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

Optimal. Leaf size=24 \[ 3-e^{\frac {e^{e^x} \left (-2+\frac {2}{x^2}\right ) x}{\log (2)}} \]

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Rubi [F]  time = 5.03, 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 {e^{e^x+\frac {e^{e^x} \left (2-2 x^2\right )}{x \log (2)}} \left (2+2 x^2+e^x \left (-2 x+2 x^3\right )\right )}{x^2 \log (2)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.95, size = 23, normalized size = 0.96 \begin {gather*} -e^{-\frac {2 e^{e^x} \left (-1+x^2\right )}{x \log (2)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^((-2*E^E^x*(-1 + x^2))/(x*Log[2]))

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fricas [A]  time = 0.69, size = 33, normalized size = 1.38 \begin {gather*} -e^{\left (\frac {x e^{x} \log \relax (2) - 2 \, {\left (x^{2} - 1\right )} e^{\left (e^{x}\right )}}{x \log \relax (2)} - e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp(x))/x/log(2))/x^2/log(2),x, algorith
m="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp(x))/x/log(2))/x^2/log(2),x, algorith
m="giac")

[Out]

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

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maple [A]  time = 0.07, size = 22, normalized size = 0.92

method	result	size
risch	\(-{\mathrm e}^{-\frac {2 \left (x -1\right ) \left (x +1\right ) {\mathrm e}^{{\mathrm e}^{x}}}{x \ln \relax (2)}}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp(x))/x/ln(2))/x^2/ln(2),x,method=_RETURNVER
BOSE)

[Out]

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

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maxima [A]  time = 0.56, size = 26, normalized size = 1.08 \begin {gather*} -e^{\left (-\frac {2 \, x e^{\left (e^{x}\right )}}{\log \relax (2)} + \frac {2 \, e^{\left (e^{x}\right )}}{x \log \relax (2)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3-2*x)*exp(x)+2*x^2+2)*exp(exp(x))*exp((-2*x^2+2)*exp(exp(x))/x/log(2))/x^2/log(2),x, algorith
m="maxima")

[Out]

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

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mupad [B]  time = 5.90, size = 25, normalized size = 1.04 \begin {gather*} -{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{{\mathrm {e}}^x}-2\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{x\,\ln \relax (2)}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.45, size = 19, normalized size = 0.79 \begin {gather*} - e^{\frac {\left (2 - 2 x^{2}\right ) e^{e^{x}}}{x \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3-2*x)*exp(x)+2*x**2+2)*exp(exp(x))*exp((-2*x**2+2)*exp(exp(x))/x/ln(2))/x**2/ln(2),x)

[Out]

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

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