\(\int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+((200-200 x) \log (x)+100 x \log ^2(x)) \log ^2(\log (4))}{(x-2 x^2+x^3) \log ^2(x)+(2 x-4 x^2+2 x^3) \log (x) \log ^2(\log (4))+(x-2 x^2+x^3) \log ^4(\log (4))} \, dx\) [2241]

Optimal result

Integrand size = 97, antiderivative size = 23 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=\frac {100 \log ^2(x)}{(1-x) \left (\log (x)+\log ^2(\log (4))\right )} \] Output:

100/(1-x)*ln(x)^2/(ln(x)+ln(2*ln(2))^2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=-\frac {100 \log ^2(x)}{(-1+x) \left (\log (x)+\log ^2(\log (4))\right )} \] Input:

Integrate[((100 - 100*x)*Log[x]^2 + 100*x*Log[x]^3 + ((200 - 200*x)*Log[x] 
 + 100*x*Log[x]^2)*Log[Log[4]]^2)/((x - 2*x^2 + x^3)*Log[x]^2 + (2*x - 4*x 
^2 + 2*x^3)*Log[x]*Log[Log[4]]^2 + (x - 2*x^2 + x^3)*Log[Log[4]]^4),x]
 

Output:

(-100*Log[x]^2)/((-1 + x)*(Log[x] + Log[Log[4]]^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.

\(\displaystyle \int \frac {100 x \log ^3(x)+(100-100 x) \log ^2(x)+\log ^2(\log (4)) \left (100 x \log ^2(x)+(200-200 x) \log (x)\right )}{\left (x^3-2 x^2+x\right ) \log ^4(\log (4))+\left (x^3-2 x^2+x\right ) \log ^2(x)+\left (2 x^3-4 x^2+2 x\right ) \log ^2(\log (4)) \log (x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {100 \log (x) \left (x \log ^2(x)+x \left (\log ^2(\log (4))-1\right ) \log (x)-2 (x-1) \log ^2(\log (4))+\log (x)\right )}{(1-x)^2 x \left (\log (x)+\log ^2(\log (4))\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 100 \int \frac {\log (x) \left (x \log ^2(x)-x \left (1-\log ^2(\log (4))\right ) \log (x)+\log (x)+2 (1-x) \log ^2(\log (4))\right )}{(1-x)^2 x \left (\log (x)+\log ^2(\log (4))\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle 100 \int \left (\frac {\log (x)}{(x-1)^2}+\frac {1-x \left (1+\log ^2(\log (4))\right )}{(1-x)^2 x}+\frac {\log ^4(\log (4))}{(x-1)^2 \left (\log (x)+\log ^2(\log (4))\right )}+\frac {\log ^4(\log (4))}{(x-1) x \left (\log (x)+\log ^2(\log (4))\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 100 \left (\log ^4(\log (4)) \int \frac {1}{(x-1) x \left (\log (x)+\log ^2(\log (4))\right )^2}dx+\log ^4(\log (4)) \int \frac {1}{(x-1)^2 \left (\log (x)+\log ^2(\log (4))\right )}dx-\frac {\log ^2(\log (4))}{1-x}+\frac {x \log (x)}{1-x}+\log (x)\right )\)

Input:

Int[((100 - 100*x)*Log[x]^2 + 100*x*Log[x]^3 + ((200 - 200*x)*Log[x] + 100 
*x*Log[x]^2)*Log[Log[4]]^2)/((x - 2*x^2 + x^3)*Log[x]^2 + (2*x - 4*x^2 + 2 
*x^3)*Log[x]*Log[Log[4]]^2 + (x - 2*x^2 + x^3)*Log[Log[4]]^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 2.37 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04

method	result	size
norman	\(-\frac {100 \ln \left (x \right )^{2}}{\left (-1+x \right ) \left (\ln \left (x \right )+\ln \left (2 \ln \left (2\right )\right )^{2}\right )}\)	\(24\)
parallelrisch	\(-\frac {100 \ln \left (x \right )^{2}}{\left (-1+x \right ) \left (\ln \left (x \right )+\ln \left (2 \ln \left (2\right )\right )^{2}\right )}\)	\(24\)
default	\(-\frac {100 \ln \left (x \right )^{2}}{\left (-1+x \right ) \left (\ln \left (2\right )^{2}+2 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )+\ln \left (\ln \left (2\right )\right )^{2}+\ln \left (x \right )\right )}\)	\(33\)
risch	\(-\frac {100 \ln \left (x \right )}{-1+x}+\frac {100 \ln \left (2\right )^{2}}{-1+x}+\frac {200 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )}{-1+x}+\frac {100 \ln \left (\ln \left (2\right )\right )^{2}}{-1+x}-\frac {100 \left (\ln \left (2\right )^{4}+4 \ln \left (2\right )^{3} \ln \left (\ln \left (2\right )\right )+6 \ln \left (2\right )^{2} \ln \left (\ln \left (2\right )\right )^{2}+4 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )^{3}+\ln \left (\ln \left (2\right )\right )^{4}\right )}{\left (-1+x \right ) \left (\ln \left (2\right )^{2}+2 \ln \left (2\right ) \ln \left (\ln \left (2\right )\right )+\ln \left (\ln \left (2\right )\right )^{2}+\ln \left (x \right )\right )}\)	\(113\)

Input:

int(((100*x*ln(x)^2+(-200*x+200)*ln(x))*ln(2*ln(2))^2+100*x*ln(x)^3+(-100* 
x+100)*ln(x)^2)/((x^3-2*x^2+x)*ln(2*ln(2))^4+(2*x^3-4*x^2+2*x)*ln(x)*ln(2* 
ln(2))^2+(x^3-2*x^2+x)*ln(x)^2),x,method=_RETURNVERBOSE)
 

Output:

-100*ln(x)^2/(-1+x)/(ln(x)+ln(2*ln(2))^2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=-\frac {100 \, \log \left (x\right )^{2}}{{\left (x - 1\right )} \log \left (2 \, \log \left (2\right )\right )^{2} + {\left (x - 1\right )} \log \left (x\right )} \] Input:

integrate(((100*x*log(x)^2+(-200*x+200)*log(x))*log(2*log(2))^2+100*x*log( 
x)^3+(-100*x+100)*log(x)^2)/((x^3-2*x^2+x)*log(2*log(2))^4+(2*x^3-4*x^2+2* 
x)*log(x)*log(2*log(2))^2+(x^3-2*x^2+x)*log(x)^2),x, algorithm="fricas")
 

Output:

-100*log(x)^2/((x - 1)*log(2*log(2))^2 + (x - 1)*log(x))
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (20) = 40\).

Time = 0.18 (sec) , antiderivative size = 148, normalized size of antiderivative = 6.43 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=\frac {- 600 \log {\left (2 \right )}^{2} \log {\left (\log {\left (2 \right )} \right )}^{2} - 100 \log {\left (2 \right )}^{4} - 100 \log {\left (\log {\left (2 \right )} \right )}^{4} - 400 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )}^{3} - 400 \log {\left (2 \right )}^{3} \log {\left (\log {\left (2 \right )} \right )}}{2 x \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )} + x \log {\left (\log {\left (2 \right )} \right )}^{2} + x \log {\left (2 \right )}^{2} + \left (x - 1\right ) \log {\left (x \right )} - \log {\left (2 \right )}^{2} - \log {\left (\log {\left (2 \right )} \right )}^{2} - 2 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )}} - \frac {100 \log {\left (x \right )}}{x - 1} - \frac {- 100 \log {\left (2 \right )}^{2} - 100 \log {\left (\log {\left (2 \right )} \right )}^{2} - 200 \log {\left (2 \right )} \log {\left (\log {\left (2 \right )} \right )}}{x - 1} \] Input:

integrate(((100*x*ln(x)**2+(-200*x+200)*ln(x))*ln(2*ln(2))**2+100*x*ln(x)* 
*3+(-100*x+100)*ln(x)**2)/((x**3-2*x**2+x)*ln(2*ln(2))**4+(2*x**3-4*x**2+2 
*x)*ln(x)*ln(2*ln(2))**2+(x**3-2*x**2+x)*ln(x)**2),x)
 

Output:

(-600*log(2)**2*log(log(2))**2 - 100*log(2)**4 - 100*log(log(2))**4 - 400* 
log(2)*log(log(2))**3 - 400*log(2)**3*log(log(2)))/(2*x*log(2)*log(log(2)) 
 + x*log(log(2))**2 + x*log(2)**2 + (x - 1)*log(x) - log(2)**2 - log(log(2 
))**2 - 2*log(2)*log(log(2))) - 100*log(x)/(x - 1) - (-100*log(2)**2 - 100 
*log(log(2))**2 - 200*log(2)*log(log(2)))/(x - 1)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (23) = 46\).

Time = 0.18 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=-\frac {100 \, \log \left (x\right )^{2}}{{\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )} x - \log \left (2\right )^{2} + {\left (x - 1\right )} \log \left (x\right ) - 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) - \log \left (\log \left (2\right )\right )^{2}} \] Input:

integrate(((100*x*log(x)^2+(-200*x+200)*log(x))*log(2*log(2))^2+100*x*log( 
x)^3+(-100*x+100)*log(x)^2)/((x^3-2*x^2+x)*log(2*log(2))^4+(2*x^3-4*x^2+2* 
x)*log(x)*log(2*log(2))^2+(x^3-2*x^2+x)*log(x)^2),x, algorithm="maxima")
 

Output:

-100*log(x)^2/((log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2)*x - log(2 
)^2 + (x - 1)*log(x) - 2*log(2)*log(log(2)) - log(log(2))^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (23) = 46\).

Time = 0.14 (sec) , antiderivative size = 127, normalized size of antiderivative = 5.52 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=-\frac {100 \, {\left (\log \left (2\right )^{4} + 4 \, \log \left (2\right )^{3} \log \left (\log \left (2\right )\right ) + 6 \, \log \left (2\right )^{2} \log \left (\log \left (2\right )\right )^{2} + 4 \, \log \left (2\right ) \log \left (\log \left (2\right )\right )^{3} + \log \left (\log \left (2\right )\right )^{4}\right )}}{x \log \left (2\right )^{2} + 2 \, x \log \left (2\right ) \log \left (\log \left (2\right )\right ) + x \log \left (\log \left (2\right )\right )^{2} - \log \left (2\right )^{2} + x \log \left (x\right ) - 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) - \log \left (\log \left (2\right )\right )^{2} - \log \left (x\right )} + \frac {100 \, {\left (\log \left (2\right )^{2} + 2 \, \log \left (2\right ) \log \left (\log \left (2\right )\right ) + \log \left (\log \left (2\right )\right )^{2}\right )}}{x - 1} - \frac {100 \, \log \left (x\right )}{x - 1} \] Input:

integrate(((100*x*log(x)^2+(-200*x+200)*log(x))*log(2*log(2))^2+100*x*log( 
x)^3+(-100*x+100)*log(x)^2)/((x^3-2*x^2+x)*log(2*log(2))^4+(2*x^3-4*x^2+2* 
x)*log(x)*log(2*log(2))^2+(x^3-2*x^2+x)*log(x)^2),x, algorithm="giac")
 

Output:

-100*(log(2)^4 + 4*log(2)^3*log(log(2)) + 6*log(2)^2*log(log(2))^2 + 4*log 
(2)*log(log(2))^3 + log(log(2))^4)/(x*log(2)^2 + 2*x*log(2)*log(log(2)) + 
x*log(log(2))^2 - log(2)^2 + x*log(x) - 2*log(2)*log(log(2)) - log(log(2)) 
^2 - log(x)) + 100*(log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2)/(x - 
1) - 100*log(x)/(x - 1)
 

Mupad [B] (verification not implemented)

Time = 2.69 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=-\frac {100\,{\ln \left (x\right )}^2}{\left (\ln \left (x\right )+{\ln \left (\ln \left (4\right )\right )}^2\right )\,\left (x-1\right )} \] Input:

int((100*x*log(x)^3 - log(x)^2*(100*x - 100) + log(2*log(2))^2*(100*x*log( 
x)^2 - log(x)*(200*x - 200)))/(log(x)^2*(x - 2*x^2 + x^3) + log(2*log(2))^ 
4*(x - 2*x^2 + x^3) + log(2*log(2))^2*log(x)*(2*x - 4*x^2 + 2*x^3)),x)
 

Output:

-(100*log(x)^2)/((log(x) + log(log(4))^2)*(x - 1))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 3.39 \[ \int \frac {(100-100 x) \log ^2(x)+100 x \log ^3(x)+\left ((200-200 x) \log (x)+100 x \log ^2(x)\right ) \log ^2(\log (4))}{\left (x-2 x^2+x^3\right ) \log ^2(x)+\left (2 x-4 x^2+2 x^3\right ) \log (x) \log ^2(\log (4))+\left (x-2 x^2+x^3\right ) \log ^4(\log (4))} \, dx=\frac {100 \mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )^{4} x -100 \mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )^{4}+100 \mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )^{2} \mathrm {log}\left (x \right ) x -100 \mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )^{2} \mathrm {log}\left (x \right )-100 \mathrm {log}\left (x \right )^{2}}{\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )^{2} x -\mathrm {log}\left (2 \,\mathrm {log}\left (2\right )\right )^{2}+\mathrm {log}\left (x \right ) x -\mathrm {log}\left (x \right )} \] Input:

int(((100*x*log(x)^2+(-200*x+200)*log(x))*log(2*log(2))^2+100*x*log(x)^3+( 
-100*x+100)*log(x)^2)/((x^3-2*x^2+x)*log(2*log(2))^4+(2*x^3-4*x^2+2*x)*log 
(x)*log(2*log(2))^2+(x^3-2*x^2+x)*log(x)^2),x)
 

Output:

(100*(log(2*log(2))**4*x - log(2*log(2))**4 + log(2*log(2))**2*log(x)*x - 
log(2*log(2))**2*log(x) - log(x)**2))/(log(2*log(2))**2*x - log(2*log(2))* 
*2 + log(x)*x - log(x))