\(\int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx\) [7878]

Optimal result

Integrand size = 14, antiderivative size = 15 \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {1}{4} (-5+\log (-1+x-\log (\log (4)))) \]

[Out]

1/4*ln(x-1-ln(2*ln(2)))-5/4

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {31} \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {1}{4} \log (-x+1+\log (\log (4))) \]

[In]

Int[-(4 - 4*x + 4*Log[Log[4]])^(-1),x]

[Out]

Log[1 - x + Log[Log[4]]]/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}& = \frac {1}{4} \log (1-x+\log (\log (4))) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {1}{4} \log (4-4 x+4 \log (\log (4))) \]

[In]

Integrate[-(4 - 4*x + 4*Log[Log[4]])^(-1),x]

[Out]

Log[4 - 4*x + 4*Log[Log[4]]]/4

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\ln \left (\ln \left (2 \ln \left (2\right )\right )-x +1\right )}{4}\)	\(14\)
parallelrisch	\(\frac {\ln \left (x -1-\ln \left (2 \ln \left (2\right )\right )\right )}{4}\)	\(14\)
norman	\(\frac {\ln \left (4 \ln \left (2 \ln \left (2\right )\right )-4 x +4\right )}{4}\)	\(16\)
risch	\(\frac {\ln \left (-\ln \left (2\right )-\ln \left (\ln \left (2\right )\right )+x -1\right )}{4}\)	\(16\)
meijerg	\(-\frac {\left (-\ln \left (2\right )-\ln \left (\ln \left (2\right )\right )-1\right ) \ln \left (1-\frac {x}{\ln \left (2\right )+\ln \left (\ln \left (2\right )\right )+1}\right )}{4 \ln \left (2 \ln \left (2\right )\right )+4}\)	\(40\)

[In]

int(-1/(4*ln(2*ln(2))-4*x+4),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(ln(2*ln(2))-x+1)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {1}{4} \, \log \left (x - \log \left (2 \, \log \left (2\right )\right ) - 1\right ) \]

[In]

integrate(-1/(4*log(2*log(2))-4*x+4),x, algorithm="fricas")

[Out]

1/4*log(x - log(2*log(2)) - 1)

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27 \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {\log {\left (4 x - 4 - 4 \log {\left (2 \right )} - 4 \log {\left (\log {\left (2 \right )} \right )} \right )}}{4} \]

[In]

integrate(-1/(4*ln(2*ln(2))-4*x+4),x)

[Out]

log(4*x - 4 - 4*log(2) - 4*log(log(2)))/4

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {1}{4} \, \log \left (x - \log \left (2 \, \log \left (2\right )\right ) - 1\right ) \]

[In]

integrate(-1/(4*log(2*log(2))-4*x+4),x, algorithm="maxima")

[Out]

1/4*log(x - log(2*log(2)) - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {1}{4} \, \log \left ({\left | x - \log \left (2 \, \log \left (2\right )\right ) - 1 \right |}\right ) \]

[In]

integrate(-1/(4*log(2*log(2))-4*x+4),x, algorithm="giac")

[Out]

1/4*log(abs(x - log(2*log(2)) - 1))

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int -\frac {1}{4-4 x+4 \log (\log (4))} \, dx=\frac {\ln \left (x-\ln \left (2\,\ln \left (2\right )\right )-1\right )}{4} \]

[In]

int(-1/(4*log(2*log(2)) - 4*x + 4),x)

[Out]

log(x - log(2*log(2)) - 1)/4