3.26.48 \(\int \frac {12 \log (\frac {\log (2)}{3})+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log (\frac {\log (2)}{3}) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=19 \[ \log \left (e^x-\frac {12 \log \left (\frac {\log (2)}{3}\right )}{\log (\log (x))}\right ) \]

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Rubi [F]  time = 3.65, 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 {12 \log \left (\frac {\log (2)}{3}\right )+e^x x \log (x) \log ^2(\log (x))}{-12 x \log (x) \log \left (\frac {\log (2)}{3}\right ) \log (\log (x))+e^x x \log (x) \log ^2(\log (x))} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(12*Log[Log[2]/3] + E^x*x*Log[x]*Log[Log[x]]^2)/(-12*x*Log[x]*Log[Log[2]/3]*Log[Log[x]] + E^x*x*Log[x]*Log
[Log[x]]^2),x]

[Out]

x - 12*Log[Log[2]/3]*Defer[Int][(12*Log[Log[2]/3] - E^x*Log[Log[x]])^(-1), x] + 12*Log[Log[2]/3]*Defer[Int][1/
(x*Log[x]*Log[Log[x]]*(-12*Log[Log[2]/3] + E^x*Log[Log[x]])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-12 \log \left (\frac {\log (2)}{3}\right )-e^x x \log (x) \log ^2(\log (x))}{x \log (x) \log (\log (x)) \left (12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))\right )} \, dx\\ &=\int \left (1-\frac {12 \log \left (\frac {\log (2)}{3}\right ) (1+x \log (x) \log (\log (x)))}{x \log (x) \log (\log (x)) \left (12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))\right )}\right ) \, dx\\ &=x-\left (12 \log \left (\frac {\log (2)}{3}\right )\right ) \int \frac {1+x \log (x) \log (\log (x))}{x \log (x) \log (\log (x)) \left (12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))\right )} \, dx\\ &=x-\left (12 \log \left (\frac {\log (2)}{3}\right )\right ) \int \left (\frac {1}{12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))}-\frac {1}{x \log (x) \log (\log (x)) \left (-12 \log \left (\frac {\log (2)}{3}\right )+e^x \log (\log (x))\right )}\right ) \, dx\\ &=x-\left (12 \log \left (\frac {\log (2)}{3}\right )\right ) \int \frac {1}{12 \log \left (\frac {\log (2)}{3}\right )-e^x \log (\log (x))} \, dx+\left (12 \log \left (\frac {\log (2)}{3}\right )\right ) \int \frac {1}{x \log (x) \log (\log (x)) \left (-12 \log \left (\frac {\log (2)}{3}\right )+e^x \log (\log (x))\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 25, normalized size = 1.32 \begin {gather*} -\log (\log (\log (x)))+\log \left (-12 \log \left (\frac {\log (2)}{3}\right )+e^x \log (\log (x))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12*Log[Log[2]/3] + E^x*x*Log[x]*Log[Log[x]]^2)/(-12*x*Log[x]*Log[Log[2]/3]*Log[Log[x]] + E^x*x*Log[
x]*Log[Log[x]]^2),x]

[Out]

-Log[Log[Log[x]]] + Log[-12*Log[Log[2]/3] + E^x*Log[Log[x]]]

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fricas [A]  time = 0.90, size = 28, normalized size = 1.47 \begin {gather*} x + \log \left ({\left (e^{x} \log \left (\log \relax (x)\right ) - 12 \, \log \left (\frac {1}{3} \, \log \relax (2)\right )\right )} e^{\left (-x\right )}\right ) - \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(x)*log(x)*log(log(x))^2+12*log(1/3*log(2)))/(x*exp(x)*log(x)*log(log(x))^2-12*x*log(x)*log(1/
3*log(2))*log(log(x))),x, algorithm="fricas")

[Out]

x + log((e^x*log(log(x)) - 12*log(1/3*log(2)))*e^(-x)) - log(log(log(x)))

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giac [A]  time = 0.24, size = 24, normalized size = 1.26 \begin {gather*} \log \left (e^{x} \log \left (\log \relax (x)\right ) + 12 \, \log \relax (3) - 12 \, \log \left (\log \relax (2)\right )\right ) - \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(x)*log(x)*log(log(x))^2+12*log(1/3*log(2)))/(x*exp(x)*log(x)*log(log(x))^2-12*x*log(x)*log(1/
3*log(2))*log(log(x))),x, algorithm="giac")

[Out]

log(e^x*log(log(x)) + 12*log(3) - 12*log(log(2))) - log(log(log(x)))

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maple [A]  time = 0.05, size = 28, normalized size = 1.47

method	result	size
risch	\(x -\ln \left (\ln \left (\ln \relax (x )\right )\right )+\ln \left (\ln \left (\ln \relax (x )\right )-12 \left (-\ln \relax (3)+\ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{-x}\right )\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(x)*ln(x)*ln(ln(x))^2+12*ln(1/3*ln(2)))/(x*exp(x)*ln(x)*ln(ln(x))^2-12*x*ln(x)*ln(1/3*ln(2))*ln(ln(x
))),x,method=_RETURNVERBOSE)

[Out]

x-ln(ln(ln(x)))+ln(ln(ln(x))-12*(-ln(3)+ln(ln(2)))*exp(-x))

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maxima [A]  time = 0.68, size = 30, normalized size = 1.58 \begin {gather*} x + \log \left ({\left (e^{x} \log \left (\log \relax (x)\right ) + 12 \, \log \relax (3) - 12 \, \log \left (\log \relax (2)\right )\right )} e^{\left (-x\right )}\right ) - \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(x)*log(x)*log(log(x))^2+12*log(1/3*log(2)))/(x*exp(x)*log(x)*log(log(x))^2-12*x*log(x)*log(1/
3*log(2))*log(log(x))),x, algorithm="maxima")

[Out]

x + log((e^x*log(log(x)) + 12*log(3) - 12*log(log(2)))*e^(-x)) - log(log(log(x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {x\,{\mathrm {e}}^x\,\ln \relax (x)\,{\ln \left (\ln \relax (x)\right )}^2+12\,\ln \left (\frac {\ln \relax (2)}{3}\right )}{12\,x\,\ln \left (\frac {\ln \relax (2)}{3}\right )\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)-x\,{\ln \left (\ln \relax (x)\right )}^2\,{\mathrm {e}}^x\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(12*log(log(2)/3) + x*log(log(x))^2*exp(x)*log(x))/(12*x*log(log(2)/3)*log(log(x))*log(x) - x*log(log(x))
^2*exp(x)*log(x)),x)

[Out]

int(-(12*log(log(2)/3) + x*log(log(x))^2*exp(x)*log(x))/(12*x*log(log(2)/3)*log(log(x))*log(x) - x*log(log(x))
^2*exp(x)*log(x)), x)

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sympy [A]  time = 0.47, size = 20, normalized size = 1.05 \begin {gather*} \log {\left (e^{x} + \frac {- 12 \log {\left (\log {\relax (2 )} \right )} + 12 \log {\relax (3 )}}{\log {\left (\log {\relax (x )} \right )}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x*exp(x)*ln(x)*ln(ln(x))**2+12*ln(1/3*ln(2)))/(x*exp(x)*ln(x)*ln(ln(x))**2-12*x*ln(x)*ln(1/3*ln(2))
*ln(ln(x))),x)

[Out]

log(exp(x) + (-12*log(log(2)) + 12*log(3))/log(log(x)))

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