\(\int \frac {e^{2 e^{-x}-x} (2 e^x-2 x \log (\frac {60}{x+4 e^2 x}))}{x \log ^3(\frac {60}{x+4 e^2 x})} \, dx\) [9256]

Optimal result

Integrand size = 54, antiderivative size = 28 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{-x}}}{\log ^2\left (\frac {15}{\frac {x}{4}+e^2 x}\right )} \]

[Out]

exp(1/exp(x))^2/ln(3/(1/5*x*exp(1)^2+1/20*x))^2

Rubi [F]

\[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 2.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{-x}}}{\log ^2\left (\frac {60}{x+4 e^2 x}\right )} \]

[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

Maple [A] (verified)

Time = 2.61 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{-x}}}{{\ln \left (\frac {60}{x \left (4 \,{\mathrm e}^{2}+1\right )}\right )}^{2}}\)	\(27\)

[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]

exp(1/exp(x))^2/ln(60/x/(4*exp(1)^2+1))^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{\left (-{\left (x e^{x} - 2\right )} e^{\left (-x\right )} + x\right )}}{\log \left (\frac {60}{4 \, x e^{2} + x}\right )^{2}} \]

[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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.68 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {e^{2 e^{- x}}}{\log {\left (\frac {60}{x + 4 x e^{2}} \right )}^{2}} \]

[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

Maxima [F]

\[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\int { -\frac {2 \, {\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\int { -\frac {2 \, {\left (x \log \left (\frac {60}{4 \, x e^{2} + x}\right ) - e^{x}\right )} e^{\left (-x + 2 \, e^{\left (-x\right )}\right )}}{x \log \left (\frac {60}{4 \, x e^{2} + x}\right )^{3}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 14.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {e^{2 e^{-x}-x} \left (2 e^x-2 x \log \left (\frac {60}{x+4 e^2 x}\right )\right )}{x \log ^3\left (\frac {60}{x+4 e^2 x}\right )} \, dx=\frac {{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}{{\ln \left (\frac {60}{x+4\,x\,{\mathrm {e}}^2}\right )}^2} \]

[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