3.41.65 \(\int \frac {1}{x+\log (\frac {\log (5)}{4})} \, dx\) [4065]

Optimal. Leaf size=17 \[ \log (11)+\log \left (\frac {2}{5} \left (x+\log \left (\frac {\log (5)}{4}\right )\right )\right ) \]

[Out]

ln(11)+ln(2/5*ln(1/4*ln(5))+2/5*x)

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Rubi [A]
time = 0.00, antiderivative size = 10, normalized size of antiderivative = 0.59, number of steps used = 1, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {31} \begin {gather*} \log \left (x+\log \left (\frac {\log (5)}{4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + Log[Log[5]/4])^(-1),x]

[Out]

Log[x + Log[Log[5]/4]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\log \left (x+\log \left (\frac {\log (5)}{4}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 10, normalized size = 0.59 \begin {gather*} \log \left (x+\log \left (\frac {\log (5)}{4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + Log[Log[5]/4])^(-1),x]

[Out]

Log[x + Log[Log[5]/4]]

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Maple [A]
time = 0.53, size = 9, normalized size = 0.53

method	result	size
default	\(\ln \left (\ln \left (\frac {\ln \left (5\right )}{4}\right )+x \right )\)	\(9\)
norman	\(\ln \left (\ln \left (\frac {\ln \left (5\right )}{4}\right )+x \right )\)	\(9\)
risch	\(\ln \left (-2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+x \right )\)	\(11\)
meijerg	\(\ln \left (1+\frac {x}{\ln \left (\frac {\ln \left (5\right )}{4}\right )}\right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(ln(1/4*ln(5))+x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(1/4*ln(5))+x)

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Maxima [A]
time = 0.29, size = 8, normalized size = 0.47 \begin {gather*} \log \left (x + \log \left (\frac {1}{4} \, \log \left (5\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(log(1/4*log(5))+x),x, algorithm="maxima")

[Out]

log(x + log(1/4*log(5)))

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Fricas [A]
time = 0.39, size = 8, normalized size = 0.47 \begin {gather*} \log \left (x + \log \left (\frac {1}{4} \, \log \left (5\right )\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(log(1/4*log(5))+x),x, algorithm="fricas")

[Out]

log(x + log(1/4*log(5)))

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Sympy [A]
time = 0.01, size = 12, normalized size = 0.71 \begin {gather*} \log {\left (x - 2 \log {\left (2 \right )} + \log {\left (\log {\left (5 \right )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(ln(1/4*ln(5))+x),x)

[Out]

log(x - 2*log(2) + log(log(5)))

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Giac [A]
time = 0.39, size = 9, normalized size = 0.53 \begin {gather*} \log \left ({\left | x + \log \left (\frac {1}{4} \, \log \left (5\right )\right ) \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(log(1/4*log(5))+x),x, algorithm="giac")

[Out]

log(abs(x + log(1/4*log(5))))

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Mupad [B]
time = 0.09, size = 8, normalized size = 0.47 \begin {gather*} \ln \left (x+\ln \left (\frac {\ln \left (5\right )}{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x + log(log(5)/4)),x)

[Out]

log(x + log(log(5)/4))

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