3.5.50 \(\int \frac {1}{3} (6-\log (\frac {1+3 \log (4)}{\log (4)})) \, dx\)

Optimal. Leaf size=19 \[ -3+2 x-\frac {1}{3} (-5+x) \log \left (3+\frac {1}{\log (4)}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.84, number of steps used = 1, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {8} \begin {gather*} \frac {1}{3} x \left (6-\log \left (3+\frac {1}{\log (4)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(6 - Log[(1 + 3*Log[4])/Log[4]])/3,x]

[Out]

(x*(6 - Log[3 + Log[4]^(-1)]))/3

Rule 8

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} x \left (6-\log \left (3+\frac {1}{\log (4)}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.11 \begin {gather*} 2 x-\frac {1}{3} x \log \left (\frac {1+3 \log (4)}{\log (4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(6 - Log[(1 + 3*Log[4])/Log[4]])/3,x]

[Out]

2*x - (x*Log[(1 + 3*Log[4])/Log[4]])/3

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fricas [A]  time = 0.82, size = 20, normalized size = 1.05 \begin {gather*} -\frac {1}{3} \, x \log \left (\frac {6 \, \log \relax (2) + 1}{2 \, \log \relax (2)}\right ) + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(1/2*(6*log(2)+1)/log(2))+2,x, algorithm="fricas")

[Out]

-1/3*x*log(1/2*(6*log(2) + 1)/log(2)) + 2*x

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giac [A]  time = 0.31, size = 18, normalized size = 0.95 \begin {gather*} -\frac {1}{3} \, x {\left (\log \left (\frac {6 \, \log \relax (2) + 1}{2 \, \log \relax (2)}\right ) - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(1/2*(6*log(2)+1)/log(2))+2,x, algorithm="giac")

[Out]

-1/3*x*(log(1/2*(6*log(2) + 1)/log(2)) - 6)

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maple [A]  time = 0.02, size = 20, normalized size = 1.05

method	result	size
default	\(\left (-\frac {\ln \left (\frac {6 \ln \relax (2)+1}{2 \ln \relax (2)}\right )}{3}+2\right ) x\)	\(20\)
norman	\(\left (\frac {\ln \relax (2)}{3}-\frac {\ln \left (6 \ln \relax (2)+1\right )}{3}+\frac {\ln \left (\ln \relax (2)\right )}{3}+2\right ) x\)	\(23\)
risch	\(\frac {x \ln \relax (2)}{3}-\frac {x \ln \left (6 \ln \relax (2)+1\right )}{3}+\frac {x \ln \left (\ln \relax (2)\right )}{3}+2 x\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/3*ln(1/2*(6*ln(2)+1)/ln(2))+2,x,method=_RETURNVERBOSE)

[Out]

(-1/3*ln(1/2*(6*ln(2)+1)/ln(2))+2)*x

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maxima [A]  time = 0.71, size = 18, normalized size = 0.95 \begin {gather*} -\frac {1}{3} \, x {\left (\log \left (\frac {6 \, \log \relax (2) + 1}{2 \, \log \relax (2)}\right ) - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*log(1/2*(6*log(2)+1)/log(2))+2,x, algorithm="maxima")

[Out]

-1/3*x*(log(1/2*(6*log(2) + 1)/log(2)) - 6)

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mupad [B]  time = 0.00, size = 19, normalized size = 1.00 \begin {gather*} -x\,\left (\frac {\ln \left (\frac {3\,\ln \relax (2)+\frac {1}{2}}{\ln \relax (2)}\right )}{3}-2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2 - log((3*log(2) + 1/2)/log(2))/3,x)

[Out]

-x*(log((3*log(2) + 1/2)/log(2))/3 - 2)

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sympy [A]  time = 0.05, size = 17, normalized size = 0.89 \begin {gather*} x \left (2 - \frac {\log {\left (\frac {\frac {1}{2} + 3 \log {\relax (2 )}}{\log {\relax (2 )}} \right )}}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-1/3*ln(1/2*(6*ln(2)+1)/ln(2))+2,x)

[Out]

x*(2 - log((1/2 + 3*log(2))/log(2))/3)

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