3.82.55 \(\int \frac {e^5-\log (10 \log (4))}{5 \log (6)} \, dx\)

Optimal. Leaf size=22 \[ 1+\frac {x \left (e^5-\log (10 \log (4))\right )}{5 \log (6)} \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 0.91, 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 {x \left (e^5-\log (10 \log (4))\right )}{5 \log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5 - Log[10*Log[4]])/(5*Log[6]),x]

[Out]

(x*(E^5 - Log[10*Log[4]]))/(5*Log[6])

Rule 8

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {x \left (e^5-\log (10 \log (4))\right )}{5 \log (6)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 27, normalized size = 1.23 \begin {gather*} \frac {e^5 x}{5 \log (6)}-\frac {x \log (10 \log (4))}{5 \log (6)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5 - Log[10*Log[4]])/(5*Log[6]),x]

[Out]

(E^5*x)/(5*Log[6]) - (x*Log[10*Log[4]])/(5*Log[6])

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fricas [A]  time = 0.70, size = 19, normalized size = 0.86 \begin {gather*} \frac {x e^{5} - x \log \left (20 \, \log \relax (2)\right )}{5 \, \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-log(20*log(2))+exp(5))/log(6),x, algorithm="fricas")

[Out]

1/5*(x*e^5 - x*log(20*log(2)))/log(6)

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giac [A]  time = 0.17, size = 17, normalized size = 0.77 \begin {gather*} \frac {x {\left (e^{5} - \log \left (20 \, \log \relax (2)\right )\right )}}{5 \, \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-log(20*log(2))+exp(5))/log(6),x, algorithm="giac")

[Out]

1/5*x*(e^5 - log(20*log(2)))/log(6)

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maple [A]  time = 0.03, size = 18, normalized size = 0.82

method	result	size
default	\(\frac {x \left (-\ln \left (20 \ln \relax (2)\right )+{\mathrm e}^{5}\right )}{5 \ln \relax (6)}\)	\(18\)
norman	\(\frac {\left (-\ln \left (20\right )-\ln \left (\ln \relax (2)\right )+{\mathrm e}^{5}\right ) x}{5 \ln \relax (6)}\)	\(20\)
risch	\(-\frac {2 x \ln \relax (2)}{5 \left (\ln \relax (2)+\ln \relax (3)\right )}-\frac {x \ln \relax (5)}{5 \left (\ln \relax (2)+\ln \relax (3)\right )}-\frac {x \ln \left (\ln \relax (2)\right )}{5 \left (\ln \relax (2)+\ln \relax (3)\right )}+\frac {x \,{\mathrm e}^{5}}{5 \ln \relax (2)+5 \ln \relax (3)}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(-ln(20*ln(2))+exp(5))/ln(6),x,method=_RETURNVERBOSE)

[Out]

1/5*x*(-ln(20*ln(2))+exp(5))/ln(6)

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maxima [A]  time = 0.38, size = 17, normalized size = 0.77 \begin {gather*} \frac {x {\left (e^{5} - \log \left (20 \, \log \relax (2)\right )\right )}}{5 \, \log \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-log(20*log(2))+exp(5))/log(6),x, algorithm="maxima")

[Out]

1/5*x*(e^5 - log(20*log(2)))/log(6)

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mupad [B]  time = 0.00, size = 19, normalized size = 0.86 \begin {gather*} -\frac {x\,\left (\frac {\ln \left (20\,\ln \relax (2)\right )}{5}-\frac {{\mathrm {e}}^5}{5}\right )}{\ln \relax (6)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(20*log(2))/5 - exp(5)/5)/log(6),x)

[Out]

-(x*(log(20*log(2))/5 - exp(5)/5))/log(6)

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sympy [A]  time = 0.06, size = 17, normalized size = 0.77 \begin {gather*} \frac {x \left (- \frac {\log {\left (20 \log {\relax (2 )} \right )}}{5} + \frac {e^{5}}{5}\right )}{\log {\relax (6 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(-ln(20*ln(2))+exp(5))/ln(6),x)

[Out]

x*(-log(20*log(2))/5 + exp(5)/5)/log(6)

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