\(\int \frac {-576 x+(-576-576 x) \log (1+x)+(180 x^2+180 x^3) \log ^2(1+x)}{2304+2304 x+(-1056 x+384 x^2+1440 x^3) \log (1+x)+(121 x^2-209 x^3-105 x^4+225 x^5) \log ^2(1+x)} \, dx\) [1203]

Optimal result

Integrand size = 87, antiderivative size = 30 \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=\frac {x}{\frac {2 x}{3}-x \left (\frac {1}{4} (-1+x)+x\right )-\frac {4}{\log (1+x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=-\frac {36 x \log (1+x)}{144+3 x (-11+15 x) \log (1+x)} \] Input:

Integrate[(-576*x + (-576 - 576*x)*Log[1 + x] + (180*x^2 + 180*x^3)*Log[1 
+ x]^2)/(2304 + 2304*x + (-1056*x + 384*x^2 + 1440*x^3)*Log[1 + x] + (121* 
x^2 - 209*x^3 - 105*x^4 + 225*x^5)*Log[1 + x]^2),x]
 

Output:

(-36*x*Log[1 + x])/(144 + 3*x*(-11 + 15*x)*Log[1 + 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.

\(\displaystyle \int \frac {\left (180 x^3+180 x^2\right ) \log ^2(x+1)-576 x+(-576 x-576) \log (x+1)}{\left (1440 x^3+384 x^2-1056 x\right ) \log (x+1)+\left (225 x^5-105 x^4-209 x^3+121 x^2\right ) \log ^2(x+1)+2304 x+2304} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \frac {36 \left (5 x^2 (x+1) \log ^2(x+1)-16 x-16 (x+1) \log (x+1)\right )}{(x+1) (x (15 x-11) \log (x+1)+48)^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle 36 \int -\frac {-5 x^2 (x+1) \log ^2(x+1)+16 (x+1) \log (x+1)+16 x}{(x+1) (48-(11-15 x) x \log (x+1))^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -36 \int \frac {-5 x^2 (x+1) \log ^2(x+1)+16 (x+1) \log (x+1)+16 x}{(x+1) (48-(11-15 x) x \log (x+1))^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -36 \int \left (\frac {16 (45 x-11)}{x (15 x-11)^2 \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )}-\frac {5}{(15 x-11)^2}+\frac {16 \left (225 x^4-330 x^3-1319 x^2-912 x+528\right )}{x (x+1) (15 x-11)^2 \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -36 \left (16 \int \frac {1}{\left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )^2}dx+\frac {768}{11} \int \frac {1}{x \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )^2}dx-16 \int \frac {1}{(x+1) \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )^2}dx-11520 \int \frac {1}{(15 x-11)^2 \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )^2}dx-\frac {11520}{11} \int \frac {1}{(15 x-11) \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )^2}dx-\frac {16}{11} \int \frac {1}{x \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )}dx+480 \int \frac {1}{(15 x-11)^2 \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )}dx+\frac {240}{11} \int \frac {1}{(15 x-11) \left (15 \log (x+1) x^2-11 \log (x+1) x+48\right )}dx-\frac {1}{3 (11-15 x)}\right )\)

Input:

Int[(-576*x + (-576 - 576*x)*Log[1 + x] + (180*x^2 + 180*x^3)*Log[1 + x]^2 
)/(2304 + 2304*x + (-1056*x + 384*x^2 + 1440*x^3)*Log[1 + x] + (121*x^2 - 
209*x^3 - 105*x^4 + 225*x^5)*Log[1 + x]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93

method	result	size
norman	\(-\frac {12 \ln \left (1+x \right ) x}{15 x^{2} \ln \left (1+x \right )-11 \ln \left (1+x \right ) x +48}\)	\(28\)
parallelrisch	\(-\frac {12 \ln \left (1+x \right ) x}{15 x^{2} \ln \left (1+x \right )-11 \ln \left (1+x \right ) x +48}\)	\(28\)
risch	\(-\frac {12}{15 x -11}+\frac {576}{\left (15 x -11\right ) \left (15 x^{2} \ln \left (1+x \right )-11 \ln \left (1+x \right ) x +48\right )}\)	\(40\)
derivativedivides	\(\frac {12 \ln \left (1+x \right )-12 \left (1+x \right ) \ln \left (1+x \right )}{15 \ln \left (1+x \right ) \left (1+x \right )^{2}-41 \left (1+x \right ) \ln \left (1+x \right )+26 \ln \left (1+x \right )+48}\)	\(49\)
default	\(\frac {12 \ln \left (1+x \right )-12 \left (1+x \right ) \ln \left (1+x \right )}{15 \ln \left (1+x \right ) \left (1+x \right )^{2}-41 \left (1+x \right ) \ln \left (1+x \right )+26 \ln \left (1+x \right )+48}\)	\(49\)

Input:

int(((180*x^3+180*x^2)*ln(1+x)^2+(-576*x-576)*ln(1+x)-576*x)/((225*x^5-105 
*x^4-209*x^3+121*x^2)*ln(1+x)^2+(1440*x^3+384*x^2-1056*x)*ln(1+x)+2304*x+2 
304),x,method=_RETURNVERBOSE)
 

Output:

-12*ln(1+x)*x/(15*x^2*ln(1+x)-11*ln(1+x)*x+48)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=-\frac {12 \, x \log \left (x + 1\right )}{{\left (15 \, x^{2} - 11 \, x\right )} \log \left (x + 1\right ) + 48} \] Input:

integrate(((180*x^3+180*x^2)*log(1+x)^2+(-576*x-576)*log(1+x)-576*x)/((225 
*x^5-105*x^4-209*x^3+121*x^2)*log(1+x)^2+(1440*x^3+384*x^2-1056*x)*log(1+x 
)+2304*x+2304),x, algorithm="fricas")
 

Output:

-12*x*log(x + 1)/((15*x^2 - 11*x)*log(x + 1) + 48)
 

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.03 \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=\frac {576}{720 x + \left (225 x^{3} - 330 x^{2} + 121 x\right ) \log {\left (x + 1 \right )} - 528} - \frac {180}{225 x - 165} \] Input:

integrate(((180*x**3+180*x**2)*ln(1+x)**2+(-576*x-576)*ln(1+x)-576*x)/((22 
5*x**5-105*x**4-209*x**3+121*x**2)*ln(1+x)**2+(1440*x**3+384*x**2-1056*x)* 
ln(1+x)+2304*x+2304),x)
 

Output:

576/(720*x + (225*x**3 - 330*x**2 + 121*x)*log(x + 1) - 528) - 180/(225*x 
- 165)
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=-\frac {12 \, x \log \left (x + 1\right )}{{\left (15 \, x^{2} - 11 \, x\right )} \log \left (x + 1\right ) + 48} \] Input:

integrate(((180*x^3+180*x^2)*log(1+x)^2+(-576*x-576)*log(1+x)-576*x)/((225 
*x^5-105*x^4-209*x^3+121*x^2)*log(1+x)^2+(1440*x^3+384*x^2-1056*x)*log(1+x 
)+2304*x+2304),x, algorithm="maxima")
 

Output:

-12*x*log(x + 1)/((15*x^2 - 11*x)*log(x + 1) + 48)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.47 \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=\frac {576}{225 \, x^{3} \log \left (x + 1\right ) - 330 \, x^{2} \log \left (x + 1\right ) + 121 \, x \log \left (x + 1\right ) + 720 \, x - 528} - \frac {12}{15 \, x - 11} \] Input:

integrate(((180*x^3+180*x^2)*log(1+x)^2+(-576*x-576)*log(1+x)-576*x)/((225 
*x^5-105*x^4-209*x^3+121*x^2)*log(1+x)^2+(1440*x^3+384*x^2-1056*x)*log(1+x 
)+2304*x+2304),x, algorithm="giac")
 

Output:

576/(225*x^3*log(x + 1) - 330*x^2*log(x + 1) + 121*x*log(x + 1) + 720*x - 
528) - 12/(15*x - 11)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=\int -\frac {\left (-180\,x^3-180\,x^2\right )\,{\ln \left (x+1\right )}^2+\left (576\,x+576\right )\,\ln \left (x+1\right )+576\,x}{\left (225\,x^5-105\,x^4-209\,x^3+121\,x^2\right )\,{\ln \left (x+1\right )}^2+\left (1440\,x^3+384\,x^2-1056\,x\right )\,\ln \left (x+1\right )+2304\,x+2304} \,d x \] Input:

int(-(576*x - log(x + 1)^2*(180*x^2 + 180*x^3) + log(x + 1)*(576*x + 576)) 
/(2304*x + log(x + 1)^2*(121*x^2 - 209*x^3 - 105*x^4 + 225*x^5) + log(x + 
1)*(384*x^2 - 1056*x + 1440*x^3) + 2304),x)
 

Output:

int(-(576*x - log(x + 1)^2*(180*x^2 + 180*x^3) + log(x + 1)*(576*x + 576)) 
/(2304*x + log(x + 1)^2*(121*x^2 - 209*x^3 - 105*x^4 + 225*x^5) + log(x + 
1)*(384*x^2 - 1056*x + 1440*x^3) + 2304), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90 \[ \int \frac {-576 x+(-576-576 x) \log (1+x)+\left (180 x^2+180 x^3\right ) \log ^2(1+x)}{2304+2304 x+\left (-1056 x+384 x^2+1440 x^3\right ) \log (1+x)+\left (121 x^2-209 x^3-105 x^4+225 x^5\right ) \log ^2(1+x)} \, dx=-\frac {12 \,\mathrm {log}\left (x +1\right ) x}{15 \,\mathrm {log}\left (x +1\right ) x^{2}-11 \,\mathrm {log}\left (x +1\right ) x +48} \] Input:

int(((180*x^3+180*x^2)*log(1+x)^2+(-576*x-576)*log(1+x)-576*x)/((225*x^5-1 
05*x^4-209*x^3+121*x^2)*log(1+x)^2+(1440*x^3+384*x^2-1056*x)*log(1+x)+2304 
*x+2304),x)
 

Output:

( - 12*log(x + 1)*x)/(15*log(x + 1)*x**2 - 11*log(x + 1)*x + 48)