\(\int \frac {-8-18 x-4 \log (5)+e^{1-x} (20 x+4 x^2+8 x \log (5))+x \log (x^2)}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+(-64 x+16 e^{1-x} x) \log (x^2)+2 x \log ^2(x^2)} \, dx\) [1771]

Optimal result

Integrand size = 92, antiderivative size = 32 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {2+\frac {x}{2}+\log (5)}{4 \left (-4+e^{1-x}+\frac {\log \left (x^2\right )}{4}\right )} \] Output:

(2+ln(5)+1/2*x)/(ln(x^2)+4*exp(1-x)-16)
 

Mathematica [A] (verified)

Time = 5.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {e^x (4+x+\log (25))}{2 \left (4 e-16 e^x+e^x \log \left (x^2\right )\right )} \] Input:

Integrate[(-8 - 18*x - 4*Log[5] + E^(1 - x)*(20*x + 4*x^2 + 8*x*Log[5]) + 
x*Log[x^2])/(512*x + 32*E^(2 - 2*x)*x - 256*E^(1 - x)*x + (-64*x + 16*E^(1 
 - x)*x)*Log[x^2] + 2*x*Log[x^2]^2),x]
 

Output:

(E^x*(4 + x + Log[25]))/(2*(4*E - 16*E^x + E^x*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.

\(\displaystyle \int \frac {x \log \left (x^2\right )+e^{1-x} \left (4 x^2+20 x+8 x \log (5)\right )-18 x-8-4 \log (5)}{2 x \log ^2\left (x^2\right )+\left (16 e^{1-x} x-64 x\right ) \log \left (x^2\right )+32 e^{2-2 x} x-256 e^{1-x} x+512 x} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {e^{2 x} \left (x \log \left (x^2\right )+e^{1-x} \left (4 x^2+20 x+8 x \log (5)\right )-18 x-8 \left (1+\frac {\log (5)}{2}\right )\right )}{2 x \left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} \int -\frac {e^{2 x} \left (-\log \left (x^2\right ) x+18 x-4 e^{1-x} \left (x^2+2 \log (5) x+5 x\right )+4 (2+\log (5))\right )}{x \left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {1}{2} \int \frac {e^{2 x} \left (-\log \left (x^2\right ) x+18 x-4 e^{1-x} \left (x^2+2 \log (5) x+5 x\right )+\log (625)+8\right )}{x \left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle -\frac {1}{2} \int \frac {e^{2 x} \left (-\log \left (x^2\right ) x+18 x-4 e^{1-x} \left (x^2+2 \log (5) x+5 x\right )+8 \left (1+\frac {\log (5)}{2}\right )\right )}{x \left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {1}{2} \int \left (-\frac {1}{4} e^{x-1} (x+\log (25)+5)+\frac {e^{2 x-1} \left (\log \left (x^2\right )-16\right ) (x+\log (25)+5)}{4 \left (e^x \log \left (x^2\right )-16 e^x+4 e\right )}+\frac {e^{2 x} (x+\log (25)+4) \left (\log \left (x^2\right ) x-16 x+2\right )}{x \left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (16 (4+\log (25)) \int \frac {e^{2 x}}{\left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx-2 \int \frac {e^{2 x}}{\left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx-2 (4+\log (25)) \int \frac {e^{2 x}}{x \left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx+16 \int \frac {e^{2 x} x}{\left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx-(4+\log (25)) \int \frac {e^{2 x} \log \left (x^2\right )}{\left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx-\int \frac {e^{2 x} x \log \left (x^2\right )}{\left (e^x \log \left (x^2\right )-16 e^x+4 e\right )^2}dx+4 (5+\log (25)) \int \frac {e^{2 x-1}}{e^x \log \left (x^2\right )-16 e^x+4 e}dx+4 \int \frac {e^{2 x-1} x}{e^x \log \left (x^2\right )-16 e^x+4 e}dx-\frac {1}{4} (5+\log (25)) \int \frac {e^{2 x-1} \log \left (x^2\right )}{e^x \log \left (x^2\right )-16 e^x+4 e}dx-\frac {1}{4} \int \frac {e^{2 x-1} x \log \left (x^2\right )}{e^x \log \left (x^2\right )-16 e^x+4 e}dx-\frac {e^{x-1}}{4}+\frac {1}{4} e^{x-1} (x+5+\log (25))\right )\)

Input:

Int[(-8 - 18*x - 4*Log[5] + E^(1 - x)*(20*x + 4*x^2 + 8*x*Log[5]) + x*Log[ 
x^2])/(512*x + 32*E^(2 - 2*x)*x - 256*E^(1 - x)*x + (-64*x + 16*E^(1 - x)* 
x)*Log[x^2] + 2*x*Log[x^2]^2),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88

method	result	size
parallelrisch	\(\frac {16+8 \ln \left (5\right )+4 x}{8 \ln \left (x^{2}\right )+32 \,{\mathrm e}^{1-x}-128}\)	\(28\)
risch	\(\frac {i \left (2 \ln \left (5\right )+x +4\right )}{\pi \operatorname {csgn}\left (i x \right )^{2} \operatorname {csgn}\left (i x^{2}\right )-2 \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x^{2}\right )^{2}+\pi \operatorname {csgn}\left (i x^{2}\right )^{3}+8 i {\mathrm e}^{1-x}+4 i \ln \left (x \right )-32 i}\)	\(74\)

Input:

int((x*ln(x^2)+(8*x*ln(5)+4*x^2+20*x)*exp(1-x)-4*ln(5)-18*x-8)/(2*x*ln(x^2 
)^2+(16*x*exp(1-x)-64*x)*ln(x^2)+32*x*exp(1-x)^2-256*x*exp(1-x)+512*x),x,m 
ethod=_RETURNVERBOSE)
 

Output:

1/8*(16+8*ln(5)+4*x)/(ln(x^2)+4*exp(1-x)-16)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {x + 2 \, \log \left (5\right ) + 4}{2 \, {\left (4 \, e^{\left (-x + 1\right )} + \log \left (x^{2}\right ) - 16\right )}} \] Input:

integrate((x*log(x^2)+(8*x*log(5)+4*x^2+20*x)*exp(1-x)-4*log(5)-18*x-8)/(2 
*x*log(x^2)^2+(16*x*exp(1-x)-64*x)*log(x^2)+32*x*exp(1-x)^2-256*x*exp(1-x) 
+512*x),x, algorithm="fricas")
 

Output:

1/2*(x + 2*log(5) + 4)/(4*e^(-x + 1) + log(x^2) - 16)
 

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {x + 2 \log {\left (5 \right )} + 4}{8 e^{1 - x} + 2 \log {\left (x^{2} \right )} - 32} \] Input:

integrate((x*ln(x**2)+(8*x*ln(5)+4*x**2+20*x)*exp(1-x)-4*ln(5)-18*x-8)/(2* 
x*ln(x**2)**2+(16*x*exp(1-x)-64*x)*ln(x**2)+32*x*exp(1-x)**2-256*x*exp(1-x 
)+512*x),x)
 

Output:

(x + 2*log(5) + 4)/(8*exp(1 - x) + 2*log(x**2) - 32)
 

Maxima [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {{\left (x + 2 \, \log \left (5\right ) + 4\right )} e^{x}}{4 \, {\left ({\left (\log \left (x\right ) - 8\right )} e^{x} + 2 \, e\right )}} \] Input:

integrate((x*log(x^2)+(8*x*log(5)+4*x^2+20*x)*exp(1-x)-4*log(5)-18*x-8)/(2 
*x*log(x^2)^2+(16*x*exp(1-x)-64*x)*log(x^2)+32*x*exp(1-x)^2-256*x*exp(1-x) 
+512*x),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {x + 2 \, \log \left (5\right ) + 4}{2 \, {\left (4 \, e^{\left (-x + 1\right )} + \log \left (x^{2}\right ) - 16\right )}} \] Input:

integrate((x*log(x^2)+(8*x*log(5)+4*x^2+20*x)*exp(1-x)-4*log(5)-18*x-8)/(2 
*x*log(x^2)^2+(16*x*exp(1-x)-64*x)*log(x^2)+32*x*exp(1-x)^2-256*x*exp(1-x) 
+512*x),x, algorithm="giac")
 

Output:

1/2*(x + 2*log(5) + 4)/(4*e^(-x + 1) + log(x^2) - 16)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=-\int \frac {18\,x+4\,\ln \left (5\right )-x\,\ln \left (x^2\right )-{\mathrm {e}}^{1-x}\,\left (20\,x+8\,x\,\ln \left (5\right )+4\,x^2\right )+8}{2\,x\,{\ln \left (x^2\right )}^2+\left (16\,x\,{\mathrm {e}}^{1-x}-64\,x\right )\,\ln \left (x^2\right )+512\,x-256\,x\,{\mathrm {e}}^{1-x}+32\,x\,{\mathrm {e}}^{2-2\,x}} \,d x \] Input:

int(-(18*x + 4*log(5) - x*log(x^2) - exp(1 - x)*(20*x + 8*x*log(5) + 4*x^2 
) + 8)/(512*x - log(x^2)*(64*x - 16*x*exp(1 - x)) - 256*x*exp(1 - x) + 32* 
x*exp(2 - 2*x) + 2*x*log(x^2)^2),x)
 

Output:

-int((18*x + 4*log(5) - x*log(x^2) - exp(1 - x)*(20*x + 8*x*log(5) + 4*x^2 
) + 8)/(512*x - log(x^2)*(64*x - 16*x*exp(1 - x)) - 256*x*exp(1 - x) + 32* 
x*exp(2 - 2*x) + 2*x*log(x^2)^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-8-18 x-4 \log (5)+e^{1-x} \left (20 x+4 x^2+8 x \log (5)\right )+x \log \left (x^2\right )}{512 x+32 e^{2-2 x} x-256 e^{1-x} x+\left (-64 x+16 e^{1-x} x\right ) \log \left (x^2\right )+2 x \log ^2\left (x^2\right )} \, dx=\frac {e^{x} \left (2 \,\mathrm {log}\left (5\right )+x +4\right )}{2 e^{x} \mathrm {log}\left (x^{2}\right )-32 e^{x}+8 e} \] Input:

int((x*log(x^2)+(8*x*log(5)+4*x^2+20*x)*exp(1-x)-4*log(5)-18*x-8)/(2*x*log 
(x^2)^2+(16*x*exp(1-x)-64*x)*log(x^2)+32*x*exp(1-x)^2-256*x*exp(1-x)+512*x 
),x)
 

Output:

(e**x*(2*log(5) + x + 4))/(2*(e**x*log(x**2) - 16*e**x + 4*e))