3.71.68 \(\int \frac {e^{\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}} (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2))))}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx\)

Optimal. Leaf size=30 \[ e^{\frac {\frac {1}{2}-e^x}{-i \pi +x-\log (4-\log (2))}} \]

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Rubi [F]  time = 5.06, 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 {\exp \left (\frac {-1+2 e^x}{-2 x+2 (i \pi +\log (4-\log (2)))}\right ) \left (-1+e^x (2-2 x)+2 e^x (i \pi +\log (4-\log (2)))\right )}{2 x^2-4 x (i \pi +\log (4-\log (2)))+2 (i \pi +\log (4-\log (2)))^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.50, size = 31, normalized size = 1.03 \begin {gather*} e^{\frac {-1+2 e^x}{2 (i \pi -x+\log (4-\log (2)))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 + 2*E^x)/(-2*x + 2*(I*Pi + Log[4 - Log[2]])))*(-1 + E^x*(2 - 2*x) + 2*E^x*(I*Pi + Log[4 - Lo
g[2]])))/(2*x^2 - 4*x*(I*Pi + Log[4 - Log[2]]) + 2*(I*Pi + Log[4 - Log[2]])^2),x]

[Out]

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

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fricas [A]  time = 0.84, size = 20, normalized size = 0.67 \begin {gather*} e^{\left (-\frac {2 \, e^{x} - 1}{2 \, {\left (x - \log \left (\log \relax (2) - 4\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.29, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (-\frac {e^{x}}{x - \log \left (\log \relax (2) - 4\right )} + \frac {1}{2 \, {\left (x - \log \left (\log \relax (2) - 4\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.30, size = 21, normalized size = 0.70

method	result	size
risch	\({\mathrm e}^{-\frac {2 \,{\mathrm e}^{x}-1}{2 \left (x -\ln \left (\ln \relax (2)-4\right )\right )}}\)	\(21\)
norman	\(\frac {\ln \left (\ln \relax (2)-4\right ) {\mathrm e}^{\frac {2 \,{\mathrm e}^{x}-1}{2 \ln \left (\ln \relax (2)-4\right )-2 x}}-x \,{\mathrm e}^{\frac {2 \,{\mathrm e}^{x}-1}{2 \ln \left (\ln \relax (2)-4\right )-2 x}}}{\ln \left (\ln \relax (2)-4\right )-x}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 30, normalized size = 1.00 \begin {gather*} e^{\left (-\frac {e^{x}}{x - \log \left (\log \relax (2) - 4\right )} + \frac {1}{2 \, {\left (x - \log \left (\log \relax (2) - 4\right )\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.69, size = 31, normalized size = 1.03 \begin {gather*} {\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{x-\ln \left (\ln \relax (2)-4\right )}}\,{\mathrm {e}}^{\frac {1}{2\,x-2\,\ln \left (\ln \relax (2)-4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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