\(\int \frac {(6 x-6 x^2) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+(3 x^2+(3 x-3 x^2) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)) \log (x^2)+(-e^5 x^2+e^5 (-x+x^2) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)) \log ^2(x^2)}{(-9 x+9 x^2) \log (2) \log ^2(1-x)+e^5 (6 x-6 x^2) \log (2) \log ^2(1-x) \log (x^2)+e^{10} (-x+x^2) \log (2) \log ^2(1-x) \log ^2(x^2)} \, dx\) [1518]

Optimal result

Integrand size = 209, antiderivative size = 34 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=1-\log (x)+\frac {x}{\log (2) \log (1-x) \left (e^5-\frac {3}{\log \left (x^2\right )}\right )} \] Output:

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

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=\frac {x}{e^5 \log (2) \log (1-x)}-\log (x)+\frac {3 x}{e^5 \log (2) \log (1-x) \left (-3+e^5 \log \left (x^2\right )\right )} \] Input:

Integrate[((6*x - 6*x^2)*Log[1 - x] + (9 - 9*x)*Log[2]*Log[1 - x]^2 + (3*x 
^2 + (3*x - 3*x^2)*Log[1 - x] + E^5*(-6 + 6*x)*Log[2]*Log[1 - x]^2)*Log[x^ 
2] + (-(E^5*x^2) + E^5*(-x + x^2)*Log[1 - x] + E^10*(1 - x)*Log[2]*Log[1 - 
 x]^2)*Log[x^2]^2)/((-9*x + 9*x^2)*Log[2]*Log[1 - x]^2 + E^5*(6*x - 6*x^2) 
*Log[2]*Log[1 - x]^2*Log[x^2] + E^10*(-x + x^2)*Log[2]*Log[1 - x]^2*Log[x^ 
2]^2),x]
 

Output:

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

\(\displaystyle \int \frac {\left (-e^5 x^2+e^5 \left (x^2-x\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (6 x-6) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)}{e^{10} \left (x^2-x\right ) \log (2) \log ^2\left (x^2\right ) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log \left (x^2\right ) \log ^2(1-x)+\left (9 x^2-9 x\right ) \log (2) \log ^2(1-x)} \, dx\)

\(\Big \downarrow \) 7239

\(\displaystyle \int \left (-\frac {x \log \left (x^2\right )}{(x-1) \log (2) \log ^2(1-x) \left (e^5 \log \left (x^2\right )-3\right )}+\frac {e^5 \log ^2\left (x^2\right )-3 \log \left (x^2\right )-6}{\log (2) \log (1-x) \left (e^5 \log \left (x^2\right )-3\right )^2}-\frac {1}{x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 \int \frac {1}{\log ^2(1-x) \left (e^5 \log \left (x^2\right )-3\right )}dx}{e^5 \log (2)}-\frac {3 \int \frac {1}{(x-1) \log ^2(1-x) \left (e^5 \log \left (x^2\right )-3\right )}dx}{e^5 \log (2)}-\frac {6 \int \frac {1}{\log (1-x) \left (e^5 \log \left (x^2\right )-3\right )^2}dx}{\log (2)}+\frac {3 \int \frac {1}{\log (1-x) \left (e^5 \log \left (x^2\right )-3\right )}dx}{e^5 \log (2)}-\frac {1-x}{e^5 \log (2) \log (1-x)}-\log (x)+\frac {1}{e^5 \log (2) \log (1-x)}\)

Input:

Int[((6*x - 6*x^2)*Log[1 - x] + (9 - 9*x)*Log[2]*Log[1 - x]^2 + (3*x^2 + ( 
3*x - 3*x^2)*Log[1 - x] + E^5*(-6 + 6*x)*Log[2]*Log[1 - x]^2)*Log[x^2] + ( 
-(E^5*x^2) + E^5*(-x + x^2)*Log[1 - x] + E^10*(1 - x)*Log[2]*Log[1 - x]^2) 
*Log[x^2]^2)/((-9*x + 9*x^2)*Log[2]*Log[1 - x]^2 + E^5*(6*x - 6*x^2)*Log[2 
]*Log[1 - x]^2*Log[x^2] + E^10*(-x + x^2)*Log[2]*Log[1 - x]^2*Log[x^2]^2), 
x]
 

Output:

$Aborted
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 2.22 (sec) , antiderivative size = 132, normalized size of antiderivative = 3.88

\[-\ln \left (x \right )+\frac {x \left (\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i \ln \left (x \right )\right )}{\left ({\mathrm e}^{5} \pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \,{\mathrm e}^{5} \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+{\mathrm e}^{5} \pi \operatorname {csgn}\left (i x^{2}\right )^{3}+4 i {\mathrm e}^{5} \ln \left (x \right )-6 i\right ) \ln \left (2\right ) \ln \left (1-x \right )}\]

Input:

int((((1-x)*exp(5)^2*ln(2)*ln(1-x)^2+(x^2-x)*exp(5)*ln(1-x)-x^2*exp(5))*ln 
(x^2)^2+((6*x-6)*exp(5)*ln(2)*ln(1-x)^2+(-3*x^2+3*x)*ln(1-x)+3*x^2)*ln(x^2 
)+(-9*x+9)*ln(2)*ln(1-x)^2+(-6*x^2+6*x)*ln(1-x))/((x^2-x)*exp(5)^2*ln(2)*l 
n(1-x)^2*ln(x^2)^2+(-6*x^2+6*x)*exp(5)*ln(2)*ln(1-x)^2*ln(x^2)+(9*x^2-9*x) 
*ln(2)*ln(1-x)^2),x)
 

Output:

-ln(x)+x*(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn( 
I*x^2)^3+4*I*ln(x))/(exp(5)*Pi*csgn(I*x)^2*csgn(I*x^2)-2*exp(5)*Pi*csgn(I* 
x)*csgn(I*x^2)^2+exp(5)*Pi*csgn(I*x^2)^3+4*I*exp(5)*ln(x)-6*I)/ln(2)/ln(1- 
x)
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=-\frac {e^{5} \log \left (2\right ) \log \left (x^{2}\right )^{2} \log \left (-x + 1\right ) - {\left (3 \, \log \left (2\right ) \log \left (-x + 1\right ) + 2 \, x\right )} \log \left (x^{2}\right )}{2 \, {\left (e^{5} \log \left (2\right ) \log \left (x^{2}\right ) \log \left (-x + 1\right ) - 3 \, \log \left (2\right ) \log \left (-x + 1\right )\right )}} \] Input:

integrate((((1-x)*exp(5)^2*log(2)*log(1-x)^2+(x^2-x)*exp(5)*log(1-x)-x^2*e 
xp(5))*log(x^2)^2+((6*x-6)*exp(5)*log(2)*log(1-x)^2+(-3*x^2+3*x)*log(1-x)+ 
3*x^2)*log(x^2)+(-9*x+9)*log(2)*log(1-x)^2+(-6*x^2+6*x)*log(1-x))/((x^2-x) 
*exp(5)^2*log(2)*log(1-x)^2*log(x^2)^2+(-6*x^2+6*x)*exp(5)*log(2)*log(1-x) 
^2*log(x^2)+(9*x^2-9*x)*log(2)*log(1-x)^2),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=\frac {x \log {\left (x^{2} \right )}}{\left (e^{5} \log {\left (2 \right )} \log {\left (x^{2} \right )} - 3 \log {\left (2 \right )}\right ) \log {\left (1 - x \right )}} - \log {\left (x \right )} \] Input:

integrate((((1-x)*exp(5)**2*ln(2)*ln(1-x)**2+(x**2-x)*exp(5)*ln(1-x)-x**2* 
exp(5))*ln(x**2)**2+((6*x-6)*exp(5)*ln(2)*ln(1-x)**2+(-3*x**2+3*x)*ln(1-x) 
+3*x**2)*ln(x**2)+(-9*x+9)*ln(2)*ln(1-x)**2+(-6*x**2+6*x)*ln(1-x))/((x**2- 
x)*exp(5)**2*ln(2)*ln(1-x)**2*ln(x**2)**2+(-6*x**2+6*x)*exp(5)*ln(2)*ln(1- 
x)**2*ln(x**2)+(9*x**2-9*x)*ln(2)*ln(1-x)**2),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=\frac {2 \, x \log \left (x\right )}{{\left (2 \, e^{5} \log \left (2\right ) \log \left (x\right ) - 3 \, \log \left (2\right )\right )} \log \left (-x + 1\right )} - \log \left (x\right ) \] Input:

integrate((((1-x)*exp(5)^2*log(2)*log(1-x)^2+(x^2-x)*exp(5)*log(1-x)-x^2*e 
xp(5))*log(x^2)^2+((6*x-6)*exp(5)*log(2)*log(1-x)^2+(-3*x^2+3*x)*log(1-x)+ 
3*x^2)*log(x^2)+(-9*x+9)*log(2)*log(1-x)^2+(-6*x^2+6*x)*log(1-x))/((x^2-x) 
*exp(5)^2*log(2)*log(1-x)^2*log(x^2)^2+(-6*x^2+6*x)*exp(5)*log(2)*log(1-x) 
^2*log(x^2)+(9*x^2-9*x)*log(2)*log(1-x)^2),x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 167 vs. \(2 (36) = 72\).

Time = 0.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 4.91 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=-\frac {3 \, e^{5} \log \left (2\right ) \log \left (x^{2}\right ) \log \left (e^{5} \log \left (x^{2}\right ) - 3\right ) \log \left (-x + 1\right ) - 3 \, e^{5} \log \left (2\right ) \log \left (x^{2}\right ) \log \left (-e^{5} \log \left (x^{2}\right ) + 3\right ) \log \left (-x + 1\right ) + e^{10} \log \left (2\right ) \log \left (x^{2}\right ) \log \left (x\right ) \log \left (-x + 1\right ) - 3 \, e^{5} \log \left (2\right ) \log \left (x\right ) \log \left (-x + 1\right ) - x e^{5} \log \left (x^{2}\right ) - 9 \, \log \left (2\right ) \log \left (e^{5} \log \left (x^{2}\right ) - 3\right ) \log \left (-x + 1\right ) + 9 \, \log \left (2\right ) \log \left (-e^{5} \log \left (x^{2}\right ) + 3\right ) \log \left (-x + 1\right )}{e^{10} \log \left (2\right ) \log \left (x^{2}\right ) \log \left (-x + 1\right ) - 3 \, e^{5} \log \left (2\right ) \log \left (-x + 1\right )} \] Input:

integrate((((1-x)*exp(5)^2*log(2)*log(1-x)^2+(x^2-x)*exp(5)*log(1-x)-x^2*e 
xp(5))*log(x^2)^2+((6*x-6)*exp(5)*log(2)*log(1-x)^2+(-3*x^2+3*x)*log(1-x)+ 
3*x^2)*log(x^2)+(-9*x+9)*log(2)*log(1-x)^2+(-6*x^2+6*x)*log(1-x))/((x^2-x) 
*exp(5)^2*log(2)*log(1-x)^2*log(x^2)^2+(-6*x^2+6*x)*exp(5)*log(2)*log(1-x) 
^2*log(x^2)+(9*x^2-9*x)*log(2)*log(1-x)^2),x, algorithm="giac")
 

Output:

-(3*e^5*log(2)*log(x^2)*log(e^5*log(x^2) - 3)*log(-x + 1) - 3*e^5*log(2)*l 
og(x^2)*log(-e^5*log(x^2) + 3)*log(-x + 1) + e^10*log(2)*log(x^2)*log(x)*l 
og(-x + 1) - 3*e^5*log(2)*log(x)*log(-x + 1) - x*e^5*log(x^2) - 9*log(2)*l 
og(e^5*log(x^2) - 3)*log(-x + 1) + 9*log(2)*log(-e^5*log(x^2) + 3)*log(-x 
+ 1))/(e^10*log(2)*log(x^2)*log(-x + 1) - 3*e^5*log(2)*log(-x + 1))
 

Mupad [B] (verification not implemented)

Time = 2.53 (sec) , antiderivative size = 407, normalized size of antiderivative = 11.97 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=\frac {{\mathrm {e}}^{10}\,\ln \left (16\right )-4\,{\mathrm {e}}^{10}\,\ln \left (2\right )+3}{4\,{\mathrm {e}}^{10}\,\ln \left (2\right )-4\,x\,{\mathrm {e}}^{10}\,\ln \left (2\right )}-\frac {\frac {3\,x^2\,{\mathrm {e}}^{-10}}{2\,\ln \left (2\right )\,\left (x-1\right )}+\frac {{\mathrm {e}}^{-10}\,{\ln \left (1-x\right )}^2\,\left (2\,{\mathrm {e}}^5+3\right )\,\left (x-1\right )}{4\,\ln \left (2\right )}-\frac {x\,{\mathrm {e}}^{-10}\,\ln \left (1-x\right )\,\left (6\,x-2\,{\mathrm {e}}^5+2\,x\,{\mathrm {e}}^5-9\right )}{4\,\ln \left (2\right )\,\left (x-1\right )}}{{\ln \left (1-x\right )}^2}-\frac {\frac {3\,x\,{\mathrm {e}}^{-15}\,\left (3\,x+3\,\ln \left (1-x\right )+2\,{\mathrm {e}}^5\,\ln \left (1-x\right )-3\,x\,\ln \left (1-x\right )-2\,x\,{\mathrm {e}}^5\,\ln \left (1-x\right )\right )}{2\,\ln \left (2\right )\,{\ln \left (1-x\right )}^2\,\left (x-1\right )}-\frac {3\,x\,\ln \left (x^2\right )\,{\mathrm {e}}^{-10}\,\left (x+\ln \left (1-x\right )-x\,\ln \left (1-x\right )\right )}{2\,\ln \left (2\right )\,{\ln \left (1-x\right )}^2\,\left (x-1\right )}}{\ln \left (x^2\right )-3\,{\mathrm {e}}^{-5}}-\frac {\frac {{\mathrm {e}}^{-10}\,\ln \left (1-x\right )\,\left (2\,{\mathrm {e}}^5\,x^2-4\,{\mathrm {e}}^5\,x+2\,{\mathrm {e}}^5-3\right )}{4\,\ln \left (2\right )\,\left (x-1\right )}-\frac {{\mathrm {e}}^{-10}\,\left (3\,x-2\,x\,{\mathrm {e}}^5+2\,x^2\,{\mathrm {e}}^5\right )}{4\,\ln \left (2\right )\,\left (x-1\right )}+\frac {{\mathrm {e}}^{-10}\,{\ln \left (1-x\right )}^2\,\left (2\,{\mathrm {e}}^5+3\right )\,\left (x-1\right )}{4\,\ln \left (2\right )}}{\ln \left (1-x\right )}-\ln \left (x\right )-\ln \left (x-1\right )\,\left (\frac {{\mathrm {e}}^{-10}\,\left (\frac {{\mathrm {e}}^5}{2}+\frac {{\mathrm {e}}^{10}\,\ln \left (16\right )}{4}+\frac {3}{4}\right )}{\ln \left (2\right )}-1\right )+\frac {x\,{\mathrm {e}}^{-10}\,\left (4\,{\mathrm {e}}^5+3\right )}{4\,\ln \left (2\right )}+\frac {x\,{\mathrm {e}}^{-10}\,\ln \left (1-x\right )\,\left (2\,{\mathrm {e}}^5+3\right )}{4\,\ln \left (2\right )} \] Input:

int(-(log(x^2)*(log(1 - x)*(3*x - 3*x^2) + 3*x^2 + exp(5)*log(2)*log(1 - x 
)^2*(6*x - 6)) + log(1 - x)*(6*x - 6*x^2) - log(x^2)^2*(x^2*exp(5) + exp(5 
)*log(1 - x)*(x - x^2) + exp(10)*log(2)*log(1 - x)^2*(x - 1)) - log(2)*log 
(1 - x)^2*(9*x - 9))/(log(2)*log(1 - x)^2*(9*x - 9*x^2) - log(x^2)*exp(5)* 
log(2)*log(1 - x)^2*(6*x - 6*x^2) + log(x^2)^2*exp(10)*log(2)*log(1 - x)^2 
*(x - x^2)),x)
 

Output:

(exp(10)*log(16) - 4*exp(10)*log(2) + 3)/(4*exp(10)*log(2) - 4*x*exp(10)*l 
og(2)) - ((3*x^2*exp(-10))/(2*log(2)*(x - 1)) + (exp(-10)*log(1 - x)^2*(2* 
exp(5) + 3)*(x - 1))/(4*log(2)) - (x*exp(-10)*log(1 - x)*(6*x - 2*exp(5) + 
 2*x*exp(5) - 9))/(4*log(2)*(x - 1)))/log(1 - x)^2 - ((3*x*exp(-15)*(3*x + 
 3*log(1 - x) + 2*exp(5)*log(1 - x) - 3*x*log(1 - x) - 2*x*exp(5)*log(1 - 
x)))/(2*log(2)*log(1 - x)^2*(x - 1)) - (3*x*log(x^2)*exp(-10)*(x + log(1 - 
 x) - x*log(1 - x)))/(2*log(2)*log(1 - x)^2*(x - 1)))/(log(x^2) - 3*exp(-5 
)) - ((exp(-10)*log(1 - x)*(2*exp(5) - 4*x*exp(5) + 2*x^2*exp(5) - 3))/(4* 
log(2)*(x - 1)) - (exp(-10)*(3*x - 2*x*exp(5) + 2*x^2*exp(5)))/(4*log(2)*( 
x - 1)) + (exp(-10)*log(1 - x)^2*(2*exp(5) + 3)*(x - 1))/(4*log(2)))/log(1 
 - x) - log(x) - log(x - 1)*((exp(-10)*(exp(5)/2 + (exp(10)*log(16))/4 + 3 
/4))/log(2) - 1) + (x*exp(-10)*(4*exp(5) + 3))/(4*log(2)) + (x*exp(-10)*lo 
g(1 - x)*(2*exp(5) + 3))/(4*log(2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.85 \[ \int \frac {\left (6 x-6 x^2\right ) \log (1-x)+(9-9 x) \log (2) \log ^2(1-x)+\left (3 x^2+\left (3 x-3 x^2\right ) \log (1-x)+e^5 (-6+6 x) \log (2) \log ^2(1-x)\right ) \log \left (x^2\right )+\left (-e^5 x^2+e^5 \left (-x+x^2\right ) \log (1-x)+e^{10} (1-x) \log (2) \log ^2(1-x)\right ) \log ^2\left (x^2\right )}{\left (-9 x+9 x^2\right ) \log (2) \log ^2(1-x)+e^5 \left (6 x-6 x^2\right ) \log (2) \log ^2(1-x) \log \left (x^2\right )+e^{10} \left (-x+x^2\right ) \log (2) \log ^2(1-x) \log ^2\left (x^2\right )} \, dx=\frac {-\mathrm {log}\left (x^{2}\right ) \mathrm {log}\left (1-x \right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (2\right ) e^{5}+\mathrm {log}\left (x^{2}\right ) x +3 \,\mathrm {log}\left (1-x \right ) \mathrm {log}\left (x \right ) \mathrm {log}\left (2\right )}{\mathrm {log}\left (1-x \right ) \mathrm {log}\left (2\right ) \left (\mathrm {log}\left (x^{2}\right ) e^{5}-3\right )} \] Input:

int((((1-x)*exp(5)^2*log(2)*log(1-x)^2+(x^2-x)*exp(5)*log(1-x)-x^2*exp(5)) 
*log(x^2)^2+((6*x-6)*exp(5)*log(2)*log(1-x)^2+(-3*x^2+3*x)*log(1-x)+3*x^2) 
*log(x^2)+(-9*x+9)*log(2)*log(1-x)^2+(-6*x^2+6*x)*log(1-x))/((x^2-x)*exp(5 
)^2*log(2)*log(1-x)^2*log(x^2)^2+(-6*x^2+6*x)*exp(5)*log(2)*log(1-x)^2*log 
(x^2)+(9*x^2-9*x)*log(2)*log(1-x)^2),x)
 

Output:

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