3.15.7 \(\int \frac {e^{\frac {2 (4 x-36 x^2+(3-108 x) \log (\log (\frac {e^x}{x})))}{x+3 \log (\log (\frac {e^x}{x}))}} (18 x^2-18 x^3+(2 x^3-72 x^4) \log (\frac {e^x}{x})+(30 x^2-432 x^3) \log (\frac {e^x}{x}) \log (\log (\frac {e^x}{x}))+(18 x-648 x^2) \log (\frac {e^x}{x}) \log ^2(\log (\frac {e^x}{x})))}{x^2 \log (\frac {e^x}{x})+6 x \log (\frac {e^x}{x}) \log (\log (\frac {e^x}{x}))+9 \log (\frac {e^x}{x}) \log ^2(\log (\frac {e^x}{x}))} \, dx\)

Optimal. Leaf size=30 \[ e^{2-2 x \left (36-\frac {3}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right )} x^2 \]

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Rubi [F]  time = 9.35, 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 {\exp \left (\frac {2 \left (4 x-36 x^2+(3-108 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) \left (18 x^2-18 x^3+\left (2 x^3-72 x^4\right ) \log \left (\frac {e^x}{x}\right )+\left (30 x^2-432 x^3\right ) \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+\left (18 x-648 x^2\right ) \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )}{x^2 \log \left (\frac {e^x}{x}\right )+6 x \log \left (\frac {e^x}{x}\right ) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 \log \left (\frac {e^x}{x}\right ) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((2*(4*x - 36*x^2 + (3 - 108*x)*Log[Log[E^x/x]]))/(x + 3*Log[Log[E^x/x]]))*(18*x^2 - 18*x^3 + (2*x^3 -
72*x^4)*Log[E^x/x] + (30*x^2 - 432*x^3)*Log[E^x/x]*Log[Log[E^x/x]] + (18*x - 648*x^2)*Log[E^x/x]*Log[Log[E^x/x
]]^2))/(x^2*Log[E^x/x] + 6*x*Log[E^x/x]*Log[Log[E^x/x]] + 9*Log[E^x/x]*Log[Log[E^x/x]]^2),x]

[Out]

2*Defer[Int][E^((8*(1 - 9*x)*x)/(x + 3*Log[Log[E^x/x]]))*x*Log[E^x/x]^((6 - 216*x)/(x + 3*Log[Log[E^x/x]])), x
] - 72*Defer[Int][E^((8*(1 - 9*x)*x)/(x + 3*Log[Log[E^x/x]]))*x^2*Log[E^x/x]^((6 - 216*x)/(x + 3*Log[Log[E^x/x
]])), x] - 6*Defer[Int][(E^((8*(1 - 9*x)*x)/(x + 3*Log[Log[E^x/x]]))*x^3*Log[E^x/x]^((6 - 216*x)/(x + 3*Log[Lo
g[E^x/x]])))/(x + 3*Log[Log[E^x/x]])^2, x] + 18*Defer[Int][(E^((8*(1 - 9*x)*x)/(x + 3*Log[Log[E^x/x]]))*x^2*Lo
g[E^x/x]^(-1 + (6 - 216*x)/(x + 3*Log[Log[E^x/x]])))/(x + 3*Log[Log[E^x/x]])^2, x] - 18*Defer[Int][(E^((8*(1 -
9*x)*x)/(x + 3*Log[Log[E^x/x]]))*x^3*Log[E^x/x]^(-1 + (6 - 216*x)/(x + 3*Log[Log[E^x/x]])))/(x + 3*Log[Log[E^
x/x]])^2, x] + 6*Defer[Int][(E^((8*(1 - 9*x)*x)/(x + 3*Log[Log[E^x/x]]))*x^2*Log[E^x/x]^((6 - 216*x)/(x + 3*Lo
g[Log[E^x/x]])))/(x + 3*Log[Log[E^x/x]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right ) \left (-9 (-1+x) x-\log \left (\frac {e^x}{x}\right ) \left (x^2 (-1+36 x)+3 x (-5+72 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 (-1+36 x) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2} \, dx\\ &=2 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right ) \left (-9 (-1+x) x-\log \left (\frac {e^x}{x}\right ) \left (x^2 (-1+36 x)+3 x (-5+72 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )+9 (-1+36 x) \log ^2\left (\log \left (\frac {e^x}{x}\right )\right )\right )\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2} \, dx\\ &=2 \int \left (-e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x (-1+36 x) \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )-\frac {3 e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right ) \left (-3+3 x+x \log \left (\frac {e^x}{x}\right )\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}+\frac {3 e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}\right ) \, dx\\ &=-\left (2 \int e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x (-1+36 x) \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right ) \, dx\right )-6 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right ) \left (-3+3 x+x \log \left (\frac {e^x}{x}\right )\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2} \, dx+6 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )} \, dx\\ &=-\left (2 \int \left (-e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )+36 e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )\right ) \, dx\right )+6 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )} \, dx-6 \int \left (\frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^3 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}-\frac {3 e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}+\frac {3 e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^3 \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2}\right ) \, dx\\ &=2 \int e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right ) \, dx-6 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^3 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2} \, dx+6 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )} \, dx+18 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2} \, dx-18 \int \frac {e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^3 \log ^{-1+\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right )}{\left (x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )\right )^2} \, dx-72 \int e^{\frac {8 (1-9 x) x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \log ^{\frac {6-216 x}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}}\left (\frac {e^x}{x}\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 46, normalized size = 1.53 \begin {gather*} e^{\frac {8 (1-9 x) x+(6-216 x) \log \left (\log \left (\frac {e^x}{x}\right )\right )}{x+3 \log \left (\log \left (\frac {e^x}{x}\right )\right )}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((2*(4*x - 36*x^2 + (3 - 108*x)*Log[Log[E^x/x]]))/(x + 3*Log[Log[E^x/x]]))*(18*x^2 - 18*x^3 + (2*
x^3 - 72*x^4)*Log[E^x/x] + (30*x^2 - 432*x^3)*Log[E^x/x]*Log[Log[E^x/x]] + (18*x - 648*x^2)*Log[E^x/x]*Log[Log
[E^x/x]]^2))/(x^2*Log[E^x/x] + 6*x*Log[E^x/x]*Log[Log[E^x/x]] + 9*Log[E^x/x]*Log[Log[E^x/x]]^2),x]

[Out]

E^((8*(1 - 9*x)*x + (6 - 216*x)*Log[Log[E^x/x]])/(x + 3*Log[Log[E^x/x]]))*x^2

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fricas [A]  time = 0.76, size = 45, normalized size = 1.50 \begin {gather*} x^{2} e^{\left (-\frac {2 \, {\left (36 \, x^{2} + 3 \, {\left (36 \, x - 1\right )} \log \left (\log \left (\frac {e^{x}}{x}\right )\right ) - 4 \, x\right )}}{x + 3 \, \log \left (\log \left (\frac {e^{x}}{x}\right )\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-648*x^2+18*x)*log(exp(x)/x)*log(log(exp(x)/x))^2+(-432*x^3+30*x^2)*log(exp(x)/x)*log(log(exp(x)/x
))+(-72*x^4+2*x^3)*log(exp(x)/x)-18*x^3+18*x^2)*exp(((-108*x+3)*log(log(exp(x)/x))-36*x^2+4*x)/(3*log(log(exp(
x)/x))+x))^2/(9*log(exp(x)/x)*log(log(exp(x)/x))^2+6*x*log(exp(x)/x)*log(log(exp(x)/x))+x^2*log(exp(x)/x)),x,
algorithm="fricas")

[Out]

x^2*e^(-2*(36*x^2 + 3*(36*x - 1)*log(log(e^x/x)) - 4*x)/(x + 3*log(log(e^x/x))))

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giac [A]  time = 71.17, size = 48, normalized size = 1.60 \begin {gather*} x^{2} e^{\left (-\frac {2 \, {\left (36 \, x^{2} + 108 \, x \log \left (x - \log \relax (x)\right ) - 4 \, x - 3 \, \log \left (x - \log \relax (x)\right )\right )}}{x + 3 \, \log \left (x - \log \relax (x)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-648*x^2+18*x)*log(exp(x)/x)*log(log(exp(x)/x))^2+(-432*x^3+30*x^2)*log(exp(x)/x)*log(log(exp(x)/x
))+(-72*x^4+2*x^3)*log(exp(x)/x)-18*x^3+18*x^2)*exp(((-108*x+3)*log(log(exp(x)/x))-36*x^2+4*x)/(3*log(log(exp(
x)/x))+x))^2/(9*log(exp(x)/x)*log(log(exp(x)/x))^2+6*x*log(exp(x)/x)*log(log(exp(x)/x))+x^2*log(exp(x)/x)),x,
algorithm="giac")

[Out]

x^2*e^(-2*(36*x^2 + 108*x*log(x - log(x)) - 4*x - 3*log(x - log(x)))/(x + 3*log(x - log(x))))

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maple [C]  time = 0.51, size = 205, normalized size = 6.83

method	result	size
risch	\(x^{2} {\mathrm e}^{-\frac {2 \left (108 \ln \left (-\ln \relax (x )+\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}\right ) x +36 x^{2}-3 \ln \left (-\ln \relax (x )+\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}\right )-4 x \right )}{3 \ln \left (-\ln \relax (x )+\ln \left ({\mathrm e}^{x}\right )-\frac {i \pi \,\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\mathrm {csgn}\left (\frac {i}{x}\right )\right ) \left (-\mathrm {csgn}\left (\frac {i {\mathrm e}^{x}}{x}\right )+\mathrm {csgn}\left (i {\mathrm e}^{x}\right )\right )}{2}\right )+x}}\)	\(205\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-648*x^2+18*x)*ln(exp(x)/x)*ln(ln(exp(x)/x))^2+(-432*x^3+30*x^2)*ln(exp(x)/x)*ln(ln(exp(x)/x))+(-72*x^4+
2*x^3)*ln(exp(x)/x)-18*x^3+18*x^2)*exp(((-108*x+3)*ln(ln(exp(x)/x))-36*x^2+4*x)/(3*ln(ln(exp(x)/x))+x))^2/(9*l
n(exp(x)/x)*ln(ln(exp(x)/x))^2+6*x*ln(exp(x)/x)*ln(ln(exp(x)/x))+x^2*ln(exp(x)/x)),x,method=_RETURNVERBOSE)

[Out]

x^2*exp(-2*(108*ln(-ln(x)+ln(exp(x))-1/2*I*Pi*csgn(I/x*exp(x))*(-csgn(I/x*exp(x))+csgn(I/x))*(-csgn(I/x*exp(x)
)+csgn(I*exp(x))))*x+36*x^2-3*ln(-ln(x)+ln(exp(x))-1/2*I*Pi*csgn(I/x*exp(x))*(-csgn(I/x*exp(x))+csgn(I/x))*(-c
sgn(I/x*exp(x))+csgn(I*exp(x))))-4*x)/(3*ln(-ln(x)+ln(exp(x))-1/2*I*Pi*csgn(I/x*exp(x))*(-csgn(I/x*exp(x))+csg
n(I/x))*(-csgn(I/x*exp(x))+csgn(I*exp(x))))+x))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-648*x^2+18*x)*log(exp(x)/x)*log(log(exp(x)/x))^2+(-432*x^3+30*x^2)*log(exp(x)/x)*log(log(exp(x)/x
))+(-72*x^4+2*x^3)*log(exp(x)/x)-18*x^3+18*x^2)*exp(((-108*x+3)*log(log(exp(x)/x))-36*x^2+4*x)/(3*log(log(exp(
x)/x))+x))^2/(9*log(exp(x)/x)*log(log(exp(x)/x))^2+6*x*log(exp(x)/x)*log(log(exp(x)/x))+x^2*log(exp(x)/x)),x,
algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{-\frac {2\,\left (\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )\,\left (108\,x-3\right )-4\,x+36\,x^2\right )}{x+3\,\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )}}\,\left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,\left (2\,x^3-72\,x^4\right )+18\,x^2-18\,x^3+\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )\,\left (30\,x^2-432\,x^3\right )+\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,{\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )}^2\,\left (18\,x-648\,x^2\right )\right )}{9\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,{\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )}^2+x^2\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )+6\,x\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\,\ln \left (\ln \left (\frac {{\mathrm {e}}^x}{x}\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-(2*(log(log(exp(x)/x))*(108*x - 3) - 4*x + 36*x^2))/(x + 3*log(log(exp(x)/x))))*(log(exp(x)/x)*(2*x^
3 - 72*x^4) + 18*x^2 - 18*x^3 + log(exp(x)/x)*log(log(exp(x)/x))*(30*x^2 - 432*x^3) + log(exp(x)/x)*log(log(ex
p(x)/x))^2*(18*x - 648*x^2)))/(9*log(exp(x)/x)*log(log(exp(x)/x))^2 + x^2*log(exp(x)/x) + 6*x*log(exp(x)/x)*lo
g(log(exp(x)/x))),x)

[Out]

int((exp(-(2*(log(log(exp(x)/x))*(108*x - 3) - 4*x + 36*x^2))/(x + 3*log(log(exp(x)/x))))*(log(exp(x)/x)*(2*x^
3 - 72*x^4) + 18*x^2 - 18*x^3 + log(exp(x)/x)*log(log(exp(x)/x))*(30*x^2 - 432*x^3) + log(exp(x)/x)*log(log(ex
p(x)/x))^2*(18*x - 648*x^2)))/(9*log(exp(x)/x)*log(log(exp(x)/x))^2 + x^2*log(exp(x)/x) + 6*x*log(exp(x)/x)*lo
g(log(exp(x)/x))), x)

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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(((-648*x**2+18*x)*ln(exp(x)/x)*ln(ln(exp(x)/x))**2+(-432*x**3+30*x**2)*ln(exp(x)/x)*ln(ln(exp(x)/x))
+(-72*x**4+2*x**3)*ln(exp(x)/x)-18*x**3+18*x**2)*exp(((-108*x+3)*ln(ln(exp(x)/x))-36*x**2+4*x)/(3*ln(ln(exp(x)
/x))+x))**2/(9*ln(exp(x)/x)*ln(ln(exp(x)/x))**2+6*x*ln(exp(x)/x)*ln(ln(exp(x)/x))+x**2*ln(exp(x)/x)),x)

[Out]

Timed out

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