3.64.82 \(\int \frac {-2 x+2 x \log (x) \log (\log (x)) \log (-\frac {5 \log (\log (x))}{e^5 (-8+4 e^4)})}{\log (x) \log (\log (x)) \log ^3(-\frac {5 \log (\log (x))}{e^5 (-8+4 e^4)})} \, dx\)

Optimal. Leaf size=26 \[ \frac {x^2}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \]

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Rubi [F]  time = 0.74, 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 {-2 x+2 x \log (x) \log (\log (x)) \log \left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )}{\log (x) \log (\log (x)) \log ^3\left (-\frac {5 \log (\log (x))}{e^5 \left (-8+4 e^4\right )}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-2*x + 2*x*Log[x]*Log[Log[x]]*Log[(-5*Log[Log[x]])/(E^5*(-8 + 4*E^4))])/(Log[x]*Log[Log[x]]*Log[(-5*Log[L
og[x]])/(E^5*(-8 + 4*E^4))]^3),x]

[Out]

-2*Defer[Int][x/(Log[x]*Log[Log[x]]*Log[(5*Log[Log[x]])/(4*E^5*(2 - E^4))]^3), x] + 2*Defer[Int][x/Log[(5*Log[
Log[x]])/(4*E^5*(2 - E^4))]^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 x \left (-1+\log (x) \log (\log (x)) \log \left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )\right )}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\\ &=2 \int \frac {x \left (-1+\log (x) \log (\log (x)) \log \left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )\right )}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\\ &=2 \int \left (-\frac {x}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}+\frac {x}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {x}{\log (x) \log (\log (x)) \log ^3\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\right )+2 \int \frac {x}{\log ^2\left (\frac {5 \log (\log (x))}{4 e^5 \left (2-e^4\right )}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 25, normalized size = 0.96 \begin {gather*} \frac {x^2}{\log ^2\left (\frac {5 \log (\log (x))}{8 e^5-4 e^9}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + 2*x*Log[x]*Log[Log[x]]*Log[(-5*Log[Log[x]])/(E^5*(-8 + 4*E^4))])/(Log[x]*Log[Log[x]]*Log[(-5
*Log[Log[x]])/(E^5*(-8 + 4*E^4))]^3),x]

[Out]

x^2/Log[(5*Log[Log[x]])/(8*E^5 - 4*E^9)]^2

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fricas [A]  time = 0.73, size = 21, normalized size = 0.81 \begin {gather*} \frac {x^{2}}{\log \left (-\frac {5 \, \log \left (\log \relax (x)\right )}{4 \, {\left (e^{9} - 2 \, e^{5}\right )}}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x)*log(log(x))*log(-5*log(log(x))/(4*exp(4)-8)/exp(5))-2*x)/log(x)/log(log(x))/log(-5*log(l
og(x))/(4*exp(4)-8)/exp(5))^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.39, size = 80, normalized size = 3.08 \begin {gather*} \frac {x^{2}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) \log \left (-5 \, \log \left (\log \relax (x)\right )\right ) + \log \left (-5 \, \log \left (\log \relax (x)\right )\right )^{2} + 4 \, \log \relax (2) \log \left (e^{4} - 2\right ) - 2 \, \log \left (-5 \, \log \left (\log \relax (x)\right )\right ) \log \left (e^{4} - 2\right ) + \log \left (e^{4} - 2\right )^{2} + 20 \, \log \relax (2) - 10 \, \log \left (-5 \, \log \left (\log \relax (x)\right )\right ) + 10 \, \log \left (e^{4} - 2\right ) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x)*log(log(x))*log(-5*log(log(x))/(4*exp(4)-8)/exp(5))-2*x)/log(x)/log(log(x))/log(-5*log(l
og(x))/(4*exp(4)-8)/exp(5))^3,x, algorithm="giac")

[Out]

x^2/(4*log(2)^2 - 4*log(2)*log(-5*log(log(x))) + log(-5*log(log(x)))^2 + 4*log(2)*log(e^4 - 2) - 2*log(-5*log(
log(x)))*log(e^4 - 2) + log(e^4 - 2)^2 + 20*log(2) - 10*log(-5*log(log(x))) + 10*log(e^4 - 2) + 25)

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maple [A]  time = 0.06, size = 23, normalized size = 0.88

method	result	size
risch	\(\frac {x^{2}}{\ln \left (-\frac {5 \ln \left (\ln \relax (x )\right ) {\mathrm e}^{-5}}{4 \,{\mathrm e}^{4}-8}\right )^{2}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*ln(x)*ln(ln(x))*ln(-5*ln(ln(x))/(4*exp(4)-8)/exp(5))-2*x)/ln(x)/ln(ln(x))/ln(-5*ln(ln(x))/(4*exp(4)-8
)/exp(5))^3,x,method=_RETURNVERBOSE)

[Out]

x^2/ln(-5*ln(ln(x))/(4*exp(4)-8)*exp(-5))^2

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maxima [B]  time = 0.65, size = 102, normalized size = 3.92 \begin {gather*} -\frac {x^{2}}{2 \, {\left (\log \left (-e^{4} + 2\right ) + 5\right )} \log \relax (5) - \log \relax (5)^{2} + 4 \, {\left (\log \relax (5) - \log \left (-e^{4} + 2\right ) - 5\right )} \log \relax (2) - 4 \, \log \relax (2)^{2} - \log \left (-e^{4} + 2\right )^{2} - 2 \, {\left (\log \relax (5) - 2 \, \log \relax (2) - \log \left (-e^{4} + 2\right ) - 5\right )} \log \left (\log \left (\log \relax (x)\right )\right ) - \log \left (\log \left (\log \relax (x)\right )\right )^{2} - 10 \, \log \left (-e^{4} + 2\right ) - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*log(x)*log(log(x))*log(-5*log(log(x))/(4*exp(4)-8)/exp(5))-2*x)/log(x)/log(log(x))/log(-5*log(l
og(x))/(4*exp(4)-8)/exp(5))^3,x, algorithm="maxima")

[Out]

-x^2/(2*(log(-e^4 + 2) + 5)*log(5) - log(5)^2 + 4*(log(5) - log(-e^4 + 2) - 5)*log(2) - 4*log(2)^2 - log(-e^4
+ 2)^2 - 2*(log(5) - 2*log(2) - log(-e^4 + 2) - 5)*log(log(log(x))) - log(log(log(x)))^2 - 10*log(-e^4 + 2) -
25)

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mupad [B]  time = 4.55, size = 149, normalized size = 5.73 \begin {gather*} {\ln \left (\ln \relax (x)\right )}^2\,\left (2\,x^2\,{\ln \relax (x)}^2+x^2\,\ln \relax (x)\right )+\frac {x^2-x^2\,\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)}{{\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )}^2}+\frac {x^2\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)-x\,\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\,\left (x+x\,\ln \left (\ln \relax (x)\right )+2\,x\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x)\right )}{\ln \left (-\frac {5\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-5}}{4\,{\mathrm {e}}^4-8}\right )}+x^2\,\ln \left (\ln \relax (x)\right )\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x - 2*x*log(-(5*log(log(x))*exp(-5))/(4*exp(4) - 8))*log(log(x))*log(x))/(log(-(5*log(log(x))*exp(-5))
/(4*exp(4) - 8))^3*log(log(x))*log(x)),x)

[Out]

log(log(x))^2*(x^2*log(x) + 2*x^2*log(x)^2) + (x^2 - x^2*log(-(5*log(log(x))*exp(-5))/(4*exp(4) - 8))*log(log(
x))*log(x))/log(-(5*log(log(x))*exp(-5))/(4*exp(4) - 8))^2 + (x^2*log(log(x))*log(x) - x*log(-(5*log(log(x))*e
xp(-5))/(4*exp(4) - 8))*log(log(x))*log(x)*(x + x*log(log(x)) + 2*x*log(log(x))*log(x)))/log(-(5*log(log(x))*e
xp(-5))/(4*exp(4) - 8)) + x^2*log(log(x))*log(x)

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sympy [A]  time = 0.35, size = 24, normalized size = 0.92 \begin {gather*} \frac {x^{2}}{\log {\left (- \frac {5 \log {\left (\log {\relax (x )} \right )}}{\left (-8 + 4 e^{4}\right ) e^{5}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x*ln(x)*ln(ln(x))*ln(-5*ln(ln(x))/(4*exp(4)-8)/exp(5))-2*x)/ln(x)/ln(ln(x))/ln(-5*ln(ln(x))/(4*ex
p(4)-8)/exp(5))**3,x)

[Out]

x**2/log(-5*exp(-5)*log(log(x))/(-8 + 4*exp(4)))**2

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