\(\int \frac {-100+100 x \log (x)+100 \log (x) \log (\frac {1}{3} e^{-x} \log (x)) \log (\log (\frac {1}{3} e^{-x} \log (x)))}{\log (x) \log (\frac {1}{3} e^{-x} \log (x)) \log ^2(\log (\frac {1}{3} e^{-x} \log (x)))} \, dx\) [8345]

Optimal result

Integrand size = 70, antiderivative size = 18 \[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=\frac {100 x}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \]

[Out]

100*x/ln(ln(1/3*ln(x)/exp(x)))

Rubi [F]

\[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=\int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=\frac {100 x}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \]

[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)]]

Maple [A] (verified)

Time = 1.07 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89

method	result	size
parallelrisch	\(\frac {100 x}{\ln \left (\ln \left (\frac {\ln \left (x \right ) {\mathrm e}^{-x}}{3}\right )\right )}\)	\(16\)
risch	\(\frac {100 x}{\ln \left (-\ln \left (3\right )-\ln \left ({\mathrm e}^{x}\right )+\ln \left (\ln \left (x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x} \ln \left (x \right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-x} \ln \left (x \right )\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-x} \ln \left (x \right )\right )+\operatorname {csgn}\left (i \ln \left (x \right )\right )\right )}{2}\right )}\)	\(74\)

[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(1/3*ln(x)/exp(x)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=\frac {100 \, x}{\log \left (\log \left (\frac {1}{3} \, e^{\left (-x\right )} \log \left (x\right )\right )\right )} \]

[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)))

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=\frac {100 x}{\log {\left (\log {\left (\frac {e^{- x} \log {\left (x \right )}}{3} \right )} \right )}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=\frac {100 \, x}{\log \left (-x - \log \left (3\right ) + \log \left (\log \left (x\right )\right )\right )} \]

[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)))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=\frac {100 \, x}{\log \left (-x - \log \left (3\right ) + \log \left (\log \left (x\right )\right )\right )} \]

[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)))

Mupad [B] (verification not implemented)

Time = 14.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 4.28 \[ \int \frac {-100+100 x \log (x)+100 \log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )} \, dx=100\,x-100\,\ln \left (\ln \left (x\right )\right )-\frac {100\,\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \left (x\right )}{3}\right )}{x\,\ln \left (x\right )-1}+\frac {100\,x+\frac {100\,x\,\ln \left (\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \left (x\right )}{3}\right )\right )\,\ln \left (x\right )\,\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \left (x\right )}{3}\right )}{x\,\ln \left (x\right )-1}}{\ln \left (\ln \left (\frac {{\mathrm {e}}^{-x}\,\ln \left (x\right )}{3}\right )\right )} \]

[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))