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

Optimal result

Integrand size = 11, antiderivative size = 17 \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\log (11)+\log \left (\frac {2}{5} \left (x+\log \left (\frac {\log (5)}{4}\right )\right )\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {31} \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\log \left (x+\log \left (\frac {\log (5)}{4}\right )\right ) \]

[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{align*} \text {integral}& = \log \left (x+\log \left (\frac {\log (5)}{4}\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.59 \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\log \left (x+\log \left (\frac {\log (5)}{4}\right )\right ) \]

[In]

Integrate[(x + Log[Log[5]/4])^(-1),x]

[Out]

Log[x + Log[Log[5]/4]]

Maple [A] (verified)

Time = 1.63 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.53

method	result	size
default	\(\ln \left (\ln \left (\frac {\ln \left (5\right )}{4}\right )+x \right )\)	\(9\)
norman	\(\ln \left (\ln \left (\frac {\ln \left (5\right )}{4}\right )+x \right )\)	\(9\)
parallelrisch	\(\ln \left (\ln \left (\frac {\ln \left (5\right )}{4}\right )+x \right )\)	\(9\)
risch	\(\ln \left (-2 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+x \right )\)	\(11\)
meijerg	\(\frac {2 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right ) \ln \left (1+\frac {x}{2 \ln \left (2\right ) \ln \left (\ln \left (5\right )\right )}\right )}{\ln \left (\frac {\ln \left (5\right )}{4}\right )}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\log \left (x + \log \left (\frac {1}{4} \, \log \left (5\right )\right )\right ) \]

[In]

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

[Out]

log(x + log(1/4*log(5)))

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\log {\left (x - 2 \log {\left (2 \right )} + \log {\left (\log {\left (5 \right )} \right )} \right )} \]

[In]

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

[Out]

log(x - 2*log(2) + log(log(5)))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\log \left (x + \log \left (\frac {1}{4} \, \log \left (5\right )\right )\right ) \]

[In]

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

[Out]

log(x + log(1/4*log(5)))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\log \left ({\left | x + \log \left (\frac {1}{4} \, \log \left (5\right )\right ) \right |}\right ) \]

[In]

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

[Out]

log(abs(x + log(1/4*log(5))))

Mupad [B] (verification not implemented)

Time = 10.53 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x+\log \left (\frac {\log (5)}{4}\right )} \, dx=\ln \left (x+\ln \left (\frac {\ln \left (5\right )}{4}\right )\right ) \]

[In]

int(1/(x + log(log(5)/4)),x)

[Out]

log(x + log(log(5)/4))