3.56.65 \(\int \frac {e^{e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}} (-1+e^{-x+e^{-x} x} (1-x) x)}{8 x^2} \, dx\)

Optimal. Leaf size=33 \[ \frac {1}{4} e^{e^{\frac {1}{4} e^{\frac {e^{e^{e^{-x} x}}+3 x}{x}}}} \]

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Rubi [F]  time = 10.05, 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 (e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}\right ) \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{8 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(E^(E^((E^E^(x/E^x) + 3*x - x*Log[2])/x)/2) + E^(x/E^x) + E^((E^E^(x/E^x) + 3*x - x*Log[2])/x)/2 + (E^E
^(x/E^x) + 3*x - x*Log[2])/x)*(-1 + E^(-x + x/E^x)*(1 - x)*x))/(8*x^2),x]

[Out]

-1/8*Defer[Int][E^(-(((-1 + E^x)*x)/E^x) + (4*E^E^(x/E^x) + 4*E^(E^(3 + E^E^(x/E^x)/x)/4)*x + E^(3 + E^E^(x/E^
x)/x)*x + 4*E^(x/E^x)*x + 12*x*(1 - Log[2]/3))/(4*x)), x] - Defer[Int][E^((4*E^E^(x/E^x) + 4*E^(E^(3 + E^E^(x/
E^x)/x)/4)*x + E^(3 + E^E^(x/E^x)/x)*x + 4*E^(x/E^x)*x + 12*x*(1 - Log[2]/3))/(4*x))/x^2, x]/8 + Defer[Int][E^
(-(((-1 + E^x)*x)/E^x) + (4*E^E^(x/E^x) + 4*E^(E^(3 + E^E^(x/E^x)/x)/4)*x + E^(3 + E^E^(x/E^x)/x)*x + 4*E^(x/E
^x)*x + 12*x*(1 - Log[2]/3))/(4*x))/x, x]/8

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{8} \int \frac {\exp \left (e^{\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}}+e^{e^{-x} x}+\frac {1}{2} e^{\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}}+\frac {e^{e^{e^{-x} x}}+3 x-x \log (2)}{x}\right ) \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{x^2} \, dx\\ &=\frac {1}{8} \int \frac {\exp \left (\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right ) \left (-1+e^{-x+e^{-x} x} (1-x) x\right )}{x^2} \, dx\\ &=\frac {1}{8} \int \left (-\frac {\exp \left (\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right )}{x^2}+\frac {\exp \left (-e^{-x} \left (-1+e^x\right ) x+\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right ) (1-x)}{x}\right ) \, dx\\ &=-\left (\frac {1}{8} \int \frac {\exp \left (\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right )}{x^2} \, dx\right )+\frac {1}{8} \int \frac {\exp \left (-e^{-x} \left (-1+e^x\right ) x+\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right ) (1-x)}{x} \, dx\\ &=\frac {1}{8} \int \left (-\exp \left (-e^{-x} \left (-1+e^x\right ) x+\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right )+\frac {\exp \left (-e^{-x} \left (-1+e^x\right ) x+\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right )}{x}\right ) \, dx-\frac {1}{8} \int \frac {\exp \left (\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right )}{x^2} \, dx\\ &=-\left (\frac {1}{8} \int \exp \left (-e^{-x} \left (-1+e^x\right ) x+\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right ) \, dx\right )-\frac {1}{8} \int \frac {\exp \left (\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right )}{x^2} \, dx+\frac {1}{8} \int \frac {\exp \left (-e^{-x} \left (-1+e^x\right ) x+\frac {4 e^{e^{e^{-x} x}}+4 e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}} x+e^{3+\frac {e^{e^{e^{-x} x}}}{x}} x+4 e^{e^{-x} x} x+12 x \left (1-\frac {\log (2)}{3}\right )}{4 x}\right )}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 31, normalized size = 0.94 \begin {gather*} \frac {1}{4} e^{e^{\frac {1}{4} e^{3+\frac {e^{e^{e^{-x} x}}}{x}}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(E^((E^E^(x/E^x) + 3*x - x*Log[2])/x)/2) + E^(x/E^x) + E^((E^E^(x/E^x) + 3*x - x*Log[2])/x)/2
+ (E^E^(x/E^x) + 3*x - x*Log[2])/x)*(-1 + E^(-x + x/E^x)*(1 - x)*x))/(8*x^2),x]

[Out]

E^E^(E^(3 + E^E^(x/E^x)/x)/4)/4

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fricas [B]  time = 0.63, size = 198, normalized size = 6.00 \begin {gather*} \frac {1}{4} \, e^{\left (\frac {{\left (x e^{\left (-x - \frac {x \log \relax (2) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{x} + \log \relax (x)\right )} + 2 \, x e^{\left (-x + e^{\left (-x + \log \relax (x)\right )} + \log \relax (x)\right )} + 2 \, x e^{\left (-x + \frac {1}{2} \, e^{\left (-\frac {x \log \relax (2) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{x}\right )} + \log \relax (x)\right )} - 2 \, {\left (x \log \relax (2) - 3 \, x\right )} e^{\left (-x + \log \relax (x)\right )} + 2 \, e^{\left (-x + e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )} + \log \relax (x)\right )}\right )} e^{\left (x - \log \relax (x)\right )}}{2 \, x} + \frac {x \log \relax (2) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{x} - \frac {1}{2} \, e^{\left (-\frac {x \log \relax (2) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{x}\right )} - e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-x+1)*exp(log(x)-x)*exp(exp(log(x)-x))-1)*exp(exp(exp(log(x)-x)))*exp((exp(exp(exp(log(x)-x)))
-x*log(2)+3*x)/x)*exp(1/2*exp((exp(exp(exp(log(x)-x)))-x*log(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(log(x)-x
)))-x*log(2)+3*x)/x)))/x^2,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {{\left ({\left (x - 1\right )} e^{\left (-x + e^{\left (-x + \log \relax (x)\right )} + \log \relax (x)\right )} + 1\right )} e^{\left (-\frac {x \log \relax (2) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{x} + e^{\left (\frac {1}{2} \, e^{\left (-\frac {x \log \relax (2) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{x}\right )}\right )} + \frac {1}{2} \, e^{\left (-\frac {x \log \relax (2) - 3 \, x - e^{\left (e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{x}\right )} + e^{\left (e^{\left (-x + \log \relax (x)\right )}\right )}\right )}}{8 \, x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-x+1)*exp(log(x)-x)*exp(exp(log(x)-x))-1)*exp(exp(exp(log(x)-x)))*exp((exp(exp(exp(log(x)-x)))
-x*log(2)+3*x)/x)*exp(1/2*exp((exp(exp(exp(log(x)-x)))-x*log(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(log(x)-x
)))-x*log(2)+3*x)/x)))/x^2,x, algorithm="giac")

[Out]

integrate(-1/8*((x - 1)*e^(-x + e^(-x + log(x)) + log(x)) + 1)*e^(-(x*log(2) - 3*x - e^(e^(e^(-x + log(x)))))/
x + e^(1/2*e^(-(x*log(2) - 3*x - e^(e^(e^(-x + log(x)))))/x)) + 1/2*e^(-(x*log(2) - 3*x - e^(e^(e^(-x + log(x)
))))/x) + e^(e^(-x + log(x))))/x^2, x)

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maple [A]  time = 0.23, size = 24, normalized size = 0.73

method	result	size
risch	\(\frac {{\mathrm e}^{{\mathrm e}^{\frac {{\mathrm e}^{\frac {{\mathrm e}^{{\mathrm e}^{x \,{\mathrm e}^{-x}}}+3 x}{x}}}{4}}}}{4}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/8*((1-x)*exp(ln(x)-x)*exp(exp(ln(x)-x))-1)*exp(exp(exp(ln(x)-x)))*exp((exp(exp(exp(ln(x)-x)))-x*ln(2)+3*
x)/x)*exp(1/2*exp((exp(exp(exp(ln(x)-x)))-x*ln(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(ln(x)-x)))-x*ln(2)+3*x
)/x)))/x^2,x,method=_RETURNVERBOSE)

[Out]

1/4*exp(exp(1/4*exp((exp(exp(x*exp(-x)))+3*x)/x)))

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maxima [A]  time = 1.88, size = 21, normalized size = 0.64 \begin {gather*} \frac {1}{4} \, e^{\left (e^{\left (\frac {1}{4} \, e^{\left (\frac {e^{\left (e^{\left (x e^{\left (-x\right )}\right )}\right )}}{x} + 3\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/8*((-x+1)*exp(log(x)-x)*exp(exp(log(x)-x))-1)*exp(exp(exp(log(x)-x)))*exp((exp(exp(exp(log(x)-x)))
-x*log(2)+3*x)/x)*exp(1/2*exp((exp(exp(exp(log(x)-x)))-x*log(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(log(x)-x
)))-x*log(2)+3*x)/x)))/x^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.72, size = 21, normalized size = 0.64 \begin {gather*} \frac {{\mathrm {e}}^{{\mathrm {e}}^{\frac {{\mathrm {e}}^3\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^{x\,{\mathrm {e}}^{-x}}}}{x}}}{4}}}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp((3*x + exp(exp(exp(log(x) - x))) - x*log(2))/x)/2)*exp(exp(exp(log(x) - x)))*exp((3*x + exp(exp(
exp(log(x) - x))) - x*log(2))/x)*exp(exp(exp((3*x + exp(exp(exp(log(x) - x))) - x*log(2))/x)/2))*(exp(log(x) -
x)*exp(exp(log(x) - x))*(x - 1) + 1))/(8*x^2),x)

[Out]

exp(exp((exp(3)*exp(exp(exp(x*exp(-x)))/x))/4))/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(1/8*((-x+1)*exp(ln(x)-x)*exp(exp(ln(x)-x))-1)*exp(exp(exp(ln(x)-x)))*exp((exp(exp(exp(ln(x)-x)))-x*l
n(2)+3*x)/x)*exp(1/2*exp((exp(exp(exp(ln(x)-x)))-x*ln(2)+3*x)/x))*exp(exp(1/2*exp((exp(exp(exp(ln(x)-x)))-x*ln
(2)+3*x)/x)))/x**2,x)

[Out]

Timed out

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