leaf size=19 \[ \frac {1}{25 e^4}-\log \left (\log \left (\frac {1}{5 x}\right )\right ) \]

[Out]

1/25/exp(2)^2-ln(ln(1/5/x))

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Rubi [A]
time = 0.01, antiderivative size = 11, normalized size of antiderivative = 0.58, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2339, 29} \begin {gather*} -\log \left (\log \left (\frac {1}{5 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x*Log[1/(5*x)]),x]

[Out]

-Log[Log[1/(5*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 {gather*} \begin {aligned} \text {integral} &=-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {1}{5 x}\right )\right )\\ &=-\log \left (\log \left (\frac {1}{5 x}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 11, normalized size = 0.58 \begin {gather*} -\log \left (\log \left (\frac {1}{5 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x*Log[1/(5*x)]),x]

[Out]

-Log[Log[1/(5*x)]]

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Maple [A]
time = 0.05, size = 10, normalized size = 0.53

method result size
derivativedivides \(-\ln \left (\ln \left (\frac {1}{5 x}\right )\right )\) \(10\)
default \(-\ln \left (\ln \left (\frac {1}{5 x}\right )\right )\) \(10\)
norman \(-\ln \left (\ln \left (\frac {1}{5 x}\right )\right )\) \(10\)
risch \(-\ln \left (\ln \left (\frac {1}{5 x}\right )\right )\) \(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/ln(1/5/x),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(1/5/x))

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Maxima [A]
time = 0.36, size = 9, normalized size = 0.47 \begin {gather*} -\log \left (\log \left (\frac {1}{5 \, x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/log(1/5/x),x, algorithm="maxima")

[Out]

-log(log(1/5/x))

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Fricas [A]
time = 0.34, size = 9, normalized size = 0.47 \begin {gather*} -\log \left (\log \left (\frac {1}{5 \, x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/log(1/5/x),x, algorithm="fricas")

[Out]

-log(log(1/5/x))

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Sympy [A]
time = 0.03, size = 8, normalized size = 0.42 \begin {gather*} - \log {\left (\log {\left (\frac {1}{5 x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/ln(1/5/x),x)

[Out]

-log(log(1/(5*x)))

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Giac [A]
time = 0.39, size = 8, normalized size = 0.42 \begin {gather*} -\log \left ({\left | \log \left (5 \, x\right ) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/log(1/5/x),x, algorithm="giac")

[Out]

-log(abs(log(5*x)))

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Mupad [B]
time = 1.14, size = 9, normalized size = 0.47 \begin {gather*} -\ln \left (\ln \left (\frac {1}{5\,x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*log(1/(5*x))),x)

[Out]

-log(log(1/(5*x)))

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