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

3.7.80.1 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)
 

3.7.80.2 Mathematica [A] (verified)

Time = 0.08 (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)
 

3.7.80.3 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
 

3.7.80.3.1 Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.7.80.4 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)
 

3.7.80.5 Fricas [A] (verification not implemented)

Time = 0.25 (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=\
 

output

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

3.7.80.6 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
 

3.7.80.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.36 (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=\
 

output

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

3.7.80.8 Giac [A] (verification not implemented)

Time = 0.27 (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=\
 

output

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

3.7.80.9 Mupad [B] (verification not implemented)

Time = 9.40 (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)