\(\int \frac {\sqrt {2} (1-e^{2 x}-2 x) (\frac {1}{\log (e^{-2 x} (2 x-2 e^{2 x} x))})^{5/4}}{-4 x+4 e^{2 x} x} \, dx\) [680]

Optimal result

Integrand size = 57, antiderivative size = 26 \[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\sqrt {2} \sqrt [4]{\frac {1}{\log \left (2 \left (-x+e^{-2 x} x\right )\right )}} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\sqrt {2} \sqrt [4]{\frac {1}{\log \left (2 \left (-1+e^{-2 x}\right ) x\right )}} \] Input:

Integrate[(Sqrt[2]*(1 - E^(2*x) - 2*x)*(Log[(2*x - 2*E^(2*x)*x)/E^(2*x)]^( 
-1))^(5/4))/(-4*x + 4*E^(2*x)*x),x]
 

Output:

Sqrt[2]*(Log[2*(-1 + E^(-2*x))*x]^(-1))^(1/4)
 

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.

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

Int[(Sqrt[2]*(1 - E^(2*x) - 2*x)*(Log[(2*x - 2*E^(2*x)*x)/E^(2*x)]^(-1))^( 
5/4))/(-4*x + 4*E^(2*x)*x),x]
 

Output:

$Aborted
 

Maple [F]

\[\int \frac {\left (-{\mathrm e}^{2 x}+1-2 x \right ) \sqrt {2}\, {\left (\frac {1}{\ln \left (\left (-2 x \,{\mathrm e}^{2 x}+2 x \right ) {\mathrm e}^{-2 x}\right )}\right )}^{\frac {1}{4}}}{\left (4 x \,{\mathrm e}^{2 x}-4 x \right ) \ln \left (\left (-2 x \,{\mathrm e}^{2 x}+2 x \right ) {\mathrm e}^{-2 x}\right )}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\frac {\sqrt {2}}{\log \left (-2 \, {\left (x e^{\left (2 \, x\right )} - x\right )} e^{\left (-2 \, x\right )}\right )^{\frac {1}{4}}} \] Input:

integrate((-exp(x)^2+1-2*x)*2^(1/2)*(1/log((-2*x*exp(x)^2+2*x)/exp(x)^2))^ 
(1/4)/(4*x*exp(x)^2-4*x)/log((-2*x*exp(x)^2+2*x)/exp(x)^2),x, algorithm="f 
ricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\text {Timed out} \] Input:

integrate((-exp(x)**2+1-2*x)*2**(1/2)*(1/ln((-2*x*exp(x)**2+2*x)/exp(x)**2 
))**(1/4)/(4*x*exp(x)**2-4*x)/ln((-2*x*exp(x)**2+2*x)/exp(x)**2),x)
 

Output:

Timed out
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.22 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\frac {\sqrt {2}}{{\left (i \, \pi - 2 \, x + \log \left (2\right ) + \log \left (x\right ) + \log \left (e^{x} + 1\right ) + \log \left (e^{x} - 1\right )\right )}^{\frac {1}{4}}} \] Input:

integrate((-exp(x)^2+1-2*x)*2^(1/2)*(1/log((-2*x*exp(x)^2+2*x)/exp(x)^2))^ 
(1/4)/(4*x*exp(x)^2-4*x)/log((-2*x*exp(x)^2+2*x)/exp(x)^2),x, algorithm="m 
axima")
 

Output:

sqrt(2)/(I*pi - 2*x + log(2) + log(x) + log(e^x + 1) + log(e^x - 1))^(1/4)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\frac {\sqrt {2}}{\log \left (-2 \, {\left (x e^{\left (2 \, x\right )} - x\right )} e^{\left (-2 \, x\right )}\right )^{\frac {1}{4}}} \] Input:

integrate((-exp(x)^2+1-2*x)*2^(1/2)*(1/log((-2*x*exp(x)^2+2*x)/exp(x)^2))^ 
(1/4)/(4*x*exp(x)^2-4*x)/log((-2*x*exp(x)^2+2*x)/exp(x)^2),x, algorithm="g 
iac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 2.94 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\sqrt {2}\,{\left (\frac {1}{\ln \left (2\,x\,{\mathrm {e}}^{-2\,x}-2\,x\right )}\right )}^{1/4} \] Input:

int((2^(1/2)*(1/log(exp(-2*x)*(2*x - 2*x*exp(2*x))))^(1/4)*(2*x + exp(2*x) 
 - 1))/(log(exp(-2*x)*(2*x - 2*x*exp(2*x)))*(4*x - 4*x*exp(2*x))),x)
 

Output:

2^(1/2)*(1/log(2*x*exp(-2*x) - 2*x))^(1/4)
 

Reduce [F]

\[ \int \frac {\sqrt {2} \left (1-e^{2 x}-2 x\right ) \left (\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right )^{5/4}}{-4 x+4 e^{2 x} x} \, dx=\frac {\sqrt {2}\, \left (-\left (\int \frac {e^{2 x}}{e^{2 x} \mathrm {log}\left (\frac {-2 e^{2 x} x +2 x}{e^{2 x}}\right )^{\frac {5}{4}} x -\mathrm {log}\left (\frac {-2 e^{2 x} x +2 x}{e^{2 x}}\right )^{\frac {5}{4}} x}d x \right )+\int \frac {1}{e^{2 x} \mathrm {log}\left (\frac {-2 e^{2 x} x +2 x}{e^{2 x}}\right )^{\frac {5}{4}} x -\mathrm {log}\left (\frac {-2 e^{2 x} x +2 x}{e^{2 x}}\right )^{\frac {5}{4}} x}d x -2 \left (\int \frac {1}{e^{2 x} \mathrm {log}\left (\frac {-2 e^{2 x} x +2 x}{e^{2 x}}\right )^{\frac {5}{4}}-\mathrm {log}\left (\frac {-2 e^{2 x} x +2 x}{e^{2 x}}\right )^{\frac {5}{4}}}d x \right )\right )}{4} \] Input:

int((-exp(x)^2+1-2*x)*2^(1/2)*(1/log((-2*x*exp(x)^2+2*x)/exp(x)^2))^(1/4)/ 
(4*x*exp(x)^2-4*x)/log((-2*x*exp(x)^2+2*x)/exp(x)^2),x)
 

Output:

(sqrt(2)*( - int(e**(2*x)/(e**(2*x)*log(( - 2*e**(2*x)*x + 2*x)/e**(2*x))* 
*(1/4)*log(( - 2*e**(2*x)*x + 2*x)/e**(2*x))*x - log(( - 2*e**(2*x)*x + 2* 
x)/e**(2*x))**(1/4)*log(( - 2*e**(2*x)*x + 2*x)/e**(2*x))*x),x) + int(1/(e 
**(2*x)*log(( - 2*e**(2*x)*x + 2*x)/e**(2*x))**(1/4)*log(( - 2*e**(2*x)*x 
+ 2*x)/e**(2*x))*x - log(( - 2*e**(2*x)*x + 2*x)/e**(2*x))**(1/4)*log(( - 
2*e**(2*x)*x + 2*x)/e**(2*x))*x),x) - 2*int(1/(e**(2*x)*log(( - 2*e**(2*x) 
*x + 2*x)/e**(2*x))**(1/4)*log(( - 2*e**(2*x)*x + 2*x)/e**(2*x)) - log(( - 
 2*e**(2*x)*x + 2*x)/e**(2*x))**(1/4)*log(( - 2*e**(2*x)*x + 2*x)/e**(2*x) 
)),x)))/4