\(\int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx\) [7099]

Optimal result

Integrand size = 49, antiderivative size = 17 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\log (3)}{x \left (e^{-2-x}+x\right )} \]

[Out]

ln(3)/x/(x+exp(-2-x))

Rubi [F]

\[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.47 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=-\left (\left (-\frac {1}{x^2}+\frac {1}{x^2 \left (1+e^{2+x} x\right )}\right ) \log (3)\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
norman	\(\frac {\ln \left (3\right )}{x \left (x +{\mathrm e}^{-2-x}\right )}\)	\(17\)
risch	\(\frac {\ln \left (3\right )}{x \left (x +{\mathrm e}^{-2-x}\right )}\)	\(17\)
parallelrisch	\(\frac {\ln \left (3\right )}{x \left (x +{\mathrm e}^{-2-x}\right )}\)	\(17\)

[In]

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

[Out]

ln(3)/x/(x+exp(-2-x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\log \left (3\right )}{x^{2} + x e^{\left (-x - 2\right )}} \]

[In]

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

[Out]

log(3)/(x^2 + x*e^(-x - 2))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\log {\left (3 \right )}}{x^{2} + x e^{- x - 2}} \]

[In]

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

[Out]

log(3)/(x**2 + x*exp(-x - 2))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {e^{\left (x + 2\right )} \log \left (3\right )}{x^{2} e^{\left (x + 2\right )} + x} \]

[In]

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

[Out]

e^(x + 2)*log(3)/(x^2*e^(x + 2) + x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {e^{2} \log \left (3\right )}{x^{2} e^{2} + x e^{\left (-x\right )}} \]

[In]

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

[Out]

e^2*log(3)/(x^2*e^2 + x*e^(-x))

Mupad [B] (verification not implemented)

Time = 12.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-2-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx=\frac {\ln \left (3\right )}{x\,{\mathrm {e}}^{-x-2}+x^2} \]

[In]

int(-(2*x*log(3) - exp(- x - 2)*log(3)*(x - 1))/(2*x^3*exp(- x - 2) + x^2*exp(- 2*x - 4) + x^4),x)

[Out]

log(3)/(x*exp(- x - 2) + x^2)