\(\int \frac {32-16 x+e^5 (12 x^2-8 x^3)}{-32+16 x+e^5 (32 x^2-16 x^3)+e^{10} (-8 x^4+4 x^5)+(e^5 (16 x-8 x^2)+e^{10} (-8 x^3+4 x^4)) \log (\frac {2-x}{2})+e^{10} (-2 x^2+x^3) \log ^2(\frac {2-x}{2})} \, dx\) [503]

Optimal result

Integrand size = 123, antiderivative size = 32 \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {x^2}{-x+\frac {1}{4} e^5 x^2 \left (2 x+\log \left (1-\frac {x}{2}\right )\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.51 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 x}{-4+2 e^5 x^2+e^5 x \log \left (1-\frac {x}{2}\right )} \] Input:

Integrate[(32 - 16*x + E^5*(12*x^2 - 8*x^3))/(-32 + 16*x + E^5*(32*x^2 - 1 
6*x^3) + E^10*(-8*x^4 + 4*x^5) + (E^5*(16*x - 8*x^2) + E^10*(-8*x^3 + 4*x^ 
4))*Log[(2 - x)/2] + E^10*(-2*x^2 + x^3)*Log[(2 - x)/2]^2),x]
 

Output:

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

\(\displaystyle \int \frac {e^5 \left (12 x^2-8 x^3\right )-16 x+32}{e^{10} \left (4 x^5-8 x^4\right )+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (x^3-2 x^2\right ) \log ^2\left (\frac {2-x}{2}\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (4 x^4-8 x^3\right )\right ) \log \left (\frac {2-x}{2}\right )+16 x-32} \, dx\)

\(\Big \downarrow \) 7239

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(32 - 16*x + E^5*(12*x^2 - 8*x^3))/(-32 + 16*x + E^5*(32*x^2 - 16*x^3) 
 + E^10*(-8*x^4 + 4*x^5) + (E^5*(16*x - 8*x^2) + E^10*(-8*x^3 + 4*x^4))*Lo 
g[(2 - x)/2] + E^10*(-2*x^2 + x^3)*Log[(2 - x)/2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78

method	result	size
norman	\(\frac {4 x}{\ln \left (1-\frac {x}{2}\right ) {\mathrm e}^{5} x +2 x^{2} {\mathrm e}^{5}-4}\)	\(25\)
risch	\(\frac {4 x}{\ln \left (1-\frac {x}{2}\right ) {\mathrm e}^{5} x +2 x^{2} {\mathrm e}^{5}-4}\)	\(25\)
derivativedivides	\(-\frac {2 x}{\ln \left (1-\frac {x}{2}\right ) {\mathrm e}^{5} \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )^{2}-{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right )+8 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5}+2}\)	\(56\)
default	\(-\frac {2 x}{\ln \left (1-\frac {x}{2}\right ) {\mathrm e}^{5} \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )^{2}-{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right )+8 \,{\mathrm e}^{5} \left (1-\frac {x}{2}\right )-4 \,{\mathrm e}^{5}+2}\)	\(56\)
parallelrisch	\(\frac {256+64 \,{\mathrm e}^{5} \ln \left (-2+x \right )+256 x -128 x^{2} {\mathrm e}^{5}-64 \,{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right )+16 x \,{\mathrm e}^{10} \ln \left (1-\frac {x}{2}\right )^{2}+32 x^{2} {\mathrm e}^{10} \ln \left (1-\frac {x}{2}\right )-64 \ln \left (1-\frac {x}{2}\right ) {\mathrm e}^{5} x -32 \,{\mathrm e}^{10} \ln \left (-2+x \right ) x^{2}-16 \,{\mathrm e}^{10} \ln \left (-2+x \right ) x \ln \left (1-\frac {x}{2}\right )}{64 \ln \left (1-\frac {x}{2}\right ) {\mathrm e}^{5} x +128 x^{2} {\mathrm e}^{5}-256}\)	\(125\)

Input:

int(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*ln(1-1/2*x)^2+( 
(4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*ln(1-1/2*x)+(4*x^5-8*x^4)*exp 
(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 \, x}{2 \, x^{2} e^{5} + x e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) - 4} \] Input:

integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2 
*x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8* 
x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 x}{2 x^{2} e^{5} + x e^{5} \log {\left (1 - \frac {x}{2} \right )} - 4} \] Input:

integrate(((-8*x**3+12*x**2)*exp(5)-16*x+32)/((x**3-2*x**2)*exp(5)**2*ln(1 
-1/2*x)**2+((4*x**4-8*x**3)*exp(5)**2+(-8*x**2+16*x)*exp(5))*ln(1-1/2*x)+( 
4*x**5-8*x**4)*exp(5)**2+(-16*x**3+32*x**2)*exp(5)+16*x-32),x)
 

Output:

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

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 \, x}{2 \, x^{2} e^{5} - x e^{5} \log \left (2\right ) + x e^{5} \log \left (-x + 2\right ) - 4} \] Input:

integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2 
*x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8* 
x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.53 \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 \, x}{2 \, {\left (x - 2\right )}^{2} e^{5} + {\left (x - 2\right )} e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) + 8 \, {\left (x - 2\right )} e^{5} + 2 \, e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) + 8 \, e^{5} - 4} \] Input:

integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2 
*x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8* 
x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm="giac")
 

Output:

4*x/(2*(x - 2)^2*e^5 + (x - 2)*e^5*log(-1/2*x + 1) + 8*(x - 2)*e^5 + 2*e^5 
*log(-1/2*x + 1) + 8*e^5 - 4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\int \frac {{\mathrm {e}}^5\,\left (12\,x^2-8\,x^3\right )-16\,x+32}{-{\mathrm {e}}^{10}\,\left (2\,x^2-x^3\right )\,{\ln \left (1-\frac {x}{2}\right )}^2+\left ({\mathrm {e}}^5\,\left (16\,x-8\,x^2\right )-{\mathrm {e}}^{10}\,\left (8\,x^3-4\,x^4\right )\right )\,\ln \left (1-\frac {x}{2}\right )+16\,x-{\mathrm {e}}^{10}\,\left (8\,x^4-4\,x^5\right )+{\mathrm {e}}^5\,\left (32\,x^2-16\,x^3\right )-32} \,d x \] Input:

int((exp(5)*(12*x^2 - 8*x^3) - 16*x + 32)/(16*x + log(1 - x/2)*(exp(5)*(16 
*x - 8*x^2) - exp(10)*(8*x^3 - 4*x^4)) - exp(10)*(8*x^4 - 4*x^5) + exp(5)* 
(32*x^2 - 16*x^3) - exp(10)*log(1 - x/2)^2*(2*x^2 - x^3) - 32),x)
 

Output:

int((exp(5)*(12*x^2 - 8*x^3) - 16*x + 32)/(16*x + log(1 - x/2)*(exp(5)*(16 
*x - 8*x^2) - exp(10)*(8*x^3 - 4*x^4)) - exp(10)*(8*x^4 - 4*x^5) + exp(5)* 
(32*x^2 - 16*x^3) - exp(10)*log(1 - x/2)^2*(2*x^2 - x^3) - 32), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.81 \[ \int \frac {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx=\frac {4 x}{\mathrm {log}\left (1-\frac {x}{2}\right ) e^{5} x +2 e^{5} x^{2}-4} \] Input:

int(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2*x)^2+ 
((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8*x^4)*e 
xp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x)
 

Output:

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