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

Optimal. Leaf size=15 \[ \frac {1}{4} (-5+\log (-1+x-\log (\log (4)))) \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 13, normalized size of antiderivative = 0.87, number of steps used = 1, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {31} \begin {gather*} \frac {1}{4} \log (-x+1+\log (\log (4))) \end {gather*}

Antiderivative was successfully verified.

[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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \log (1-x+\log (\log (4)))\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 15, normalized size = 1.00 \begin {gather*} \frac {1}{4} \log (4-4 x+4 \log (\log (4))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Log[4 - 4*x + 4*Log[Log[4]]]/4

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Maple [A]
time = 0.84, size = 14, normalized size = 0.93

method result size
default \(\frac {\ln \left (\ln \left (2 \ln \left (2\right )\right )-x +1\right )}{4}\) \(14\)
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 \(-\frac {\left (-\ln \left (2 \ln \left (2\right )\right )-1\right ) \ln \left (1-\frac {x}{\ln \left (2 \ln \left (2\right )\right )+1}\right )}{4 \ln \left (2 \ln \left (2\right )\right )+4}\) \(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{4} \, \log \left (x - \log \left (2 \, \log \left (2\right )\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(x - log(2*log(2)) - 1)

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Fricas [A]
time = 0.36, size = 13, normalized size = 0.87 \begin {gather*} \frac {1}{4} \, \log \left (x - \log \left (2 \, \log \left (2\right )\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*log(x - log(2*log(2)) - 1)

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Sympy [A]
time = 0.04, size = 19, normalized size = 1.27 \begin {gather*} \frac {\log {\left (4 x - 4 - 4 \log {\left (2 \right )} - 4 \log {\left (\log {\left (2 \right )} \right )} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]
time = 0.40, size = 14, normalized size = 0.93 \begin {gather*} \frac {1}{4} \, \log \left ({\left | x - \log \left (2 \, \log \left (2\right )\right ) - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Mupad [B]
time = 7.53, size = 13, normalized size = 0.87 \begin {gather*} \frac {\ln \left (x-\ln \left (2\,\ln \left (2\right )\right )-1\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(4*log(2*log(2)) - 4*x + 4),x)

[Out]

log(x - log(2*log(2)) - 1)/4

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