3.22.98 \(\int \frac {e^x (9 e^3 x-9 x^2)+(e^x (-18 e^3+27 x)+(18 e^3-27 x) \log (\log (4))) \log (e^x-\log (\log (4)))}{e^x (-e^6 x^3+2 e^3 x^4-x^5)+(e^6 x^3-2 e^3 x^4+x^5) \log (\log (4))} \, dx\)

Optimal. Leaf size=26 \[ 2+\frac {9 \log \left (e^x-\log (\log (4))\right )}{x^2 \left (-e^3+x\right )} \]

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Rubi [F]  time = 2.75, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {e^x \left (9 e^3 x-9 x^2\right )+\left (e^x \left (-18 e^3+27 x\right )+\left (18 e^3-27 x\right ) \log (\log (4))\right ) \log \left (e^x-\log (\log (4))\right )}{e^x \left (-e^6 x^3+2 e^3 x^4-x^5\right )+\left (e^6 x^3-2 e^3 x^4+x^5\right ) \log (\log (4))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(9*E^3*x - 9*x^2) + (E^x*(-18*E^3 + 27*x) + (18*E^3 - 27*x)*Log[Log[4]])*Log[E^x - Log[Log[4]]])/(E^x
*(-(E^6*x^3) + 2*E^3*x^4 - x^5) + (E^6*x^3 - 2*E^3*x^4 + x^5)*Log[Log[4]]),x]

[Out]

9/(E^3*x) + (9*Log[E^3 - x])/E^6 - (9*Log[x])/E^6 - (9*Log[E^x - Log[Log[4]]])/((E^3 - x)*x^2) - (9*Log[Log[4]
]*Defer[Int][1/((E^3 - x)*(E^x - Log[Log[4]])), x])/E^6 + (9*Defer[Int][E^x/((E^3 - x)*(E^x - Log[Log[4]])), x
])/E^6 - (9*Log[Log[4]]*Defer[Int][1/(x^2*(E^x - Log[Log[4]])), x])/E^3 + (9*Defer[Int][E^x/(x^2*(E^x - Log[Lo
g[4]])), x])/E^3 - (9*Log[Log[4]]*Defer[Int][1/(x*(E^x - Log[Log[4]])), x])/E^6 + (9*Defer[Int][E^x/(x*(E^x -
Log[Log[4]])), x])/E^6

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {9 \left (-\frac {e^x \left (e^3-x\right ) x}{e^x-\log (\log (4))}+\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )\right )}{\left (e^3-x\right )^2 x^3} \, dx\\ &=9 \int \frac {-\frac {e^x \left (e^3-x\right ) x}{e^x-\log (\log (4))}+\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3} \, dx\\ &=9 \int \left (-\frac {\log (\log (4))}{\left (e^3-x\right ) x^2 \left (e^x-\log (\log (4))\right )}+\frac {-e^3 x+x^2+2 e^3 \log \left (e^x-\log (\log (4))\right )-3 x \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3}\right ) \, dx\\ &=9 \int \frac {-e^3 x+x^2+2 e^3 \log \left (e^x-\log (\log (4))\right )-3 x \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3} \, dx-(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) x^2 \left (e^x-\log (\log (4))\right )} \, dx\\ &=9 \int \left (-\frac {1}{\left (e^3-x\right ) x^2}+\frac {\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3}\right ) \, dx-(9 \log (\log (4))) \int \left (\frac {1}{e^6 \left (e^3-x\right ) \left (e^x-\log (\log (4))\right )}+\frac {1}{e^3 x^2 \left (e^x-\log (\log (4))\right )}+\frac {1}{e^6 x \left (e^x-\log (\log (4))\right )}\right ) \, dx\\ &=-\left (9 \int \frac {1}{\left (e^3-x\right ) x^2} \, dx\right )+9 \int \frac {\left (2 e^3-3 x\right ) \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right )^2 x^3} \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}-9 \int \left (\frac {1}{e^6 \left (e^3-x\right )}+\frac {1}{e^3 x^2}+\frac {1}{e^6 x}\right ) \, dx+9 \int \frac {e^x}{\left (e^3-x\right ) x^2 \left (e^x-\log (\log (4))\right )} \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=\frac {9}{e^3 x}+\frac {9 \log \left (e^3-x\right )}{e^6}-\frac {9 \log (x)}{e^6}-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}+9 \int \left (\frac {e^{-6+x}}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )}+\frac {e^{-3+x}}{x^2 \left (e^x-\log (\log (4))\right )}+\frac {e^{-6+x}}{x \left (e^x-\log (\log (4))\right )}\right ) \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=\frac {9}{e^3 x}+\frac {9 \log \left (e^3-x\right )}{e^6}-\frac {9 \log (x)}{e^6}-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}+9 \int \frac {e^{-6+x}}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx+9 \int \frac {e^{-3+x}}{x^2 \left (e^x-\log (\log (4))\right )} \, dx+9 \int \frac {e^{-6+x}}{x \left (e^x-\log (\log (4))\right )} \, dx-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ &=\frac {9}{e^3 x}+\frac {9 \log \left (e^3-x\right )}{e^6}-\frac {9 \log (x)}{e^6}-\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2}+\frac {9 \int \frac {e^x}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}+\frac {9 \int \frac {e^x}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}+\frac {9 \int \frac {e^x}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}-\frac {(9 \log (\log (4))) \int \frac {1}{\left (e^3-x\right ) \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x \left (e^x-\log (\log (4))\right )} \, dx}{e^6}-\frac {(9 \log (\log (4))) \int \frac {1}{x^2 \left (e^x-\log (\log (4))\right )} \, dx}{e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.52, size = 24, normalized size = 0.92 \begin {gather*} -\frac {9 \log \left (e^x-\log (\log (4))\right )}{\left (e^3-x\right ) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(9*E^3*x - 9*x^2) + (E^x*(-18*E^3 + 27*x) + (18*E^3 - 27*x)*Log[Log[4]])*Log[E^x - Log[Log[4]]]
)/(E^x*(-(E^6*x^3) + 2*E^3*x^4 - x^5) + (E^6*x^3 - 2*E^3*x^4 + x^5)*Log[Log[4]]),x]

[Out]

(-9*Log[E^x - Log[Log[4]]])/((E^3 - x)*x^2)

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fricas [A]  time = 1.06, size = 26, normalized size = 1.00 \begin {gather*} \frac {9 \, \log \left (e^{x} - \log \left (2 \, \log \relax (2)\right )\right )}{x^{3} - x^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*exp(3)-27*x)*log(2*log(2))+(-18*exp(3)+27*x)*exp(x))*log(-log(2*log(2))+exp(x))+(9*x*exp(3)-9*
x^2)*exp(x))/((x^3*exp(3)^2-2*x^4*exp(3)+x^5)*log(2*log(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x, algori
thm="fricas")

[Out]

9*log(e^x - log(2*log(2)))/(x^3 - x^2*e^3)

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giac [A]  time = 1.09, size = 28, normalized size = 1.08 \begin {gather*} \frac {9 \, \log \left (e^{x} - \log \relax (2) - \log \left (\log \relax (2)\right )\right )}{x^{3} - x^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*exp(3)-27*x)*log(2*log(2))+(-18*exp(3)+27*x)*exp(x))*log(-log(2*log(2))+exp(x))+(9*x*exp(3)-9*
x^2)*exp(x))/((x^3*exp(3)^2-2*x^4*exp(3)+x^5)*log(2*log(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x, algori
thm="giac")

[Out]

9*log(e^x - log(2) - log(log(2)))/(x^3 - x^2*e^3)

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maple [A]  time = 0.16, size = 27, normalized size = 1.04

method	result	size
risch	\(-\frac {9 \ln \left (-\ln \relax (2)-\ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}{x^{2} \left (-x +{\mathrm e}^{3}\right )}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((18*exp(3)-27*x)*ln(2*ln(2))+(-18*exp(3)+27*x)*exp(x))*ln(-ln(2*ln(2))+exp(x))+(9*x*exp(3)-9*x^2)*exp(x)
)/((x^3*exp(3)^2-2*x^4*exp(3)+x^5)*ln(2*ln(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x,method=_RETURNVERBOS
E)

[Out]

-9/x^2/(-x+exp(3))*ln(-ln(2)-ln(ln(2))+exp(x))

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maxima [A]  time = 0.69, size = 28, normalized size = 1.08 \begin {gather*} \frac {9 \, \log \left (e^{x} - \log \relax (2) - \log \left (\log \relax (2)\right )\right )}{x^{3} - x^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*exp(3)-27*x)*log(2*log(2))+(-18*exp(3)+27*x)*exp(x))*log(-log(2*log(2))+exp(x))+(9*x*exp(3)-9*
x^2)*exp(x))/((x^3*exp(3)^2-2*x^4*exp(3)+x^5)*log(2*log(2))+(-x^3*exp(3)^2+2*x^4*exp(3)-x^5)*exp(x)),x, algori
thm="maxima")

[Out]

9*log(e^x - log(2) - log(log(2)))/(x^3 - x^2*e^3)

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mupad [B]  time = 0.65, size = 27, normalized size = 1.04 \begin {gather*} -\frac {9\,\ln \left ({\mathrm {e}}^x-\ln \left (2\,\ln \relax (2)\right )\right )}{x^2\,{\mathrm {e}}^3-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(9*x*exp(3) - 9*x^2) + log(exp(x) - log(2*log(2)))*(exp(x)*(27*x - 18*exp(3)) - log(2*log(2))*(27*
x - 18*exp(3))))/(log(2*log(2))*(x^3*exp(6) - 2*x^4*exp(3) + x^5) - exp(x)*(x^3*exp(6) - 2*x^4*exp(3) + x^5)),
x)

[Out]

-(9*log(exp(x) - log(2*log(2))))/(x^2*exp(3) - x^3)

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sympy [A]  time = 0.35, size = 22, normalized size = 0.85 \begin {gather*} \frac {9 \log {\left (e^{x} - \log {\left (2 \log {\relax (2 )} \right )} \right )}}{x^{3} - x^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((18*exp(3)-27*x)*ln(2*ln(2))+(-18*exp(3)+27*x)*exp(x))*ln(-ln(2*ln(2))+exp(x))+(9*x*exp(3)-9*x**2)
*exp(x))/((x**3*exp(3)**2-2*x**4*exp(3)+x**5)*ln(2*ln(2))+(-x**3*exp(3)**2+2*x**4*exp(3)-x**5)*exp(x)),x)

[Out]

9*log(exp(x) - log(2*log(2)))/(x**3 - x**2*exp(3))

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