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

Optimal. Leaf size=17 \[ \frac {\log (3)}{x \left (e^{-2-x}+x\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 1.55, 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-x} (-1+x) \log (3)-2 x \log (3)}{e^{-4-2 x} x^2+2 e^{-2-x} x^3+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 25, normalized size = 1.47 \begin {gather*} -\left (\left (-\frac {1}{x^2}+\frac {1}{x^2 \left (1+e^{2+x} x\right )}\right ) \log (3)\right ) \end {gather*}

Antiderivative was successfully verified.

[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])

________________________________________________________________________________________

fricas [A]  time = 0.56, size = 17, normalized size = 1.00 \begin {gather*} \frac {\log \relax (3)}{x^{2} + x e^{\left (-x - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3)/(x^2 + x*e^(-x - 2))

________________________________________________________________________________________

giac [A]  time = 0.21, size = 19, normalized size = 1.12 \begin {gather*} \frac {e^{\left (x + 2\right )} \log \relax (3)}{x^{2} e^{\left (x + 2\right )} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x + 2)*log(3)/(x^2*e^(x + 2) + x)

________________________________________________________________________________________

maple [A]  time = 0.24, size = 17, normalized size = 1.00




method result size



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



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.53, size = 19, normalized size = 1.12 \begin {gather*} \frac {e^{\left (x + 2\right )} \log \relax (3)}{x^{2} e^{\left (x + 2\right )} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(x + 2)*log(3)/(x^2*e^(x + 2) + x)

________________________________________________________________________________________

mupad [B]  time = 4.16, size = 17, normalized size = 1.00 \begin {gather*} \frac {\ln \relax (3)}{x\,{\mathrm {e}}^{-x-2}+x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 14, normalized size = 0.82 \begin {gather*} \frac {\log {\relax (3 )}}{x^{2} + x e^{- x - 2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(3)/(x**2 + x*exp(-x - 2))

________________________________________________________________________________________