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

Optimal. Leaf size=23 \[ 1-x+x \left (e^{e^{e^{\frac {1}{x}-x}}}+\log (2)\right ) \]

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Rubi [F]  time = 0.77, 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 {-x+e^{e^{e^{\frac {1-x^2}{x}}}} \left (x+e^{e^{\frac {1-x^2}{x}}+\frac {1-x^2}{x}} \left (-1-x^2\right )\right )+x \log (2)}{x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 19, normalized size = 0.83 \begin {gather*} x \left (-1+e^{e^{e^{\frac {1}{x}-x}}}+\log (2)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x*(-1 + E^E^E^(x^(-1) - x) + Log[2])

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fricas [A]  time = 0.69, size = 23, normalized size = 1.00 \begin {gather*} x e^{\left (e^{\left (e^{\left (-\frac {x^{2} - 1}{x}\right )}\right )}\right )} + x \log \relax (2) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.16, size = 24, normalized size = 1.04

method	result	size
norman	\(x \left (\ln \relax (2)-1\right )+x \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{\frac {-x^{2}+1}{x}}}}\)	\(24\)
risch	\(x \ln \relax (2)-x +x \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{-\frac {\left (x -1\right ) \left (x +1\right )}{x}}}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.59, size = 20, normalized size = 0.87 \begin {gather*} x e^{\left (e^{\left (e^{\left (-x + \frac {1}{x}\right )}\right )}\right )} + x \log \relax (2) - x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(e^(e^(-x + 1/x))) + x*log(2) - x

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mupad [B]  time = 1.22, size = 20, normalized size = 0.87 \begin {gather*} x\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{-x}\,{\mathrm {e}}^{1/x}}}+x\,\left (\ln \relax (2)-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(exp(exp(-x)*exp(1/x))) + x*(log(2) - 1)

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sympy [A]  time = 132.07, size = 19, normalized size = 0.83 \begin {gather*} x e^{e^{e^{\frac {1 - x^{2}}{x}}}} + x \left (-1 + \log {\relax (2 )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(exp(exp((1 - x**2)/x))) + x*(-1 + log(2))

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