3.24.45 \(\int \frac {-100+100 x \log (x)+100 \log (x) \log (\frac {1}{3} e^{-x} \log (x)) \log (\log (\frac {1}{3} e^{-x} \log (x)))}{\log (x) \log (\frac {1}{3} e^{-x} \log (x)) \log ^2(\log (\frac {1}{3} e^{-x} \log (x)))} \, dx\) [2345]

3.24.45.1 Optimal result

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

output

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

3.24.45.2 Mathematica [A] (verified)

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

input

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

output

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

3.24.45.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7293

\(\displaystyle -100 \int \left (\frac {1-x \log (x)}{\log (x) \log \left (\frac {1}{3} e^{-x} \log (x)\right ) \log ^2\left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}-\frac {1}{\log \left (\log \left (\frac {1}{3} e^{-x} \log (x)\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

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

input

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]
 

output

$Aborted
 

3.24.45.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.24.45.4 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	\(\frac {100 x}{\ln \left (-\ln \left (3\right )-\ln \left ({\mathrm e}^{x}\right )+\ln \left (\ln \left (x \right )\right )-\frac {i \pi \,\operatorname {csgn}\left (i {\mathrm e}^{-x} \ln \left (x \right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-x} \ln \left (x \right )\right )+\operatorname {csgn}\left (i {\mathrm e}^{-x}\right )\right ) \left (-\operatorname {csgn}\left (i {\mathrm e}^{-x} \ln \left (x \right )\right )+\operatorname {csgn}\left (i \ln \left (x \right )\right )\right )}{2}\right )}\)	\(74\)

input

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

output

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

3.24.45.5 Fricas [A] (verification not implemented)

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

input

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

output

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

3.24.45.6 Sympy [A] (verification not implemented)

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

input

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

output

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

3.24.45.7 Maxima [A] (verification not implemented)

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

input

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

output

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

3.24.45.8 Giac [A] (verification not implemented)

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

input

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

output

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

3.24.45.9 Mupad [B] (verification not implemented)

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

input

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

output

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))