3.56.57 \(\int \frac {1}{5} (12-i \pi -\log (-\log (\log (2)))) \, dx\) [5557]

Optimal. Leaf size=31 \[ 2 x+\frac {1}{5} (-5+2 x-(x+\log (2)) (i \pi +\log (-\log (\log (2))))) \]

[Out]

12/5*x-1/5*(ln(2)+x)*ln(ln(ln(2)))-1

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Rubi [A]
time = 0.00, antiderivative size = 20, normalized size of antiderivative = 0.65, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {8} \begin {gather*} \frac {1}{5} x (12-i \pi -\log (-\log (\log (2)))) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 - I*Pi - Log[-Log[Log[2]]])/5,x]

[Out]

(x*(12 - I*Pi - Log[-Log[Log[2]]]))/5

Rule 8

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} x (12-i \pi -\log (-\log (\log (2))))\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.00, size = 25, normalized size = 0.81 \begin {gather*} \frac {12 x}{5}-\frac {i \pi x}{5}-\frac {1}{5} x \log (-\log (\log (2))) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 - I*Pi - Log[-Log[Log[2]]])/5,x]

[Out]

(12*x)/5 - (I/5)*Pi*x - (x*Log[-Log[Log[2]]])/5

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Maple [A]
time = 0.18, size = 11, normalized size = 0.35

method	result	size
default	\(x \left (-\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )}{5}+\frac {12}{5}\right )\)	\(11\)
norman	\(x \left (-\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )}{5}+\frac {12}{5}\right )\)	\(11\)
risch	\(-\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right ) x}{5}+\frac {12 x}{5}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/5*ln(ln(ln(2)))+12/5,x,method=_RETURNVERBOSE)

[Out]

x*(-1/5*ln(ln(ln(2)))+12/5)

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Maxima [A]
time = 0.28, size = 9, normalized size = 0.29 \begin {gather*} -\frac {1}{5} \, x {\left (\log \left (\log \left (\log \left (2\right )\right )\right ) - 12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/5*log(log(log(2)))+12/5,x, algorithm="maxima")

[Out]

-1/5*x*(log(log(log(2))) - 12)

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Fricas [A]
time = 0.34, size = 11, normalized size = 0.35 \begin {gather*} -\frac {1}{5} \, x \log \left (\log \left (\log \left (2\right )\right )\right ) + \frac {12}{5} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/5*log(log(log(2)))+12/5,x, algorithm="fricas")

[Out]

-1/5*x*log(log(log(2))) + 12/5*x

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Sympy [A]
time = 0.01, size = 19, normalized size = 0.61 \begin {gather*} x \left (- \frac {\log {\left (- \log {\left (\log {\left (2 \right )} \right )} \right )}}{5} + \frac {12}{5} - \frac {i \pi }{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/5*ln(ln(ln(2)))+12/5,x)

[Out]

x*(-log(-log(log(2)))/5 + 12/5 - I*pi/5)

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Giac [A]
time = 0.40, size = 9, normalized size = 0.29 \begin {gather*} -\frac {1}{5} \, x {\left (\log \left (\log \left (\log \left (2\right )\right )\right ) - 12\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/5*log(log(log(2)))+12/5,x, algorithm="giac")

[Out]

-1/5*x*(log(log(log(2))) - 12)

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Mupad [B]
time = 0.00, size = 11, normalized size = 0.35 \begin {gather*} -x\,\left (\frac {\ln \left (\ln \left (\ln \left (2\right )\right )\right )}{5}-\frac {12}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(12/5 - log(log(log(2)))/5,x)

[Out]

-x*(log(log(log(2)))/5 - 12/5)

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