3.17.73 \(\int \frac {-9+5 \log (\frac {1}{4} (6+e^4-\log (2)))}{\log (\frac {1}{4} (6+e^4-\log (2)))} \, dx\)

Optimal. Leaf size=22 \[ x \left (5-\frac {9}{\log \left (\frac {1}{4} \left (6+e^4-\log (2)\right )\right )}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {8} \begin {gather*} x \left (5-\frac {9}{\log \left (\frac {1}{4} \left (6+e^4-\log (2)\right )\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9 + 5*Log[(6 + E^4 - Log[2])/4])/Log[(6 + E^4 - Log[2])/4],x]

[Out]

x*(5 - 9/Log[(6 + E^4 - Log[2])/4])

Rule 8

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x \left (5-\frac {9}{\log \left (\frac {1}{4} \left (6+e^4-\log (2)\right )\right )}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 1.05 \begin {gather*} 5 x-\frac {9 x}{\log \left (\frac {1}{4} \left (6+e^4-\log (2)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9 + 5*Log[(6 + E^4 - Log[2])/4])/Log[(6 + E^4 - Log[2])/4],x]

[Out]

5*x - (9*x)/Log[(6 + E^4 - Log[2])/4]

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fricas [A]  time = 0.71, size = 32, normalized size = 1.45 \begin {gather*} \frac {5 \, x \log \left (\frac {1}{4} \, e^{4} - \frac {1}{4} \, \log \relax (2) + \frac {3}{2}\right ) - 9 \, x}{\log \left (\frac {1}{4} \, e^{4} - \frac {1}{4} \, \log \relax (2) + \frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(-1/4*log(2)+1/4*exp(4)+3/2)-9)/log(-1/4*log(2)+1/4*exp(4)+3/2),x, algorithm="fricas")

[Out]

(5*x*log(1/4*e^4 - 1/4*log(2) + 3/2) - 9*x)/log(1/4*e^4 - 1/4*log(2) + 3/2)

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giac [A]  time = 0.16, size = 30, normalized size = 1.36 \begin {gather*} \frac {x {\left (5 \, \log \left (\frac {1}{4} \, e^{4} - \frac {1}{4} \, \log \relax (2) + \frac {3}{2}\right ) - 9\right )}}{\log \left (\frac {1}{4} \, e^{4} - \frac {1}{4} \, \log \relax (2) + \frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(-1/4*log(2)+1/4*exp(4)+3/2)-9)/log(-1/4*log(2)+1/4*exp(4)+3/2),x, algorithm="giac")

[Out]

x*(5*log(1/4*e^4 - 1/4*log(2) + 3/2) - 9)/log(1/4*e^4 - 1/4*log(2) + 3/2)

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maple [A]  time = 0.03, size = 31, normalized size = 1.41

method	result	size
default	\(\frac {\left (5 \ln \left (-\frac {\ln \relax (2)}{4}+\frac {{\mathrm e}^{4}}{4}+\frac {3}{2}\right )-9\right ) x}{\ln \left (-\frac {\ln \relax (2)}{4}+\frac {{\mathrm e}^{4}}{4}+\frac {3}{2}\right )}\)	\(31\)
norman	\(\frac {\left (10 \ln \relax (2)-5 \ln \left (-\ln \relax (2)+{\mathrm e}^{4}+6\right )+9\right ) x}{2 \ln \relax (2)-\ln \left (-\ln \relax (2)+{\mathrm e}^{4}+6\right )}\)	\(38\)
risch	\(-\frac {10 x \ln \relax (2)}{-2 \ln \relax (2)+\ln \left (-\ln \relax (2)+{\mathrm e}^{4}+6\right )}+\frac {5 x \ln \left (-\ln \relax (2)+{\mathrm e}^{4}+6\right )}{-2 \ln \relax (2)+\ln \left (-\ln \relax (2)+{\mathrm e}^{4}+6\right )}-\frac {9 x}{-2 \ln \relax (2)+\ln \left (-\ln \relax (2)+{\mathrm e}^{4}+6\right )}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*ln(-1/4*ln(2)+1/4*exp(4)+3/2)-9)/ln(-1/4*ln(2)+1/4*exp(4)+3/2),x,method=_RETURNVERBOSE)

[Out]

(5*ln(-1/4*ln(2)+1/4*exp(4)+3/2)-9)/ln(-1/4*ln(2)+1/4*exp(4)+3/2)*x

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maxima [A]  time = 0.51, size = 30, normalized size = 1.36 \begin {gather*} \frac {x {\left (5 \, \log \left (\frac {1}{4} \, e^{4} - \frac {1}{4} \, \log \relax (2) + \frac {3}{2}\right ) - 9\right )}}{\log \left (\frac {1}{4} \, e^{4} - \frac {1}{4} \, \log \relax (2) + \frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*log(-1/4*log(2)+1/4*exp(4)+3/2)-9)/log(-1/4*log(2)+1/4*exp(4)+3/2),x, algorithm="maxima")

[Out]

x*(5*log(1/4*e^4 - 1/4*log(2) + 3/2) - 9)/log(1/4*e^4 - 1/4*log(2) + 3/2)

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mupad [B]  time = 0.00, size = 30, normalized size = 1.36 \begin {gather*} \frac {x\,\left (5\,\ln \left (\frac {{\mathrm {e}}^4}{4}-\frac {\ln \relax (2)}{4}+\frac {3}{2}\right )-9\right )}{\ln \left (\frac {{\mathrm {e}}^4}{4}-\frac {\ln \relax (2)}{4}+\frac {3}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*log(exp(4)/4 - log(2)/4 + 3/2) - 9)/log(exp(4)/4 - log(2)/4 + 3/2),x)

[Out]

(x*(5*log(exp(4)/4 - log(2)/4 + 3/2) - 9))/log(exp(4)/4 - log(2)/4 + 3/2)

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sympy [A]  time = 0.06, size = 34, normalized size = 1.55 \begin {gather*} \frac {x \left (-9 + 5 \log {\left (- \frac {\log {\relax (2 )}}{4} + \frac {3}{2} + \frac {e^{4}}{4} \right )}\right )}{\log {\left (- \frac {\log {\relax (2 )}}{4} + \frac {3}{2} + \frac {e^{4}}{4} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*ln(-1/4*ln(2)+1/4*exp(4)+3/2)-9)/ln(-1/4*ln(2)+1/4*exp(4)+3/2),x)

[Out]

x*(-9 + 5*log(-log(2)/4 + 3/2 + exp(4)/4))/log(-log(2)/4 + 3/2 + exp(4)/4)

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