3.4.27 \(\int \frac {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log (\frac {\log (\log (\log (x)))}{x})}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2(\frac {\log (\log (\log (x)))}{x})} \, dx\)

Optimal. Leaf size=23 \[ -2-16 \left (5+\frac {e^{e^x}}{\log \left (\frac {\log (\log (\log (x)))}{x}\right )}\right ) \]

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Rubi [F]  time = 9.06, 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 {16 e^{e^x}-16 e^{e^x} \log (x) \log (\log (x)) \log (\log (\log (x)))-16 e^{e^x+x} x \log (x) \log (\log (x)) \log (\log (\log (x))) \log \left (\frac {\log (\log (\log (x)))}{x}\right )}{x \log (x) \log (\log (x)) \log (\log (\log (x))) \log ^2\left (\frac {\log (\log (\log (x)))}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 18, normalized size = 0.78 \begin {gather*} -\frac {16 e^{e^x}}{\log \left (\frac {\log (\log (\log (x)))}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16*E^E^x - 16*E^E^x*Log[x]*Log[Log[x]]*Log[Log[Log[x]]] - 16*E^(E^x + x)*x*Log[x]*Log[Log[x]]*Log[L
og[Log[x]]]*Log[Log[Log[Log[x]]]/x])/(x*Log[x]*Log[Log[x]]*Log[Log[Log[x]]]*Log[Log[Log[Log[x]]]/x]^2),x]

[Out]

(-16*E^E^x)/Log[Log[Log[Log[x]]]/x]

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fricas [A]  time = 0.71, size = 16, normalized size = 0.70 \begin {gather*} -\frac {16 \, e^{\left (e^{x}\right )}}{\log \left (\frac {\log \left (\log \left (\log \relax (x)\right )\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(x)*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))*log(log(log(log(x)))/x)-16*log(x)*exp(
exp(x))*log(log(x))*log(log(log(x)))+16*exp(exp(x)))/x/log(x)/log(log(x))/log(log(log(x)))/log(log(log(log(x))
)/x)^2,x, algorithm="fricas")

[Out]

-16*e^(e^x)/log(log(log(log(x)))/x)

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giac [A]  time = 0.39, size = 17, normalized size = 0.74 \begin {gather*} \frac {16 \, e^{\left (e^{x}\right )}}{\log \relax (x) - \log \left (\log \left (\log \left (\log \relax (x)\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(x)*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))*log(log(log(log(x)))/x)-16*log(x)*exp(
exp(x))*log(log(x))*log(log(log(x)))+16*exp(exp(x)))/x/log(x)/log(log(x))/log(log(log(x)))/log(log(log(log(x))
)/x)^2,x, algorithm="giac")

[Out]

16*e^(e^x)/(log(x) - log(log(log(log(x)))))

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maple [C]  time = 0.14, size = 113, normalized size = 4.91

method	result	size
risch	\(-\frac {32 i {\mathrm e}^{{\mathrm e}^{x}}}{\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \ln \left (\ln \left (\ln \relax (x )\right )\right )\right ) \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\ln \relax (x )\right )\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\ln \relax (x )\right )\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \ln \left (\ln \left (\ln \relax (x )\right )\right )\right ) \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\ln \relax (x )\right )\right )}{x}\right )^{2}+\pi \mathrm {csgn}\left (\frac {i \ln \left (\ln \left (\ln \relax (x )\right )\right )}{x}\right )^{3}-2 i \ln \relax (x )+2 i \ln \left (\ln \left (\ln \left (\ln \relax (x )\right )\right )\right )}\)	\(113\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-16*x*exp(x)*ln(x)*exp(exp(x))*ln(ln(x))*ln(ln(ln(x)))*ln(ln(ln(ln(x)))/x)-16*ln(x)*exp(exp(x))*ln(ln(x))
*ln(ln(ln(x)))+16*exp(exp(x)))/x/ln(x)/ln(ln(x))/ln(ln(ln(x)))/ln(ln(ln(ln(x)))/x)^2,x,method=_RETURNVERBOSE)

[Out]

-32*I*exp(exp(x))/(Pi*csgn(I/x)*csgn(I*ln(ln(ln(x))))*csgn(I/x*ln(ln(ln(x))))-Pi*csgn(I/x)*csgn(I/x*ln(ln(ln(x
))))^2-Pi*csgn(I*ln(ln(ln(x))))*csgn(I/x*ln(ln(ln(x))))^2+Pi*csgn(I/x*ln(ln(ln(x))))^3-2*I*ln(x)+2*I*ln(ln(ln(
ln(x)))))

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maxima [A]  time = 0.52, size = 17, normalized size = 0.74 \begin {gather*} \frac {16 \, e^{\left (e^{x}\right )}}{\log \relax (x) - \log \left (\log \left (\log \left (\log \relax (x)\right )\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-16*x*exp(x)*log(x)*exp(exp(x))*log(log(x))*log(log(log(x)))*log(log(log(log(x)))/x)-16*log(x)*exp(
exp(x))*log(log(x))*log(log(log(x)))+16*exp(exp(x)))/x/log(x)/log(log(x))/log(log(log(x)))/log(log(log(log(x))
)/x)^2,x, algorithm="maxima")

[Out]

16*e^(e^x)/(log(x) - log(log(log(log(x)))))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {16\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{{\mathrm {e}}^x}\,\ln \left (\ln \left (\ln \relax (x)\right )\right )\,\ln \relax (x)-16\,{\mathrm {e}}^{{\mathrm {e}}^x}+16\,x\,\ln \left (\ln \relax (x)\right )\,\ln \left (\frac {\ln \left (\ln \left (\ln \relax (x)\right )\right )}{x}\right )\,{\mathrm {e}}^{{\mathrm {e}}^x}\,{\mathrm {e}}^x\,\ln \left (\ln \left (\ln \relax (x)\right )\right )\,\ln \relax (x)}{x\,\ln \left (\ln \relax (x)\right )\,{\ln \left (\frac {\ln \left (\ln \left (\ln \relax (x)\right )\right )}{x}\right )}^2\,\ln \left (\ln \left (\ln \relax (x)\right )\right )\,\ln \relax (x)} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*log(log(x))*exp(exp(x))*log(log(log(x)))*log(x) - 16*exp(exp(x)) + 16*x*log(log(x))*log(log(log(log(x
)))/x)*exp(exp(x))*exp(x)*log(log(log(x)))*log(x))/(x*log(log(x))*log(log(log(log(x)))/x)^2*log(log(log(x)))*l
og(x)),x)

[Out]

int(-(16*log(log(x))*exp(exp(x))*log(log(log(x)))*log(x) - 16*exp(exp(x)) + 16*x*log(log(x))*log(log(log(log(x
)))/x)*exp(exp(x))*exp(x)*log(log(log(x)))*log(x))/(x*log(log(x))*log(log(log(log(x)))/x)^2*log(log(log(x)))*l
og(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((-16*x*exp(x)*ln(x)*exp(exp(x))*ln(ln(x))*ln(ln(ln(x)))*ln(ln(ln(ln(x)))/x)-16*ln(x)*exp(exp(x))*ln(
ln(x))*ln(ln(ln(x)))+16*exp(exp(x)))/x/ln(x)/ln(ln(x))/ln(ln(ln(x)))/ln(ln(ln(ln(x)))/x)**2,x)

[Out]

Timed out

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