\(\int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (-25 x-100 x \log (x)-200 x \log ^2(x))}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+(100-40 \log (2)+4 \log ^2(2)) \log (x)} \, dx\) [2076]

Optimal result

Integrand size = 124, antiderivative size = 34 \[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=\frac {x^2 \log (x)}{-1-e^{\frac {\frac {3}{4}+\log (x)}{3 \log (x)}}+\frac {\log (2)}{5}} \] Output:

x^2*ln(x)/(1/5*ln(2)-1-exp(1/3*(3/4+ln(x))/ln(x)))
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=-\frac {5 x^2 \log (x)}{5+5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-\log (2)} \] Input:

Integrate[((-100*x + 20*x*Log[2])*Log[x] + (-200*x + 40*x*Log[2])*Log[x]^2 
 + E^((3 + 4*Log[x])/(12*Log[x]))*(-25*x - 100*x*Log[x] - 200*x*Log[x]^2)) 
/(100*E^((3 + 4*Log[x])/(6*Log[x]))*Log[x] + E^((3 + 4*Log[x])/(12*Log[x]) 
)*(200 - 40*Log[2])*Log[x] + (100 - 40*Log[2] + 4*Log[2]^2)*Log[x]),x]
 

Output:

(-5*x^2*Log[x])/(5 + 5*E^(1/3 + 1/(4*Log[x])) - Log[2])
 

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.

\(\displaystyle \int \frac {(40 x \log (2)-200 x) \log ^2(x)+e^{\frac {4 \log (x)+3}{12 \log (x)}} \left (-25 x-200 x \log ^2(x)-100 x \log (x)\right )+(20 x \log (2)-100 x) \log (x)}{\left (100+4 \log ^2(2)-40 \log (2)\right ) \log (x)+100 e^{\frac {4 \log (x)+3}{6 \log (x)}} \log (x)+(200-40 \log (2)) e^{\frac {4 \log (x)+3}{12 \log (x)}} \log (x)} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {(40 x \log (2)-200 x) \log ^2(x)+e^{\frac {4 \log (x)+3}{12 \log (x)}} \left (-25 x-200 x \log ^2(x)-100 x \log (x)\right )+(20 x \log (2)-100 x) \log (x)}{4 \left (5 e^{\frac {1}{4 \log (x)}+\frac {1}{3}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \int -\frac {5 \left (5 e^{\frac {1}{4 \log (x)}} \left (8 x \log ^2(x)+4 x \log (x)+x\right ) x^{\frac {1}{3 \log (x)}}+8 (5-\log (2)) \log ^2(x) x+4 (5-\log (2)) \log (x) x\right )}{\left (5+5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-\log (2)\right )^2 \log (x)}dx\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {5}{4} \int \frac {5 e^{\frac {1}{4 \log (x)}} \left (8 x \log ^2(x)+4 x \log (x)+x\right ) x^{\frac {1}{3 \log (x)}}+8 (5-\log (2)) \log ^2(x) x+4 (5-\log (2)) \log (x) x}{\left (5+5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}-\log (2)\right )^2 \log (x)}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle -\frac {5}{4} \int \frac {5 e^{\frac {1}{4 \log (x)}} \left (8 x \log ^2(x)+4 x \log (x)+x\right ) x^{\frac {1}{3 \log (x)}}+8 (5-\log (2)) \log ^2(x) x+4 (5-\log (2)) \log (x) x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {5}{4} \int \left (\frac {40 \left (1-\frac {\log (2)}{5}\right ) \log (x) x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2}+\frac {5 e^{\frac {4 \log (x)+3}{12 \log (x)}} \left (8 \log ^2(x)+4 \log (x)+1\right ) x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)}+\frac {20 \left (1-\frac {\log (2)}{5}\right ) x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {5}{4} \left (4 (5-\log (2)) \int \frac {x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2}dx+20 \int \frac {e^{\frac {4 \log (x)+3}{12 \log (x)}} x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2}dx+5 \int \frac {e^{\frac {4 \log (x)+3}{12 \log (x)}} x}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2 \log (x)}dx+8 (5-\log (2)) \int \frac {x \log (x)}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2}dx+40 \int \frac {e^{\frac {4 \log (x)+3}{12 \log (x)}} x \log (x)}{\left (5 e^{\frac {1}{3}+\frac {1}{4 \log (x)}}+5 \left (1-\frac {\log (2)}{5}\right )\right )^2}dx\right )\)

Input:

Int[((-100*x + 20*x*Log[2])*Log[x] + (-200*x + 40*x*Log[2])*Log[x]^2 + E^( 
(3 + 4*Log[x])/(12*Log[x]))*(-25*x - 100*x*Log[x] - 200*x*Log[x]^2))/(100* 
E^((3 + 4*Log[x])/(6*Log[x]))*Log[x] + E^((3 + 4*Log[x])/(12*Log[x]))*(200 
 - 40*Log[2])*Log[x] + (100 - 40*Log[2] + 4*Log[2]^2)*Log[x]),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85

method	result	size
norman	\(\frac {5 x^{2} \ln \left (x \right )}{\ln \left (2\right )-5 \,{\mathrm e}^{\frac {4 \ln \left (x \right )+3}{12 \ln \left (x \right )}}-5}\)	\(29\)
risch	\(\frac {5 x^{2} \ln \left (x \right )}{\ln \left (2\right )-5 \,{\mathrm e}^{\frac {4 \ln \left (x \right )+3}{12 \ln \left (x \right )}}-5}\)	\(29\)
parallelrisch	\(\frac {5 x^{2} \ln \left (x \right )}{\ln \left (2\right )-5 \,{\mathrm e}^{\frac {4 \ln \left (x \right )+3}{12 \ln \left (x \right )}}-5}\)	\(29\)

Input:

int(((-200*x*ln(x)^2-100*x*ln(x)-25*x)*exp(1/12*(4*ln(x)+3)/ln(x))+(40*x*l 
n(2)-200*x)*ln(x)^2+(20*x*ln(2)-100*x)*ln(x))/(100*ln(x)*exp(1/12*(4*ln(x) 
+3)/ln(x))^2+(-40*ln(2)+200)*ln(x)*exp(1/12*(4*ln(x)+3)/ln(x))+(4*ln(2)^2- 
40*ln(2)+100)*ln(x)),x,method=_RETURNVERBOSE)
 

Output:

5*x^2*ln(x)/(ln(2)-5*exp(1/12*(4*ln(x)+3)/ln(x))-5)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=-\frac {5 \, x^{2} \log \left (x\right )}{5 \, e^{\left (\frac {4 \, \log \left (x\right ) + 3}{12 \, \log \left (x\right )}\right )} - \log \left (2\right ) + 5} \] Input:

integrate(((-200*x*log(x)^2-100*x*log(x)-25*x)*exp(1/12*(4*log(x)+3)/log(x 
))+(40*x*log(2)-200*x)*log(x)^2+(20*x*log(2)-100*x)*log(x))/(100*log(x)*ex 
p(1/12*(4*log(x)+3)/log(x))^2+(-40*log(2)+200)*log(x)*exp(1/12*(4*log(x)+3 
)/log(x))+(4*log(2)^2-40*log(2)+100)*log(x)),x, algorithm="fricas")
 

Output:

-5*x^2*log(x)/(5*e^(1/12*(4*log(x) + 3)/log(x)) - log(2) + 5)
 

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=- \frac {5 x^{2} \log {\left (x \right )}}{5 e^{\frac {\frac {\log {\left (x \right )}}{3} + \frac {1}{4}}{\log {\left (x \right )}}} - \log {\left (2 \right )} + 5} \] Input:

integrate(((-200*x*ln(x)**2-100*x*ln(x)-25*x)*exp(1/12*(4*ln(x)+3)/ln(x))+ 
(40*x*ln(2)-200*x)*ln(x)**2+(20*x*ln(2)-100*x)*ln(x))/(100*ln(x)*exp(1/12* 
(4*ln(x)+3)/ln(x))**2+(-40*ln(2)+200)*ln(x)*exp(1/12*(4*ln(x)+3)/ln(x))+(4 
*ln(2)**2-40*ln(2)+100)*ln(x)),x)
 

Output:

-5*x**2*log(x)/(5*exp((log(x)/3 + 1/4)/log(x)) - log(2) + 5)
 

Maxima [F]

\[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=\int { -\frac {5 \, {\left (8 \, {\left (x \log \left (2\right ) - 5 \, x\right )} \log \left (x\right )^{2} - 5 \, {\left (8 \, x \log \left (x\right )^{2} + 4 \, x \log \left (x\right ) + x\right )} e^{\left (\frac {4 \, \log \left (x\right ) + 3}{12 \, \log \left (x\right )}\right )} + 4 \, {\left (x \log \left (2\right ) - 5 \, x\right )} \log \left (x\right )\right )}}{4 \, {\left (10 \, {\left (\log \left (2\right ) - 5\right )} e^{\left (\frac {4 \, \log \left (x\right ) + 3}{12 \, \log \left (x\right )}\right )} \log \left (x\right ) - {\left (\log \left (2\right )^{2} - 10 \, \log \left (2\right ) + 25\right )} \log \left (x\right ) - 25 \, e^{\left (\frac {4 \, \log \left (x\right ) + 3}{6 \, \log \left (x\right )}\right )} \log \left (x\right )\right )}} \,d x } \] Input:

integrate(((-200*x*log(x)^2-100*x*log(x)-25*x)*exp(1/12*(4*log(x)+3)/log(x 
))+(40*x*log(2)-200*x)*log(x)^2+(20*x*log(2)-100*x)*log(x))/(100*log(x)*ex 
p(1/12*(4*log(x)+3)/log(x))^2+(-40*log(2)+200)*log(x)*exp(1/12*(4*log(x)+3 
)/log(x))+(4*log(2)^2-40*log(2)+100)*log(x)),x, algorithm="maxima")
 

Output:

-5/4*integrate((8*(x*log(2) - 5*x)*log(x)^2 - 5*(8*x*log(x)^2 + 4*x*log(x) 
 + x)*e^(1/12*(4*log(x) + 3)/log(x)) + 4*(x*log(2) - 5*x)*log(x))/(10*(log 
(2) - 5)*e^(1/12*(4*log(x) + 3)/log(x))*log(x) - (log(2)^2 - 10*log(2) + 2 
5)*log(x) - 25*e^(1/6*(4*log(x) + 3)/log(x))*log(x)), x)
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.76 \[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=-\frac {5 \, x^{2} \log \left (x\right )}{5 \, e^{\left (\frac {1}{4 \, \log \left (x\right )} + \frac {1}{3}\right )} - \log \left (2\right ) + 5} \] Input:

integrate(((-200*x*log(x)^2-100*x*log(x)-25*x)*exp(1/12*(4*log(x)+3)/log(x 
))+(40*x*log(2)-200*x)*log(x)^2+(20*x*log(2)-100*x)*log(x))/(100*log(x)*ex 
p(1/12*(4*log(x)+3)/log(x))^2+(-40*log(2)+200)*log(x)*exp(1/12*(4*log(x)+3 
)/log(x))+(4*log(2)^2-40*log(2)+100)*log(x)),x, algorithm="giac")
 

Output:

-5*x^2*log(x)/(5*e^(1/4/log(x) + 1/3) - log(2) + 5)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=\int -\frac {{\mathrm {e}}^{\frac {\frac {\ln \left (x\right )}{3}+\frac {1}{4}}{\ln \left (x\right )}}\,\left (200\,x\,{\ln \left (x\right )}^2+100\,x\,\ln \left (x\right )+25\,x\right )+{\ln \left (x\right )}^2\,\left (200\,x-40\,x\,\ln \left (2\right )\right )+\ln \left (x\right )\,\left (100\,x-20\,x\,\ln \left (2\right )\right )}{\ln \left (x\right )\,\left (4\,{\ln \left (2\right )}^2-40\,\ln \left (2\right )+100\right )+100\,{\mathrm {e}}^{\frac {2\,\left (\frac {\ln \left (x\right )}{3}+\frac {1}{4}\right )}{\ln \left (x\right )}}\,\ln \left (x\right )-{\mathrm {e}}^{\frac {\frac {\ln \left (x\right )}{3}+\frac {1}{4}}{\ln \left (x\right )}}\,\ln \left (x\right )\,\left (40\,\ln \left (2\right )-200\right )} \,d x \] Input:

int(-(exp((log(x)/3 + 1/4)/log(x))*(25*x + 200*x*log(x)^2 + 100*x*log(x)) 
+ log(x)^2*(200*x - 40*x*log(2)) + log(x)*(100*x - 20*x*log(2)))/(log(x)*( 
4*log(2)^2 - 40*log(2) + 100) + 100*exp((2*(log(x)/3 + 1/4))/log(x))*log(x 
) - exp((log(x)/3 + 1/4)/log(x))*log(x)*(40*log(2) - 200)),x)
 

Output:

int(-(exp((log(x)/3 + 1/4)/log(x))*(25*x + 200*x*log(x)^2 + 100*x*log(x)) 
+ log(x)^2*(200*x - 40*x*log(2)) + log(x)*(100*x - 20*x*log(2)))/(log(x)*( 
4*log(2)^2 - 40*log(2) + 100) + 100*exp((2*(log(x)/3 + 1/4))/log(x))*log(x 
) - exp((log(x)/3 + 1/4)/log(x))*log(x)*(40*log(2) - 200)), x)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {(-100 x+20 x \log (2)) \log (x)+(-200 x+40 x \log (2)) \log ^2(x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} \left (-25 x-100 x \log (x)-200 x \log ^2(x)\right )}{100 e^{\frac {3+4 \log (x)}{6 \log (x)}} \log (x)+e^{\frac {3+4 \log (x)}{12 \log (x)}} (200-40 \log (2)) \log (x)+\left (100-40 \log (2)+4 \log ^2(2)\right ) \log (x)} \, dx=-\frac {5 \,\mathrm {log}\left (x \right ) x^{2}}{5 e^{\frac {4 \,\mathrm {log}\left (x \right )+3}{12 \,\mathrm {log}\left (x \right )}}-\mathrm {log}\left (2\right )+5} \] Input:

int(((-200*x*log(x)^2-100*x*log(x)-25*x)*exp(1/12*(4*log(x)+3)/log(x))+(40 
*x*log(2)-200*x)*log(x)^2+(20*x*log(2)-100*x)*log(x))/(100*log(x)*exp(1/12 
*(4*log(x)+3)/log(x))^2+(-40*log(2)+200)*log(x)*exp(1/12*(4*log(x)+3)/log( 
x))+(4*log(2)^2-40*log(2)+100)*log(x)),x)
 

Output:

( - 5*log(x)*x**2)/(5*e**((4*log(x) + 3)/(12*log(x))) - log(2) + 5)