\(\int \frac {e^4 x-e^5 x+e^3 \log (2 e^{-2 x}) \log (x)+(-2 e^2 x+e^3 x) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx\) [938]

Optimal result

Integrand size = 76, antiderivative size = 34 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=2+\frac {1}{4} \left (2 x+\frac {e^3 \log \left (2 e^{-2 x}\right )}{e^2-\log ^2(x)}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=\frac {1}{2} \left (x+\frac {e^3 \log \left (2 e^{-2 x}\right )}{2 \left (e^2-\log ^2(x)\right )}\right ) \] Input:

Integrate[(E^4*x - E^5*x + E^3*Log[2/E^(2*x)]*Log[x] + (-2*E^2*x + E^3*x)* 
Log[x]^2 + x*Log[x]^4)/(2*E^4*x - 4*E^2*x*Log[x]^2 + 2*x*Log[x]^4),x]
 

Output:

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

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

\(\Big \downarrow \) 6

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(E^4*x - E^5*x + E^3*Log[2/E^(2*x)]*Log[x] + (-2*E^2*x + E^3*x)*Log[x] 
^2 + x*Log[x]^4)/(2*E^4*x - 4*E^2*x*Log[x]^2 + 2*x*Log[x]^4),x]
 

Output:

$Aborted
 

Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09

method	result	size
parallelrisch	\(\frac {-2 x \ln \left (x \right )^{2}+{\mathrm e}^{3} \ln \left (2 \,{\mathrm e}^{-2 x}\right )+2 \,{\mathrm e}^{2} x}{4 \,{\mathrm e}^{2}-4 \ln \left (x \right )^{2}}\)	\(37\)
risch	\(-\frac {{\mathrm e}^{3} \ln \left ({\mathrm e}^{x}\right )}{2 \left ({\mathrm e}^{2}-\ln \left (x \right )^{2}\right )}+\frac {-i {\mathrm e}^{3} \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{3}+2 i {\mathrm e}^{3} \pi \operatorname {csgn}\left (i {\mathrm e}^{2 x}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-i {\mathrm e}^{3} \pi \,\operatorname {csgn}\left (i {\mathrm e}^{2 x}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )^{2}-2 \,{\mathrm e}^{3} \ln \left (2\right )-4 \,{\mathrm e}^{2} x +4 x \ln \left (x \right )^{2}}{8 \ln \left (x \right )^{2}-8 \,{\mathrm e}^{2}}\)	\(112\)
default	\(-\frac {{\mathrm e}^{3} x}{2 \left ({\mathrm e}^{2}-\ln \left (x \right )^{2}\right )}-\frac {{\mathrm e}^{3} \left (\ln \left ({\mathrm e}^{x}\right )-x \right )}{2 \left ({\mathrm e}^{2}-\ln \left (x \right )^{2}\right )}+\frac {{\mathrm e}^{3} \left (\ln \left (2 \,{\mathrm e}^{-2 x}\right )+2 \ln \left ({\mathrm e}^{x}\right )\right )}{4 \,{\mathrm e}^{2}-4 \ln \left (x \right )^{2}}+\frac {{\mathrm e}^{2} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{3} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{3} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{2} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {x}{2}-\frac {{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{3} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{2} {\mathrm e}^{3} {\mathrm e}^{-2} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}+\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )+{\mathrm e}\right )}{8}-\frac {{\mathrm e}^{4} {\mathrm e}^{-2} {\mathrm e}^{-1} {\mathrm e}^{-{\mathrm e}} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-{\mathrm e}\right )}{8}\)	\(423\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=\frac {2 \, x \log \left (x\right )^{2} + 2 \, x e^{3} - 2 \, x e^{2} - e^{3} \log \left (2\right )}{4 \, {\left (\log \left (x\right )^{2} - e^{2}\right )}} \] Input:

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

Output:

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=\frac {x}{2} + \frac {2 x e^{3} - e^{3} \log {\left (2 \right )}}{4 \log {\left (x \right )}^{2} - 4 e^{2}} \] Input:

integrate((exp(3)*ln(x)*ln(2/exp(x)**2)+x*ln(x)**4+(x*exp(3)-2*exp(2)*x)*l 
n(x)**2-x*exp(2)*exp(3)+x*exp(2)**2)/(2*x*ln(x)**4-4*x*exp(2)*ln(x)**2+2*x 
*exp(2)**2),x)
 

Output:

x/2 + (2*x*exp(3) - exp(3)*log(2))/(4*log(x)**2 - 4*exp(2))
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=\frac {2 \, x \log \left (x\right )^{2} + 2 \, x {\left (e^{3} - e^{2}\right )} - e^{3} \log \left (2\right )}{4 \, {\left (\log \left (x\right )^{2} - e^{2}\right )}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=\frac {2 \, x \log \left (x\right )^{2} + 2 \, x e^{3} - 2 \, x e^{2} - e^{3} \log \left (2\right )}{4 \, {\left (\log \left (x\right )^{2} - e^{2}\right )}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 7.68 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=\frac {x}{2}-\frac {{\mathrm {e}}^3\,\left (2\,x-\ln \left (2\right )\right )}{4\,\left ({\mathrm {e}}^2-{\ln \left (x\right )}^2\right )} \] Input:

int((x*log(x)^4 + x*exp(4) - x*exp(5) - log(x)^2*(2*x*exp(2) - x*exp(3)) + 
 log(2*exp(-2*x))*exp(3)*log(x))/(2*x*log(x)^4 + 2*x*exp(4) - 4*x*exp(2)*l 
og(x)^2),x)
 

Output:

x/2 - (exp(3)*(2*x - log(2)))/(4*(exp(2) - log(x)^2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.29 \[ \int \frac {e^4 x-e^5 x+e^3 \log \left (2 e^{-2 x}\right ) \log (x)+\left (-2 e^2 x+e^3 x\right ) \log ^2(x)+x \log ^4(x)}{2 e^4 x-4 e^2 x \log ^2(x)+2 x \log ^4(x)} \, dx=\frac {-\mathrm {log}\left (\frac {2}{e^{2 x}}\right ) e^{3}+2 \mathrm {log}\left (x \right )^{2} x -2 e^{2} x}{4 \mathrm {log}\left (x \right )^{2}-4 e^{2}} \] Input:

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

Output:

( - log(2/e**(2*x))*e**3 + 2*log(x)**2*x - 2*e**2*x)/(4*(log(x)**2 - e**2) 
)