3.21.70 \(\int \frac {\log ^{\frac {2}{x}}(\frac {1}{9} x \log ^2(e^{-x} x)) (32-32 x+16 \log (e^{-x} x)-16 \log (e^{-x} x) \log (\frac {1}{9} x \log ^2(e^{-x} x)) \log (\log (\frac {1}{9} x \log ^2(e^{-x} x))))+\log ^{\frac {4}{x}}(\frac {1}{9} x \log ^2(e^{-x} x)) (32-32 x+16 \log (e^{-x} x)-16 \log (e^{-x} x) \log (\frac {1}{9} x \log ^2(e^{-x} x)) \log (\log (\frac {1}{9} x \log ^2(e^{-x} x))))}{x^2 \log (e^{-x} x) \log (\frac {1}{9} x \log ^2(e^{-x} x))} \, dx\)

Optimal. Leaf size=30 \[ 4 \left (4+\left (1+\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )^2\right ) \]

________________________________________________________________________________________

Rubi [F]  time = 4.01, 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 {\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (32-32 x+16 \log \left (e^{-x} x\right )-16 \log \left (e^{-x} x\right ) \log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )+\log ^{\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (32-32 x+16 \log \left (e^{-x} x\right )-16 \log \left (e^{-x} x\right ) \log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )}{x^2 \log \left (e^{-x} x\right ) \log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(Log[(x*Log[x/E^x]^2)/9]^(2/x)*(32 - 32*x + 16*Log[x/E^x] - 16*Log[x/E^x]*Log[(x*Log[x/E^x]^2)/9]*Log[Log[
(x*Log[x/E^x]^2)/9]]) + Log[(x*Log[x/E^x]^2)/9]^(4/x)*(32 - 32*x + 16*Log[x/E^x] - 16*Log[x/E^x]*Log[(x*Log[x/
E^x]^2)/9]*Log[Log[(x*Log[x/E^x]^2)/9]]))/(x^2*Log[x/E^x]*Log[(x*Log[x/E^x]^2)/9]),x]

[Out]

16*Defer[Int][Log[(x*Log[x/E^x]^2)/9]^(-1 + 2/x)/x^2, x] + 32*Defer[Int][Log[(x*Log[x/E^x]^2)/9]^(-1 + 2/x)/(x
^2*Log[x/E^x]), x] - 32*Defer[Int][Log[(x*Log[x/E^x]^2)/9]^(-1 + 2/x)/(x*Log[x/E^x]), x] + 16*Defer[Int][Log[(
x*Log[x/E^x]^2)/9]^(-1 + 4/x)/x^2, x] + 32*Defer[Int][Log[(x*Log[x/E^x]^2)/9]^(-1 + 4/x)/(x^2*Log[x/E^x]), x]
- 32*Defer[Int][Log[(x*Log[x/E^x]^2)/9]^(-1 + 4/x)/(x*Log[x/E^x]), x] - 16*Defer[Int][(Log[(x*Log[x/E^x]^2)/9]
^(2/x)*Log[Log[(x*Log[x/E^x]^2)/9]])/x^2, x] - 16*Defer[Int][(Log[(x*Log[x/E^x]^2)/9]^(4/x)*Log[Log[(x*Log[x/E
^x]^2)/9]])/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 \log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (1+\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right ) \left (-2 (-1+x)-\log \left (e^{-x} x\right ) \left (-1+\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx\\ &=16 \int \frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (1+\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right ) \left (-2 (-1+x)-\log \left (e^{-x} x\right ) \left (-1+\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx\\ &=16 \int \left (\frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (2-2 x+\log \left (e^{-x} x\right )-\log \left (e^{-x} x\right ) \log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )}{x^2 \log \left (e^{-x} x\right )}-\frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (-2+2 x-\log \left (e^{-x} x\right )+\log \left (e^{-x} x\right ) \log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )}{x^2 \log \left (e^{-x} x\right )}\right ) \, dx\\ &=16 \int \frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (2-2 x+\log \left (e^{-x} x\right )-\log \left (e^{-x} x\right ) \log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx-16 \int \frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (-2+2 x-\log \left (e^{-x} x\right )+\log \left (e^{-x} x\right ) \log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx\\ &=16 \int \left (\frac {\left (2-2 x+\log \left (e^{-x} x\right )\right ) \log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )}-\frac {\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2}\right ) \, dx-16 \int \left (\frac {\left (-2+2 x-\log \left (e^{-x} x\right )\right ) \log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )}+\frac {\log ^{\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2}\right ) \, dx\\ &=16 \int \frac {\left (2-2 x+\log \left (e^{-x} x\right )\right ) \log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx-16 \int \frac {\left (-2+2 x-\log \left (e^{-x} x\right )\right ) \log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx-16 \int \frac {\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx\\ &=16 \int \left (\frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2}-\frac {2 (-1+x) \log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )}\right ) \, dx-16 \int \left (-\frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2}+\frac {2 (-1+x) \log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )}\right ) \, dx-16 \int \frac {\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx\\ &=16 \int \frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2} \, dx+16 \int \frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx-32 \int \frac {(-1+x) \log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx-32 \int \frac {(-1+x) \log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx\\ &=16 \int \frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2} \, dx+16 \int \frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx-32 \int \left (-\frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )}+\frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x \log \left (e^{-x} x\right )}\right ) \, dx-32 \int \left (-\frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )}+\frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x \log \left (e^{-x} x\right )}\right ) \, dx\\ &=16 \int \frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2} \, dx+16 \int \frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx-16 \int \frac {\log ^{\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \log \left (\log \left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right )}{x^2} \, dx+32 \int \frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx-32 \int \frac {\log ^{-1+\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x \log \left (e^{-x} x\right )} \, dx+32 \int \frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x^2 \log \left (e^{-x} x\right )} \, dx-32 \int \frac {\log ^{-1+\frac {4}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )}{x \log \left (e^{-x} x\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 48, normalized size = 1.60 \begin {gather*} 4 \log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right ) \left (2+\log ^{\frac {2}{x}}\left (\frac {1}{9} x \log ^2\left (e^{-x} x\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Log[(x*Log[x/E^x]^2)/9]^(2/x)*(32 - 32*x + 16*Log[x/E^x] - 16*Log[x/E^x]*Log[(x*Log[x/E^x]^2)/9]*Lo
g[Log[(x*Log[x/E^x]^2)/9]]) + Log[(x*Log[x/E^x]^2)/9]^(4/x)*(32 - 32*x + 16*Log[x/E^x] - 16*Log[x/E^x]*Log[(x*
Log[x/E^x]^2)/9]*Log[Log[(x*Log[x/E^x]^2)/9]]))/(x^2*Log[x/E^x]*Log[(x*Log[x/E^x]^2)/9]),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 43, normalized size = 1.43 \begin {gather*} 4 \, \log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right )^{\frac {4}{x}} + 8 \, \log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right )^{\frac {2}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(x/exp(x))*log(1/9*x*log(x/exp(x))^2)*log(log(1/9*x*log(x/exp(x))^2))+16*log(x/exp(x))-32*x
+32)*exp(2*log(log(1/9*x*log(x/exp(x))^2))/x)^2+(-16*log(x/exp(x))*log(1/9*x*log(x/exp(x))^2)*log(log(1/9*x*lo
g(x/exp(x))^2))+16*log(x/exp(x))-32*x+32)*exp(2*log(log(1/9*x*log(x/exp(x))^2))/x))/x^2/log(x/exp(x))/log(1/9*
x*log(x/exp(x))^2),x, algorithm="fricas")

[Out]

4*log(1/9*x*log(x*e^(-x))^2)^(4/x) + 8*log(1/9*x*log(x*e^(-x))^2)^(2/x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {16 \, {\left ({\left (\log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right ) \log \left (x e^{\left (-x\right )}\right ) \log \left (\log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right )\right ) + 2 \, x - \log \left (x e^{\left (-x\right )}\right ) - 2\right )} \log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right )^{\frac {4}{x}} + {\left (\log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right ) \log \left (x e^{\left (-x\right )}\right ) \log \left (\log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right )\right ) + 2 \, x - \log \left (x e^{\left (-x\right )}\right ) - 2\right )} \log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right )^{\frac {2}{x}}\right )}}{x^{2} \log \left (\frac {1}{9} \, x \log \left (x e^{\left (-x\right )}\right )^{2}\right ) \log \left (x e^{\left (-x\right )}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(x/exp(x))*log(1/9*x*log(x/exp(x))^2)*log(log(1/9*x*log(x/exp(x))^2))+16*log(x/exp(x))-32*x
+32)*exp(2*log(log(1/9*x*log(x/exp(x))^2))/x)^2+(-16*log(x/exp(x))*log(1/9*x*log(x/exp(x))^2)*log(log(1/9*x*lo
g(x/exp(x))^2))+16*log(x/exp(x))-32*x+32)*exp(2*log(log(1/9*x*log(x/exp(x))^2))/x))/x^2/log(x/exp(x))/log(1/9*
x*log(x/exp(x))^2),x, algorithm="giac")

[Out]

integrate(-16*((log(1/9*x*log(x*e^(-x))^2)*log(x*e^(-x))*log(log(1/9*x*log(x*e^(-x))^2)) + 2*x - log(x*e^(-x))
- 2)*log(1/9*x*log(x*e^(-x))^2)^(4/x) + (log(1/9*x*log(x*e^(-x))^2)*log(x*e^(-x))*log(log(1/9*x*log(x*e^(-x))
^2)) + 2*x - log(x*e^(-x)) - 2)*log(1/9*x*log(x*e^(-x))^2)^(2/x))/(x^2*log(1/9*x*log(x*e^(-x))^2)*log(x*e^(-x)
)), x)

________________________________________________________________________________________

maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (-16 \ln \left (x \,{\mathrm e}^{-x}\right ) \ln \left (\frac {x \ln \left (x \,{\mathrm e}^{-x}\right )^{2}}{9}\right ) \ln \left (\ln \left (\frac {x \ln \left (x \,{\mathrm e}^{-x}\right )^{2}}{9}\right )\right )+16 \ln \left (x \,{\mathrm e}^{-x}\right )-32 x +32\right ) {\mathrm e}^{\frac {4 \ln \left (\ln \left (\frac {x \ln \left (x \,{\mathrm e}^{-x}\right )^{2}}{9}\right )\right )}{x}}+\left (-16 \ln \left (x \,{\mathrm e}^{-x}\right ) \ln \left (\frac {x \ln \left (x \,{\mathrm e}^{-x}\right )^{2}}{9}\right ) \ln \left (\ln \left (\frac {x \ln \left (x \,{\mathrm e}^{-x}\right )^{2}}{9}\right )\right )+16 \ln \left (x \,{\mathrm e}^{-x}\right )-32 x +32\right ) {\mathrm e}^{\frac {2 \ln \left (\ln \left (\frac {x \ln \left (x \,{\mathrm e}^{-x}\right )^{2}}{9}\right )\right )}{x}}}{x^{2} \ln \left (x \,{\mathrm e}^{-x}\right ) \ln \left (\frac {x \ln \left (x \,{\mathrm e}^{-x}\right )^{2}}{9}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-16*ln(x/exp(x))*ln(1/9*x*ln(x/exp(x))^2)*ln(ln(1/9*x*ln(x/exp(x))^2))+16*ln(x/exp(x))-32*x+32)*exp(2*ln
(ln(1/9*x*ln(x/exp(x))^2))/x)^2+(-16*ln(x/exp(x))*ln(1/9*x*ln(x/exp(x))^2)*ln(ln(1/9*x*ln(x/exp(x))^2))+16*ln(
x/exp(x))-32*x+32)*exp(2*ln(ln(1/9*x*ln(x/exp(x))^2))/x))/x^2/ln(x/exp(x))/ln(1/9*x*ln(x/exp(x))^2),x)

[Out]

int(((-16*ln(x/exp(x))*ln(1/9*x*ln(x/exp(x))^2)*ln(ln(1/9*x*ln(x/exp(x))^2))+16*ln(x/exp(x))-32*x+32)*exp(2*ln
(ln(1/9*x*ln(x/exp(x))^2))/x)^2+(-16*ln(x/exp(x))*ln(1/9*x*ln(x/exp(x))^2)*ln(ln(1/9*x*ln(x/exp(x))^2))+16*ln(
x/exp(x))-32*x+32)*exp(2*ln(ln(1/9*x*ln(x/exp(x))^2))/x))/x^2/ln(x/exp(x))/ln(1/9*x*ln(x/exp(x))^2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, {\left (-2 \, \log \relax (3) + \log \relax (x) + 2 \, \log \left (-x + \log \relax (x)\right )\right )}^{\frac {4}{x}} + 8 \, {\left (-2 \, \log \relax (3) + \log \relax (x) + 2 \, \log \left (-x + \log \relax (x)\right )\right )}^{\frac {2}{x}} + 16 \, \int -\frac {2 \, {\left ({\left (3 \, x - \log \relax (x) - 2\right )} \log \left (x - \log \relax (x)\right ) - {\left (3 \, x - \log \relax (x) - 2\right )} \log \left (-x + \log \relax (x)\right )\right )} {\left (-2 \, \log \relax (3) + \log \relax (x) + 2 \, \log \left (-x + \log \relax (x)\right )\right )}^{\frac {4}{x}}}{4 \, x^{3} \log \relax (3)^{2} - x^{2} \log \relax (x)^{3} + {\left (x^{3} + 4 \, x^{2} \log \relax (3)\right )} \log \relax (x)^{2} - 2 \, {\left (2 \, x^{3} \log \relax (3) + x^{2} \log \relax (x)^{2} - {\left (x^{3} + 2 \, x^{2} \log \relax (3)\right )} \log \relax (x)\right )} \log \left (x - \log \relax (x)\right ) - 4 \, {\left (x^{3} \log \relax (3) + x^{2} \log \relax (3)^{2}\right )} \log \relax (x) - 2 \, {\left (2 \, x^{3} \log \relax (3) + x^{2} \log \relax (x)^{2} - 2 \, {\left (x^{3} - x^{2} \log \relax (x)\right )} \log \left (x - \log \relax (x)\right ) - {\left (x^{3} + 2 \, x^{2} \log \relax (3)\right )} \log \relax (x)\right )} \log \left (-x + \log \relax (x)\right )}\,{d x} + 16 \, \int -\frac {2 \, {\left ({\left (3 \, x - \log \relax (x) - 2\right )} \log \left (x - \log \relax (x)\right ) - {\left (3 \, x - \log \relax (x) - 2\right )} \log \left (-x + \log \relax (x)\right )\right )} {\left (-2 \, \log \relax (3) + \log \relax (x) + 2 \, \log \left (-x + \log \relax (x)\right )\right )}^{\frac {2}{x}}}{4 \, x^{3} \log \relax (3)^{2} - x^{2} \log \relax (x)^{3} + {\left (x^{3} + 4 \, x^{2} \log \relax (3)\right )} \log \relax (x)^{2} - 2 \, {\left (2 \, x^{3} \log \relax (3) + x^{2} \log \relax (x)^{2} - {\left (x^{3} + 2 \, x^{2} \log \relax (3)\right )} \log \relax (x)\right )} \log \left (x - \log \relax (x)\right ) - 4 \, {\left (x^{3} \log \relax (3) + x^{2} \log \relax (3)^{2}\right )} \log \relax (x) - 2 \, {\left (2 \, x^{3} \log \relax (3) + x^{2} \log \relax (x)^{2} - 2 \, {\left (x^{3} - x^{2} \log \relax (x)\right )} \log \left (x - \log \relax (x)\right ) - {\left (x^{3} + 2 \, x^{2} \log \relax (3)\right )} \log \relax (x)\right )} \log \left (-x + \log \relax (x)\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*log(x/exp(x))*log(1/9*x*log(x/exp(x))^2)*log(log(1/9*x*log(x/exp(x))^2))+16*log(x/exp(x))-32*x
+32)*exp(2*log(log(1/9*x*log(x/exp(x))^2))/x)^2+(-16*log(x/exp(x))*log(1/9*x*log(x/exp(x))^2)*log(log(1/9*x*lo
g(x/exp(x))^2))+16*log(x/exp(x))-32*x+32)*exp(2*log(log(1/9*x*log(x/exp(x))^2))/x))/x^2/log(x/exp(x))/log(1/9*
x*log(x/exp(x))^2),x, algorithm="maxima")

[Out]

4*(-2*log(3) + log(x) + 2*log(-x + log(x)))^(4/x) + 8*(-2*log(3) + log(x) + 2*log(-x + log(x)))^(2/x) + 16*int
egrate(-2*((3*x - log(x) - 2)*log(x - log(x)) - (3*x - log(x) - 2)*log(-x + log(x)))*(-2*log(3) + log(x) + 2*l
og(-x + log(x)))^(4/x)/(4*x^3*log(3)^2 - x^2*log(x)^3 + (x^3 + 4*x^2*log(3))*log(x)^2 - 2*(2*x^3*log(3) + x^2*
log(x)^2 - (x^3 + 2*x^2*log(3))*log(x))*log(x - log(x)) - 4*(x^3*log(3) + x^2*log(3)^2)*log(x) - 2*(2*x^3*log(
3) + x^2*log(x)^2 - 2*(x^3 - x^2*log(x))*log(x - log(x)) - (x^3 + 2*x^2*log(3))*log(x))*log(-x + log(x))), x)
+ 16*integrate(-2*((3*x - log(x) - 2)*log(x - log(x)) - (3*x - log(x) - 2)*log(-x + log(x)))*(-2*log(3) + log(
x) + 2*log(-x + log(x)))^(2/x)/(4*x^3*log(3)^2 - x^2*log(x)^3 + (x^3 + 4*x^2*log(3))*log(x)^2 - 2*(2*x^3*log(3
) + x^2*log(x)^2 - (x^3 + 2*x^2*log(3))*log(x))*log(x - log(x)) - 4*(x^3*log(3) + x^2*log(3)^2)*log(x) - 2*(2*
x^3*log(3) + x^2*log(x)^2 - 2*(x^3 - x^2*log(x))*log(x - log(x)) - (x^3 + 2*x^2*log(3))*log(x))*log(-x + log(x
))), x)

________________________________________________________________________________________

mupad [B]  time = 1.65, size = 58, normalized size = 1.93 \begin {gather*} 4\,{\ln \left (\frac {x^3}{9}-\frac {2\,x^2\,\ln \relax (x)}{9}+\frac {x\,{\ln \relax (x)}^2}{9}\right )}^{2/x}\,\left ({\ln \left (\frac {x^3}{9}-\frac {2\,x^2\,\ln \relax (x)}{9}+\frac {x\,{\ln \relax (x)}^2}{9}\right )}^{2/x}+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*log(log((x*log(x*exp(-x))^2)/9)))/x)*(32*x - 16*log(x*exp(-x)) + 16*log((x*log(x*exp(-x))^2)/9)*l
og(x*exp(-x))*log(log((x*log(x*exp(-x))^2)/9)) - 32) + exp((4*log(log((x*log(x*exp(-x))^2)/9)))/x)*(32*x - 16*
log(x*exp(-x)) + 16*log((x*log(x*exp(-x))^2)/9)*log(x*exp(-x))*log(log((x*log(x*exp(-x))^2)/9)) - 32))/(x^2*lo
g((x*log(x*exp(-x))^2)/9)*log(x*exp(-x))),x)

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-16*ln(x/exp(x))*ln(1/9*x*ln(x/exp(x))**2)*ln(ln(1/9*x*ln(x/exp(x))**2))+16*ln(x/exp(x))-32*x+32)*
exp(2*ln(ln(1/9*x*ln(x/exp(x))**2))/x)**2+(-16*ln(x/exp(x))*ln(1/9*x*ln(x/exp(x))**2)*ln(ln(1/9*x*ln(x/exp(x))
**2))+16*ln(x/exp(x))-32*x+32)*exp(2*ln(ln(1/9*x*ln(x/exp(x))**2))/x))/x**2/ln(x/exp(x))/ln(1/9*x*ln(x/exp(x))
**2),x)

[Out]

Timed out

________________________________________________________________________________________