3.73.44 \(\int \frac {(\frac {1}{3} (-20+4 e^2))^{\frac {1}{x^2}} (\frac {\log (x)}{x})^{\frac {1}{x^2}} (1-\log (x)-2 \log (x) \log (\frac {(-20+4 e^2) \log (x)}{3 x}))}{x^3 \log (x)} \, dx\)

Optimal. Leaf size=24 \[ \left (\frac {4}{3} \left (-5+e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}} \]

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Rubi [F]  time = 2.22, 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 {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}} \left (1-\log (x)-2 \log (x) \log \left (\frac {\left (-20+4 e^2\right ) \log (x)}{3 x}\right )\right )}{x^3 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(((-20 + 4*E^2)/3)^x^(-2)*(Log[x]/x)^x^(-2)*(1 - Log[x] - 2*Log[x]*Log[((-20 + 4*E^2)*Log[x])/(3*x)]))/(x^
3*Log[x]),x]

[Out]

-Defer[Int][(((-20 + 4*E^2)/3)^x^(-2)*(Log[x]/x)^x^(-2))/x^3, x] + Defer[Int][(((-20 + 4*E^2)/3)^x^(-2)*(Log[x
]/x)^x^(-2))/(x^3*Log[x]), x] - 2*Defer[Int][(((-20 + 4*E^2)/3)^x^(-2)*(Log[x]/x)^x^(-2)*Log[(4*(-5 + E^2)*Log
[x])/(3*x)])/x^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} (1-\log (x)) \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}}}{x^3 \log (x)}-\frac {2 \left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}} \log \left (\frac {4 \left (-5+e^2\right ) \log (x)}{3 x}\right )}{x^3}\right ) \, dx\\ &=-\left (2 \int \frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}} \log \left (\frac {4 \left (-5+e^2\right ) \log (x)}{3 x}\right )}{x^3} \, dx\right )+\int \frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} (1-\log (x)) \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}}}{x^3 \log (x)} \, dx\\ &=-\left (2 \int \frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}} \log \left (\frac {4 \left (-5+e^2\right ) \log (x)}{3 x}\right )}{x^3} \, dx\right )+\int \left (-\frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}}}{x^3}+\frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}}}{x^3 \log (x)}\right ) \, dx\\ &=-\left (2 \int \frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}} \log \left (\frac {4 \left (-5+e^2\right ) \log (x)}{3 x}\right )}{x^3} \, dx\right )-\int \frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}}}{x^3} \, dx+\int \frac {\left (\frac {1}{3} \left (-20+4 e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}}}{x^3 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.76, size = 24, normalized size = 1.00 \begin {gather*} \left (\frac {4}{3} \left (-5+e^2\right )\right )^{\frac {1}{x^2}} \left (\frac {\log (x)}{x}\right )^{\frac {1}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(((-20 + 4*E^2)/3)^x^(-2)*(Log[x]/x)^x^(-2)*(1 - Log[x] - 2*Log[x]*Log[((-20 + 4*E^2)*Log[x])/(3*x)]
))/(x^3*Log[x]),x]

[Out]

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

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fricas [A]  time = 0.71, size = 15, normalized size = 0.62 \begin {gather*} \left (\frac {4 \, {\left (e^{2} - 5\right )} \log \relax (x)}{3 \, x}\right )^{\left (\frac {1}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)*log(1/3*(4*exp(2)-20)*log(x)/x)+1-log(x))*exp(log(1/3*(4*exp(2)-20)*log(x)/x)/x^2)/x^3/lo
g(x),x, algorithm="fricas")

[Out]

(4/3*(e^2 - 5)*log(x)/x)^(x^(-2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left (2 \, \log \relax (x) \log \left (\frac {4 \, {\left (e^{2} - 5\right )} \log \relax (x)}{3 \, x}\right ) + \log \relax (x) - 1\right )} \left (\frac {4 \, {\left (e^{2} - 5\right )} \log \relax (x)}{3 \, x}\right )^{\left (\frac {1}{x^{2}}\right )}}{x^{3} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)*log(1/3*(4*exp(2)-20)*log(x)/x)+1-log(x))*exp(log(1/3*(4*exp(2)-20)*log(x)/x)/x^2)/x^3/lo
g(x),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.11, size = 82, normalized size = 3.42

method	result	size
risch	\(x^{-\frac {1}{x^{2}}} \ln \relax (x )^{\frac {1}{x^{2}}} \left (\frac {4}{3}\right )^{\frac {1}{x^{2}}} \left ({\mathrm e}^{2}-5\right )^{\frac {1}{x^{2}}} {\mathrm e}^{-\frac {i \pi \,\mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right )+\mathrm {csgn}\left (i \ln \relax (x )\right )\right ) \left (-\mathrm {csgn}\left (\frac {i \ln \relax (x )}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right )}{2 x^{2}}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*ln(x)*ln(1/3*(4*exp(2)-20)*ln(x)/x)+1-ln(x))*exp(ln(1/3*(4*exp(2)-20)*ln(x)/x)/x^2)/x^3/ln(x),x,method
=_RETURNVERBOSE)

[Out]

x^(-1/x^2)*ln(x)^(1/x^2)*(4/3)^(1/x^2)*(exp(2)-5)^(1/x^2)*exp(-1/2*I*Pi*csgn(I/x*ln(x))*(-csgn(I/x*ln(x))+csgn
(I*ln(x)))*(-csgn(I/x*ln(x))+csgn(I/x))/x^2)

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maxima [B]  time = 0.51, size = 39, normalized size = 1.62 \begin {gather*} e^{\left (-\frac {\log \relax (3)}{x^{2}} + \frac {2 \, \log \relax (2)}{x^{2}} - \frac {\log \relax (x)}{x^{2}} + \frac {\log \left (e^{2} - 5\right )}{x^{2}} + \frac {\log \left (\log \relax (x)\right )}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*log(x)*log(1/3*(4*exp(2)-20)*log(x)/x)+1-log(x))*exp(log(1/3*(4*exp(2)-20)*log(x)/x)/x^2)/x^3/lo
g(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.58, size = 20, normalized size = 0.83 \begin {gather*} {\left (-\frac {20\,\ln \relax (x)-4\,{\mathrm {e}}^2\,\ln \relax (x)}{3\,x}\right )}^{\frac {1}{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log((log(x)*(4*exp(2) - 20))/(3*x))/x^2)*(log(x) + 2*log((log(x)*(4*exp(2) - 20))/(3*x))*log(x) - 1)
)/(x^3*log(x)),x)

[Out]

(-(20*log(x) - 4*exp(2)*log(x))/(3*x))^(1/x^2)

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sympy [A]  time = 0.55, size = 20, normalized size = 0.83 \begin {gather*} e^{\frac {\log {\left (\frac {\left (- \frac {20}{3} + \frac {4 e^{2}}{3}\right ) \log {\relax (x )}}{x} \right )}}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*ln(x)*ln(1/3*(4*exp(2)-20)*ln(x)/x)+1-ln(x))*exp(ln(1/3*(4*exp(2)-20)*ln(x)/x)/x**2)/x**3/ln(x),
x)

[Out]

exp(log((-20/3 + 4*exp(2)/3)*log(x)/x)/x**2)

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