\(\int \frac {1}{8 \log (\log (\log (6)))} \, dx\) [1851]

Optimal result

Integrand size = 10, antiderivative size = 13 \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=6+\frac {x}{8 \log (\log (\log (6)))} \]

[Out]

1/8*x/ln(ln(ln(6)))+6

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {8} \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=\frac {x}{8 \log (\log (\log (6)))} \]

[In]

Int[1/(8*Log[Log[Log[6]]]),x]

[Out]

x/(8*Log[Log[Log[6]]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps \begin{align*} \text {integral}& = \frac {x}{8 \log (\log (\log (6)))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=\frac {x}{8 \log (\log (\log (6)))} \]

[In]

Integrate[1/(8*Log[Log[Log[6]]]),x]

[Out]

x/(8*Log[Log[Log[6]]])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.77

method	result	size
default	\(\frac {x}{8 \ln \left (\ln \left (\ln \left (6\right )\right )\right )}\)	\(10\)
norman	\(\frac {x}{8 \ln \left (\ln \left (\ln \left (6\right )\right )\right )}\)	\(10\)
parallelrisch	\(\frac {x}{8 \ln \left (\ln \left (\ln \left (6\right )\right )\right )}\)	\(10\)
risch	\(\frac {x}{8 \ln \left (\ln \left (\ln \left (2\right )+\ln \left (3\right )\right )\right )}\)	\(13\)

[In]

int(1/8/ln(ln(ln(6))),x,method=_RETURNVERBOSE)

[Out]

1/8*x/ln(ln(ln(6)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=\frac {x}{8 \, \log \left (\log \left (\log \left (6\right )\right )\right )} \]

[In]

integrate(1/8/log(log(log(6))),x, algorithm="fricas")

[Out]

1/8*x/log(log(log(6)))

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=\frac {x}{8 \log {\left (\log {\left (\log {\left (6 \right )} \right )} \right )}} \]

[In]

integrate(1/8/ln(ln(ln(6))),x)

[Out]

x/(8*log(log(log(6))))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=\frac {x}{8 \, \log \left (\log \left (\log \left (6\right )\right )\right )} \]

[In]

integrate(1/8/log(log(log(6))),x, algorithm="maxima")

[Out]

1/8*x/log(log(log(6)))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=\frac {x}{8 \, \log \left (\log \left (\log \left (6\right )\right )\right )} \]

[In]

integrate(1/8/log(log(log(6))),x, algorithm="giac")

[Out]

1/8*x/log(log(log(6)))

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.69 \[ \int \frac {1}{8 \log (\log (\log (6)))} \, dx=\frac {x}{8\,\ln \left (\ln \left (\ln \left (6\right )\right )\right )} \]

[In]

int(1/(8*log(log(log(6)))),x)

[Out]

x/(8*log(log(log(6))))