3.84.45 \(\int \frac {-100+100 x \log (x)+100 \log (x) \log (\frac {1}{3} e^{-x} \log (x)) \log (\log (\frac {1}{3} e^{-x} \log (x)))}{\log (x) \log (\frac {1}{3} e^{-x} \log (x)) \log ^2(\log (\frac {1}{3} e^{-x} \log (x)))} \, dx\)

Optimal. Leaf size=18 \[ \frac {100 x}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \]

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Rubi [F]  time = 0.64, 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 {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-100 + 100*x*Log[x] + 100*Log[x]*Log[Log[x]/(3*E^x)]*Log[Log[Log[x]/(3*E^x)]])/(Log[x]*Log[Log[x]/(3*E^x)
]*Log[Log[Log[x]/(3*E^x)]]^2),x]

[Out]

100*Defer[Int][x/(Log[Log[x]/(3*E^x)]*Log[Log[Log[x]/(3*E^x)]]^2), x] - 100*Defer[Int][1/(Log[x]*Log[Log[x]/(3
*E^x)]*Log[Log[Log[x]/(3*E^x)]]^2), x] + 100*Defer[Int][Log[Log[Log[x]/(3*E^x)]]^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {100 \left (-1+x \log (x)+\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx\\ &=100 \int \frac {-1+x \log (x)+\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx\\ &=100 \int \left (\frac {-1+x \log (x)}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}+\frac {1}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}\right ) \, dx\\ &=100 \int \frac {-1+x \log (x)}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx+100 \int \frac {1}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx\\ &=100 \int \left (\frac {x}{\log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}-\frac {1}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}\right ) \, dx+100 \int \frac {1}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx\\ &=100 \int \frac {x}{\log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx-100 \int \frac {1}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx+100 \int \frac {1}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.18, size = 18, normalized size = 1.00 \begin {gather*} \frac {100 x}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100 + 100*x*Log[x] + 100*Log[x]*Log[Log[x]/(3*E^x)]*Log[Log[Log[x]/(3*E^x)]])/(Log[x]*Log[Log[x]/(
3*E^x)]*Log[Log[Log[x]/(3*E^x)]]^2),x]

[Out]

(100*x)/Log[Log[Log[x]/(3*E^x)]]

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fricas [A]  time = 0.84, size = 15, normalized size = 0.83 \begin {gather*} \frac {100 \, x}{\log \left (\log \left (\frac {1}{3} \, e^{\left (-x\right )} \log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*log(x)*log(1/3*log(x)/exp(x))*log(log(1/3*log(x)/exp(x)))+100*x*log(x)-100)/log(x)/log(1/3*log(
x)/exp(x))/log(log(1/3*log(x)/exp(x)))^2,x, algorithm="fricas")

[Out]

100*x/log(log(1/3*e^(-x)*log(x)))

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giac [A]  time = 0.17, size = 17, normalized size = 0.94 \begin {gather*} \frac {100 \, x}{\log \left (-x - \log \relax (3) + \log \left (\log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*log(x)*log(1/3*log(x)/exp(x))*log(log(1/3*log(x)/exp(x)))+100*x*log(x)-100)/log(x)/log(1/3*log(
x)/exp(x))/log(log(1/3*log(x)/exp(x)))^2,x, algorithm="giac")

[Out]

100*x/log(-x - log(3) + log(log(x)))

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maple [C]  time = 0.18, size = 74, normalized size = 4.11

method	result	size
risch	\(\frac {100 x}{\ln \left (-\ln \relax (3)-\ln \left ({\mathrm e}^{x}\right )+\ln \left (\ln \relax (x )\right )-\frac {i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x} \ln \relax (x )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-x} \ln \relax (x )\right )+\mathrm {csgn}\left (i {\mathrm e}^{-x}\right )\right ) \left (-\mathrm {csgn}\left (i {\mathrm e}^{-x} \ln \relax (x )\right )+\mathrm {csgn}\left (i \ln \relax (x )\right )\right )}{2}\right )}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*ln(x)*ln(1/3*ln(x)/exp(x))*ln(ln(1/3*ln(x)/exp(x)))+100*x*ln(x)-100)/ln(x)/ln(1/3*ln(x)/exp(x))/ln(ln
(1/3*ln(x)/exp(x)))^2,x,method=_RETURNVERBOSE)

[Out]

100*x/ln(-ln(3)-ln(exp(x))+ln(ln(x))-1/2*I*Pi*csgn(I*exp(-x)*ln(x))*(-csgn(I*exp(-x)*ln(x))+csgn(I*exp(-x)))*(
-csgn(I*exp(-x)*ln(x))+csgn(I*ln(x))))

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maxima [A]  time = 0.49, size = 17, normalized size = 0.94 \begin {gather*} \frac {100 \, x}{\log \left (-x - \log \relax (3) + \log \left (\log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*log(x)*log(1/3*log(x)/exp(x))*log(log(1/3*log(x)/exp(x)))+100*x*log(x)-100)/log(x)/log(1/3*log(
x)/exp(x))/log(log(1/3*log(x)/exp(x)))^2,x, algorithm="maxima")

[Out]

100*x/log(-x - log(3) + log(log(x)))

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mupad [B]  time = 5.67, size = 77, normalized size = 4.28 \begin {gather*} 100\,x-100\,\ln \left (\ln \relax (x)\right )-\frac {100\,\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \relax (x)}{3}\right )}{x\,\ln \relax (x)-1}+\frac {100\,x+\frac {100\,x\,\ln \left (\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \relax (x)}{3}\right )\right )\,\ln \relax (x)\,\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \relax (x)}{3}\right )}{x\,\ln \relax (x)-1}}{\ln \left (\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \relax (x)}{3}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*x*log(x) + 100*log(log((exp(-x)*log(x))/3))*log(x)*log((exp(-x)*log(x))/3) - 100)/(log(log((exp(-x)*l
og(x))/3))^2*log(x)*log((exp(-x)*log(x))/3)),x)

[Out]

100*x - 100*log(log(x)) - (100*log((exp(-x)*log(x))/3))/(x*log(x) - 1) + (100*x + (100*x*log(log((exp(-x)*log(
x))/3))*log(x)*log((exp(-x)*log(x))/3))/(x*log(x) - 1))/log(log((exp(-x)*log(x))/3))

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sympy [A]  time = 0.70, size = 14, normalized size = 0.78 \begin {gather*} \frac {100 x}{\log {\left (\log {\left (\frac {e^{- x} \log {\relax (x )}}{3} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*ln(x)*ln(1/3*ln(x)/exp(x))*ln(ln(1/3*ln(x)/exp(x)))+100*x*ln(x)-100)/ln(x)/ln(1/3*ln(x)/exp(x))
/ln(ln(1/3*ln(x)/exp(x)))**2,x)

[Out]

100*x/log(log(exp(-x)*log(x)/3))

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