\(\int \frac {x \log ^2(\frac {x}{2})+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log (\frac {x}{2})}{\log (2) \log (\frac {x}{2})}} (-27+9 x-9 x \log (\frac {x}{2}))}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log (\frac {x}{2})}{\log (2) \log (\frac {x}{2})}} x \log ^2(\frac {x}{2})+x^2 \log ^2(\frac {x}{2})} \, dx\) [706]

Optimal result

Integrand size = 126, antiderivative size = 33 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\log \left (\left (e^{3 \left (5+\frac {2}{\log (2)}+\frac {3 (3-x)}{\log \left (\frac {x}{2}\right )}\right )}+x\right ) \log (4)\right ) \] Output:

ln(2*(exp(9*(3-x)/ln(1/2*x)+15+6/ln(2))+x)*ln(2))
 

Mathematica [A] (verified)

Time = 1.31 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\log \left (e^{15+\frac {6}{\log (2)}-\frac {9 (-3+x)}{\log \left (\frac {x}{2}\right )}}+x\right ) \] Input:

Integrate[(x*Log[x/2]^2 + E^(((27 - 9*x)*Log[2] + (6 + 15*Log[2])*Log[x/2] 
)/(Log[2]*Log[x/2]))*(-27 + 9*x - 9*x*Log[x/2]))/(E^(((27 - 9*x)*Log[2] + 
(6 + 15*Log[2])*Log[x/2])/(Log[2]*Log[x/2]))*x*Log[x/2]^2 + x^2*Log[x/2]^2 
),x]
 

Output:

Log[E^(15 + 6/Log[2] - (9*(-3 + x))/Log[x/2]) + x]
 

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.

\(\displaystyle \int \frac {\left (9 x-9 x \log \left (\frac {x}{2}\right )-27\right ) \exp \left (\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}\right )+x \log ^2\left (\frac {x}{2}\right )}{x \log ^2\left (\frac {x}{2}\right ) \exp \left (\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} \left (\left (9 x-9 x \log \left (\frac {x}{2}\right )-27\right ) \exp \left (\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}\right )+x \log ^2\left (\frac {x}{2}\right )\right )}{x \left (x e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}}+e^{\frac {27}{\log \left (\frac {x}{2}\right )}+15+\frac {6}{\log (2)}}\right ) \log ^2\left (\frac {x}{2}\right )}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle \int \left (\frac {e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} \left (-9 x+\log ^2\left (\frac {x}{2}\right )+9 x \log \left (\frac {x}{2}\right )+27\right )}{\left (x e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}}+e^{\frac {27}{\log \left (\frac {x}{2}\right )}+15+\frac {6}{\log (2)}}\right ) \log ^2\left (\frac {x}{2}\right )}+\frac {9 (x (-\log (x))+x (1+\log (2))-3)}{x \log ^2\left (\frac {x}{2}\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 27 \int \frac {e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}}}{\left (e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} x+e^{\frac {6}{\log (2)}+15+\frac {27}{\log \left (\frac {x}{2}\right )}}\right ) \log ^2\left (\frac {x}{2}\right )}dx-9 \int \frac {e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} x}{\left (e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} x+e^{\frac {6}{\log (2)}+15+\frac {27}{\log \left (\frac {x}{2}\right )}}\right ) \log ^2\left (\frac {x}{2}\right )}dx+\int \frac {e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}}}{e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} x+e^{\frac {6}{\log (2)}+15+\frac {27}{\log \left (\frac {x}{2}\right )}}}dx+9 \int \frac {e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} x}{\left (e^{\frac {9 x}{\log \left (\frac {x}{2}\right )}} x+e^{\frac {6}{\log (2)}+15+\frac {27}{\log \left (\frac {x}{2}\right )}}\right ) \log \left (\frac {x}{2}\right )}dx-18 \operatorname {ExpIntegralEi}(\log (x)-\log (2))+18 \operatorname {LogIntegral}\left (\frac {x}{2}\right ) \log \left (\frac {x}{2}\right )-18 \operatorname {LogIntegral}\left (\frac {x}{2}\right ) \log (x)+18 (1+\log (2)) \operatorname {LogIntegral}\left (\frac {x}{2}\right )-9 x+\frac {9 x \log (x)}{\log \left (\frac {x}{2}\right )}-\frac {9 x (1+\log (2))}{\log \left (\frac {x}{2}\right )}+\frac {27}{\log \left (\frac {x}{2}\right )}\)

Input:

Int[(x*Log[x/2]^2 + E^(((27 - 9*x)*Log[2] + (6 + 15*Log[2])*Log[x/2])/(Log 
[2]*Log[x/2]))*(-27 + 9*x - 9*x*Log[x/2]))/(E^(((27 - 9*x)*Log[2] + (6 + 1 
5*Log[2])*Log[x/2])/(Log[2]*Log[x/2]))*x*Log[x/2]^2 + x^2*Log[x/2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09

method	result	size
norman	\(\ln \left (x +{\mathrm e}^{\frac {\left (15 \ln \left (2\right )+6\right ) \ln \left (\frac {x}{2}\right )+\left (-9 x +27\right ) \ln \left (2\right )}{\ln \left (2\right ) \ln \left (\frac {x}{2}\right )}}\right )\)	\(36\)
parallelrisch	\(\ln \left (x +{\mathrm e}^{\frac {\left (15 \ln \left (2\right )+6\right ) \ln \left (\frac {x}{2}\right )+\left (-9 x +27\right ) \ln \left (2\right )}{\ln \left (2\right ) \ln \left (\frac {x}{2}\right )}}\right )\)	\(36\)
risch	\(-\frac {9 \left (-3+x \right )}{\ln \left (\frac {x}{2}\right )}-\frac {\left (15 \ln \left (2\right )+6\right ) \ln \left (\frac {x}{2}\right )+\left (-9 x +27\right ) \ln \left (2\right )}{\ln \left (2\right ) \ln \left (\frac {x}{2}\right )}+\ln \left (x +{\mathrm e}^{-\frac {3 \left (-5 \ln \left (\frac {x}{2}\right ) \ln \left (2\right )+3 x \ln \left (2\right )-2 \ln \left (\frac {x}{2}\right )-9 \ln \left (2\right )\right )}{\ln \left (\frac {x}{2}\right ) \ln \left (2\right )}}\right )\)	\(85\)

Input:

int(((-9*x*ln(1/2*x)+9*x-27)*exp(((15*ln(2)+6)*ln(1/2*x)+(-9*x+27)*ln(2))/ 
ln(2)/ln(1/2*x))+x*ln(1/2*x)^2)/(x*ln(1/2*x)^2*exp(((15*ln(2)+6)*ln(1/2*x) 
+(-9*x+27)*ln(2))/ln(2)/ln(1/2*x))+x^2*ln(1/2*x)^2),x,method=_RETURNVERBOS 
E)
 

Output:

ln(x+exp(((15*ln(2)+6)*ln(1/2*x)+(-9*x+27)*ln(2))/ln(2)/ln(1/2*x)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.09 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\log \left (x + e^{\left (-\frac {3 \, {\left (3 \, {\left (x - 3\right )} \log \left (2\right ) - {\left (5 \, \log \left (2\right ) + 2\right )} \log \left (\frac {1}{2} \, x\right )\right )}}{\log \left (2\right ) \log \left (\frac {1}{2} \, x\right )}\right )}\right ) \] Input:

integrate(((-9*x*log(1/2*x)+9*x-27)*exp(((15*log(2)+6)*log(1/2*x)+(-9*x+27 
)*log(2))/log(2)/log(1/2*x))+x*log(1/2*x)^2)/(x*log(1/2*x)^2*exp(((15*log( 
2)+6)*log(1/2*x)+(-9*x+27)*log(2))/log(2)/log(1/2*x))+x^2*log(1/2*x)^2),x, 
 algorithm="fricas")
 

Output:

log(x + e^(-3*(3*(x - 3)*log(2) - (5*log(2) + 2)*log(1/2*x))/(log(2)*log(1 
/2*x))))
 

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\log {\left (x + e^{\frac {\left (27 - 9 x\right ) \log {\left (2 \right )} + \left (6 + 15 \log {\left (2 \right )}\right ) \log {\left (\frac {x}{2} \right )}}{\log {\left (2 \right )} \log {\left (\frac {x}{2} \right )}}} \right )} \] Input:

integrate(((-9*x*ln(1/2*x)+9*x-27)*exp(((15*ln(2)+6)*ln(1/2*x)+(-9*x+27)*l 
n(2))/ln(2)/ln(1/2*x))+x*ln(1/2*x)**2)/(x*ln(1/2*x)**2*exp(((15*ln(2)+6)*l 
n(1/2*x)+(-9*x+27)*ln(2))/ln(2)/ln(1/2*x))+x**2*ln(1/2*x)**2),x)
 

Output:

log(x + exp(((27 - 9*x)*log(2) + (6 + 15*log(2))*log(x/2))/(log(2)*log(x/2 
))))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (27) = 54\).

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.70 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\frac {9 \, x}{\log \left (2\right ) - \log \left (x\right )} + \log \left (x\right ) + \log \left (\frac {x e^{\left (-\frac {9 \, x}{\log \left (2\right ) - \log \left (x\right )}\right )} + e^{\left (-\frac {27}{\log \left (2\right ) - \log \left (x\right )} + \frac {6}{\log \left (2\right )} + 15\right )}}{x}\right ) \] Input:

integrate(((-9*x*log(1/2*x)+9*x-27)*exp(((15*log(2)+6)*log(1/2*x)+(-9*x+27 
)*log(2))/log(2)/log(1/2*x))+x*log(1/2*x)^2)/(x*log(1/2*x)^2*exp(((15*log( 
2)+6)*log(1/2*x)+(-9*x+27)*log(2))/log(2)/log(1/2*x))+x^2*log(1/2*x)^2),x, 
 algorithm="maxima")
 

Output:

9*x/(log(2) - log(x)) + log(x) + log((x*e^(-9*x/(log(2) - log(x))) + e^(-2 
7/(log(2) - log(x)) + 6/log(2) + 15))/x)
 

Giac [A] (verification not implemented)

Time = 0.57 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\log \left (x + e^{\left (\frac {9 \, {\left (x \log \left (2\right ) - 3 \, \log \left (x\right )\right )}}{\log \left (2\right )^{2} - \log \left (2\right ) \log \left (x\right )} + \frac {3 \, {\left (5 \, \log \left (2\right ) - 7\right )}}{\log \left (2\right )}\right )}\right ) \] Input:

integrate(((-9*x*log(1/2*x)+9*x-27)*exp(((15*log(2)+6)*log(1/2*x)+(-9*x+27 
)*log(2))/log(2)/log(1/2*x))+x*log(1/2*x)^2)/(x*log(1/2*x)^2*exp(((15*log( 
2)+6)*log(1/2*x)+(-9*x+27)*log(2))/log(2)/log(1/2*x))+x^2*log(1/2*x)^2),x, 
 algorithm="giac")
 

Output:

log(x + e^(9*(x*log(2) - 3*log(x))/(log(2)^2 - log(2)*log(x)) + 3*(5*log(2 
) - 7)/log(2)))
 

Mupad [B] (verification not implemented)

Time = 4.78 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.94 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\ln \left (x+\frac {{\mathrm {e}}^{\frac {3\,\ln \left (x\right )\,\left (2\,\ln \left (\frac {x}{2}\right )+5\,\ln \left (2\right )\,\ln \left (x\right )-5\,{\ln \left (2\right )}^2\right )}{{\ln \left (\frac {x}{2}\right )}^2\,\ln \left (2\right )}-\frac {9\,x-21}{\ln \left (\frac {x}{2}\right )}}}{2^{\frac {15}{\ln \left (\frac {x}{2}\right )}}}\right ) \] Input:

int(-(exp(-(log(2)*(9*x - 27) - log(x/2)*(15*log(2) + 6))/(log(x/2)*log(2) 
))*(9*x*log(x/2) - 9*x + 27) - x*log(x/2)^2)/(x^2*log(x/2)^2 + x*log(x/2)^ 
2*exp(-(log(2)*(9*x - 27) - log(x/2)*(15*log(2) + 6))/(log(x/2)*log(2)))), 
x)
 

Output:

log(x + exp((3*log(x)*(2*log(x/2) + 5*log(2)*log(x) - 5*log(2)^2))/(log(x/ 
2)^2*log(2)) - (9*x - 21)/log(x/2))/2^(15/log(x/2)))
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.79 \[ \int \frac {x \log ^2\left (\frac {x}{2}\right )+e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} \left (-27+9 x-9 x \log \left (\frac {x}{2}\right )\right )}{e^{\frac {(27-9 x) \log (2)+(6+15 \log (2)) \log \left (\frac {x}{2}\right )}{\log (2) \log \left (\frac {x}{2}\right )}} x \log ^2\left (\frac {x}{2}\right )+x^2 \log ^2\left (\frac {x}{2}\right )} \, dx=\frac {\mathrm {log}\left (e^{\frac {9 x}{\mathrm {log}\left (\frac {x}{2}\right )}} x +e^{\frac {6 \,\mathrm {log}\left (\frac {x}{2}\right )+27 \,\mathrm {log}\left (2\right )}{\mathrm {log}\left (\frac {x}{2}\right ) \mathrm {log}\left (2\right )}} e^{15}\right ) \mathrm {log}\left (\frac {x}{2}\right )-9 x}{\mathrm {log}\left (\frac {x}{2}\right )} \] Input:

int(((-9*x*log(1/2*x)+9*x-27)*exp(((15*log(2)+6)*log(1/2*x)+(-9*x+27)*log( 
2))/log(2)/log(1/2*x))+x*log(1/2*x)^2)/(x*log(1/2*x)^2*exp(((15*log(2)+6)* 
log(1/2*x)+(-9*x+27)*log(2))/log(2)/log(1/2*x))+x^2*log(1/2*x)^2),x)
 

Output:

(log(e**((9*x)/log(x/2))*x + e**((6*log(x/2) + 27*log(2))/(log(x/2)*log(2) 
))*e**15)*log(x/2) - 9*x)/log(x/2)