3.2.64 \(\int \frac {e^x+e^x \log (\frac {x}{2})-18 \log (\frac {x}{2}) \log (\frac {x \log (\frac {x}{2})}{-4+\log (2)})+e^x (-1+x) \log (\frac {x}{2}) \log (\frac {x \log (\frac {x}{2})}{-4+\log (2)}) \log (\log (\frac {x \log (\frac {x}{2})}{-4+\log (2)}))}{6 x^2 \log (\frac {x}{2}) \log (\frac {x \log (\frac {x}{2})}{-4+\log (2)})} \, dx\) [164]

3.2.64.1 Optimal result

Integrand size = 113, antiderivative size = 32 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {6+\frac {1}{3} e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{2 x} \]

output

1/2*(6+1/3*ln(ln(x*ln(1/2*x)/(ln(2)-4)))*exp(x))/x
 

3.2.64.2 Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {18+e^x \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x} \]

input

Integrate[(E^x + E^x*Log[x/2] - 18*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2]) 
] + E^x*(-1 + x)*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]*Log[Log[(x*Log[x 
/2])/(-4 + Log[2])]])/(6*x^2*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]),x]
 

output

(18 + E^x*Log[Log[(x*Log[x/2])/(-4 + Log[2])]])/(6*x)
 

3.2.64.3 Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {27, 7293, 2009}

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.

\(\displaystyle \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right ) \log \left (\frac {x}{2}\right )+e^x (x-1) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right )\right ) \log \left (\frac {x}{2}\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{\log (2)-4}\right )} \, dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{6} \int \frac {e^x \log \left (\frac {x}{2}\right )-18 \log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right ) \log \left (\frac {x}{2}\right )-e^x (1-x) \log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right ) \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right ) \log \left (\frac {x}{2}\right )+e^x}{x^2 \log \left (\frac {x}{2}\right ) \log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \frac {1}{6} \int \left (\frac {e^x \left (x \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right ) \log \left (\frac {x}{2}\right )-\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right ) \log \left (\frac {x}{2}\right )+\log \left (\frac {x}{2}\right )+1\right )}{x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )}-\frac {18}{x^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{6} \left (\frac {18}{x}+\frac {e^x \log \left (\log \left (-\frac {x \log \left (\frac {x}{2}\right )}{4-\log (2)}\right )\right )}{x}\right )\)

input

Int[(E^x + E^x*Log[x/2] - 18*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])] + E^ 
x*(-1 + x)*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]*Log[Log[(x*Log[x/2])/( 
-4 + Log[2])]])/(6*x^2*Log[x/2]*Log[(x*Log[x/2])/(-4 + Log[2])]),x]
 

output

(18/x + (E^x*Log[Log[-((x*Log[x/2])/(4 - Log[2]))]])/x)/6
 

3.2.64.3.1 Defintions of rubi rules used

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 7293

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

3.2.64.4 Maple [A] (verified)

Time = 1.54 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
parallelrisch	\(\frac {18+\ln \left (\ln \left (\frac {x \ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\right ) {\mathrm e}^{x}}{6 x}\)	\(25\)

input

int(1/6*((-1+x)*exp(x)*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))*ln(ln(x*ln(1/2* 
x)/(ln(2)-4)))-18*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))+exp(x)*ln(1/2*x)+exp 
(x))/x^2/ln(1/2*x)/ln(x*ln(1/2*x)/(ln(2)-4)),x,method=_RETURNVERBOSE)
 

output

1/6/x*(18+ln(ln(x*ln(1/2*x)/(ln(2)-4)))*exp(x))
 

3.2.64.5 Fricas [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (\frac {x \log \left (\frac {1}{2} \, x\right )}{\log \left (2\right ) - 4}\right )\right ) + 18}{6 \, x} \]

input

integrate(1/6*((-1+x)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(l 
og(x*log(1/2*x)/(log(2)-4)))-18*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))+ex 
p(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, alg 
orithm=\
 

output

1/6*(e^x*log(log(x*log(1/2*x)/(log(2) - 4))) + 18)/x
 

3.2.64.6 Sympy [A] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log {\left (\log {\left (\frac {x \log {\left (\frac {x}{2} \right )}}{-4 + \log {\left (2 \right )}} \right )} \right )}}{6 x} + \frac {3}{x} \]

input

integrate(1/6*((-1+x)*exp(x)*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))*ln(ln(x*l 
n(1/2*x)/(ln(2)-4)))-18*ln(1/2*x)*ln(x*ln(1/2*x)/(ln(2)-4))+exp(x)*ln(1/2* 
x)+exp(x))/x**2/ln(1/2*x)/ln(x*ln(1/2*x)/(ln(2)-4)),x)
 

output

exp(x)*log(log(x*log(x/2)/(-4 + log(2))))/(6*x) + 3/x
 

3.2.64.7 Maxima [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (x\right ) - \log \left (\log \left (2\right ) - 4\right ) + \log \left (-\log \left (2\right ) + \log \left (x\right )\right )\right )}{6 \, x} + \frac {3}{x} \]

input

integrate(1/6*((-1+x)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(l 
og(x*log(1/2*x)/(log(2)-4)))-18*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))+ex 
p(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, alg 
orithm=\
 

output

1/6*e^x*log(log(x) - log(log(2) - 4) + log(-log(2) + log(x)))/x + 3/x
 

3.2.64.8 Giac [A] (verification not implemented)

Time = 0.70 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\frac {e^{x} \log \left (\log \left (x\right ) - \log \left (\log \left (2\right ) - 4\right ) + \log \left (-\log \left (2\right ) + \log \left (x\right )\right )\right ) + 18}{6 \, x} \]

input

integrate(1/6*((-1+x)*exp(x)*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))*log(l 
og(x*log(1/2*x)/(log(2)-4)))-18*log(1/2*x)*log(x*log(1/2*x)/(log(2)-4))+ex 
p(x)*log(1/2*x)+exp(x))/x^2/log(1/2*x)/log(x*log(1/2*x)/(log(2)-4)),x, alg 
orithm=\
 

output

1/6*(e^x*log(log(x) - log(log(2) - 4) + log(-log(2) + log(x))) + 18)/x
 

3.2.64.9 Mupad [F(-1)]

Timed out. \[ \int \frac {e^x+e^x \log \left (\frac {x}{2}\right )-18 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )+e^x (-1+x) \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right ) \log \left (\log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )\right )}{6 x^2 \log \left (\frac {x}{2}\right ) \log \left (\frac {x \log \left (\frac {x}{2}\right )}{-4+\log (2)}\right )} \, dx=\int \frac {\frac {{\mathrm {e}}^x}{6}-3\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )+\frac {\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x}{6}+\frac {\ln \left (\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\right )\,\ln \left (\frac {x}{2}\right )\,{\mathrm {e}}^x\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )\,\left (x-1\right )}{6}}{x^2\,\ln \left (\frac {x}{2}\right )\,\ln \left (\frac {x\,\ln \left (\frac {x}{2}\right )}{\ln \left (2\right )-4}\right )} \,d x \]

input

int((exp(x)/6 - 3*log(x/2)*log((x*log(x/2))/(log(2) - 4)) + (log(x/2)*exp( 
x))/6 + (log(log((x*log(x/2))/(log(2) - 4)))*log(x/2)*exp(x)*log((x*log(x/ 
2))/(log(2) - 4))*(x - 1))/6)/(x^2*log(x/2)*log((x*log(x/2))/(log(2) - 4)) 
),x)
 

output

int((exp(x)/6 - 3*log(x/2)*log((x*log(x/2))/(log(2) - 4)) + (log(x/2)*exp( 
x))/6 + (log(log((x*log(x/2))/(log(2) - 4)))*log(x/2)*exp(x)*log((x*log(x/ 
2))/(log(2) - 4))*(x - 1))/6)/(x^2*log(x/2)*log((x*log(x/2))/(log(2) - 4)) 
), x)