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

Optimal. Leaf size=26 \[ \sqrt {2} \sqrt [4]{\frac {1}{\log \left (2 \left (-x+e^{-2 x} x\right )\right )}} \]

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Rubi [F]  time = 1.37, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\sqrt {2} \int \frac {\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\\ &=\left (\sqrt {2} \sqrt [4]{\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}} \sqrt [4]{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right ) \int \frac {1-e^{2 x}-2 x}{\left (-4 x+4 e^{2 x} x\right ) \log ^{\frac {5}{4}}\left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )} \, dx\\ &=\left (\sqrt {2} \sqrt [4]{\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}} \sqrt [4]{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right ) \int \frac {-1+e^{2 x}+2 x}{4 \left (1-e^{2 x}\right ) x \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )} \, dx\\ &=\frac {\left (\sqrt [4]{\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}} \sqrt [4]{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right ) \int \frac {-1+e^{2 x}+2 x}{\left (1-e^{2 x}\right ) x \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )} \, dx}{2 \sqrt {2}}\\ &=\frac {\left (\sqrt [4]{\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}} \sqrt [4]{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right ) \int \left (-\frac {1}{\left (-1+e^x\right ) \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )}+\frac {1}{\left (1+e^x\right ) \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )}-\frac {1}{x \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )}\right ) \, dx}{2 \sqrt {2}}\\ &=-\frac {\left (\sqrt [4]{\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}} \sqrt [4]{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right ) \int \frac {1}{\left (-1+e^x\right ) \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )} \, dx}{2 \sqrt {2}}+\frac {\left (\sqrt [4]{\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}} \sqrt [4]{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right ) \int \frac {1}{\left (1+e^x\right ) \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )} \, dx}{2 \sqrt {2}}-\frac {\left (\sqrt [4]{\frac {1}{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}} \sqrt [4]{\log \left (e^{-2 x} \left (2 x-2 e^{2 x} x\right )\right )}\right ) \int \frac {1}{x \log ^{\frac {5}{4}}\left (-2 e^{-2 x} \left (-1+e^{2 x}\right ) x\right )} \, dx}{2 \sqrt {2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 23, normalized size = 0.88 \begin {gather*} \sqrt {2} \sqrt [4]{\frac {1}{\log \left (2 \left (-1+e^{-2 x}\right ) x\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

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

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fricas [A]  time = 1.03, size = 23, normalized size = 0.88 \begin {gather*} \frac {\sqrt {2}}{\log \left (-2 \, {\left (x e^{\left (2 \, x\right )} - x\right )} e^{\left (-2 \, x\right )}\right )^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*e
xp(x)^2+2*x)/exp(x)^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 23, normalized size = 0.88 \begin {gather*} \frac {\sqrt {2}}{\log \left (-2 \, {\left (x e^{\left (2 \, x\right )} - x\right )} e^{\left (-2 \, x\right )}\right )^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*e
xp(x)^2+2*x)/exp(x)^2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (-{\mathrm e}^{2 x}+1-2 x \right ) \sqrt {2}\, \left (\frac {1}{\ln \left (\left (-2 x \,{\mathrm e}^{2 x}+2 x \right ) {\mathrm e}^{-2 x}\right )}\right )^{\frac {1}{4}}}{\left (4 x \,{\mathrm e}^{2 x}-4 x \right ) \ln \left (\left (-2 x \,{\mathrm e}^{2 x}+2 x \right ) {\mathrm e}^{-2 x}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

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maxima [C]  time = 0.60, size = 27, normalized size = 1.04 \begin {gather*} \frac {\sqrt {2}}{{\left (i \, \pi - 2 \, x + \log \relax (2) + \log \relax (x) + \log \left (e^{x} + 1\right ) + \log \left (e^{x} - 1\right )\right )}^{\frac {1}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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*e
xp(x)^2+2*x)/exp(x)^2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.40, size = 20, normalized size = 0.77 \begin {gather*} \sqrt {2}\,{\left (\frac {1}{\ln \left (2\,x\,{\mathrm {e}}^{-2\,x}-2\,x\right )}\right )}^{1/4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

Timed out

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