3.93.56 \(\int \frac {e^{2 e^{-x}-x} (2 e^x-2 x \log (\frac {60}{x+4 e^2 x}))}{x \log ^3(\frac {60}{x+4 e^2 x})} \, dx\)

Optimal. Leaf size=28 \[ \frac {e^{2 e^{-x}}}{\log ^2\left (\frac {15}{\frac {x}{4}+e^2 x}\right )} \]

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Rubi [F]  time = 1.92, 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^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2/E^x - x)*(2*E^x - 2*x*Log[60/(x + 4*E^2*x)]))/(x*Log[60/(x + 4*E^2*x)]^3),x]

[Out]

2*Defer[Int][E^(2/E^x)/(x*Log[60/((1 + 4*E^2)*x)]^3), x] - 2*Defer[Int][E^(2/E^x - x)/Log[60/((1 + 4*E^2)*x)]^
2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )} \, dx\\ &=\int \frac {2 e^{2 e^{-x}-x} \left (e^x-x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )} \, dx\\ &=2 \int \frac {e^{2 e^{-x}-x} \left (e^x-x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )} \, dx\\ &=2 \int \left (\frac {e^{2 e^{-x}}}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}-\frac {e^{2 e^{-x}-x}}{\log ^2\left (\frac {60}{\left (1+4 e^2\right ) x}\right )}\right ) \, dx\\ &=2 \int \frac {e^{2 e^{-x}}}{x \log ^3\left (\frac {60}{\left (1+4 e^2\right ) x}\right )} \, dx-2 \int \frac {e^{2 e^{-x}-x}}{\log ^2\left (\frac {60}{\left (1+4 e^2\right ) x}\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.85, size = 25, normalized size = 0.89 \begin {gather*} \frac {e^{2 e^{-x}}}{\log ^2\left (\frac {60}{x+4 e^2 x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2/E^x - x)*(2*E^x - 2*x*Log[60/(x + 4*E^2*x)]))/(x*Log[60/(x + 4*E^2*x)]^3),x]

[Out]

E^(2/E^x)/Log[60/(x + 4*E^2*x)]^2

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fricas [A]  time = 0.69, size = 30, normalized size = 1.07 \begin {gather*} \frac {e^{\left (-{\left (x e^{x} - 2\right )} e^{\left (-x\right )} + x\right )}}{\log \left (\frac {60}{4 \, x e^{2} + x}\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x)/log(60/(4*x*exp(1)^2+x))^3,x, algo
rithm="fricas")

[Out]

e^(-(x*e^x - 2)*e^(-x) + x)/log(60/(4*x*e^2 + x))^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {2 \, {\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x)/log(60/(4*x*exp(1)^2+x))^3,x, algo
rithm="giac")

[Out]

integrate(-2*(x*log(60/(4*x*e^2 + x)) - e^x)*e^(-x + 2*e^(-x))/(x*log(60/(4*x*e^2 + x))^3), x)

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maple [C]  time = 15.04, size = 43, normalized size = 1.54

method	result	size
risch	\(-\frac {4 \,{\mathrm e}^{2 \,{\mathrm e}^{-x}}}{\left (2 i \ln \relax (5)+2 i \ln \relax (3)+4 i \ln \relax (2)-2 i \ln \left (4 \,{\mathrm e}^{2}+1\right )-2 i \ln \relax (x )\right )^{2}}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x)/ln(60/(4*x*exp(1)^2+x))^3,x,method=_RETUR
NVERBOSE)

[Out]

-4/(2*I*ln(5)+2*I*ln(3)+4*I*ln(2)-2*I*ln(4*exp(2)+1)-2*I*ln(x))^2*exp(2*exp(-x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -2 \, \int \frac {{\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(60/(4*x*exp(1)^2+x))+2*exp(x))*exp(1/exp(x))^2/x/exp(x)/log(60/(4*x*exp(1)^2+x))^3,x, algo
rithm="maxima")

[Out]

-2*integrate((x*log(60/(4*x*e^2 + x)) - e^x)*e^(-x + 2*e^(-x))/(x*log(60/(4*x*e^2 + x))^3), x)

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mupad [B]  time = 7.63, size = 22, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}{{\ln \left (\frac {60}{x+4\,x\,{\mathrm {e}}^2}\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*exp(-x))*exp(-x)*(2*exp(x) - 2*x*log(60/(x + 4*x*exp(2)))))/(x*log(60/(x + 4*x*exp(2)))^3),x)

[Out]

exp(2*exp(-x))/log(60/(x + 4*x*exp(2)))^2

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sympy [A]  time = 0.31, size = 19, normalized size = 0.68 \begin {gather*} \frac {e^{2 e^{- x}}}{\log {\left (\frac {60}{x + 4 x e^{2}} \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(60/(4*x*exp(1)**2+x))+2*exp(x))*exp(1/exp(x))**2/x/exp(x)/ln(60/(4*x*exp(1)**2+x))**3,x)

[Out]

exp(2*exp(-x))/log(60/(x + 4*x*exp(2)))**2

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