\(\int \frac {e^{-3-\frac {x}{e^3}} (e^3 (1-x)-2 x+x^2-x \log (x))}{x} \, dx\) [6471]

Optimal result

Integrand size = 35, antiderivative size = 16 \[ \int \frac {e^{-3-\frac {x}{e^3}} \left (e^3 (1-x)-2 x+x^2-x \log (x)\right )}{x} \, dx=e^{-\frac {x}{e^3}} (2-x+\log (x)) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.56, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2326} \[ \int \frac {e^{-3-\frac {x}{e^3}} \left (e^3 (1-x)-2 x+x^2-x \log (x)\right )}{x} \, dx=\frac {e^{-\frac {x}{e^3}} \left (-x^2+2 x+x \log (x)\right )}{x} \]

[In]

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

[Out]

(2*x - x^2 + x*Log[x])/(E^(x/E^3)*x)

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-\frac {x}{e^3}} \left (e^3 (1-x)-2 x+x^2-x \log (x)\right )}{x} \, dx=e^{-\frac {x}{e^3}} (2-x+\log (x)) \]

[In]

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

[Out]

(2 - x + Log[x])/E^(x/E^3)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.12

method	result	size
norman	\(\left (2+\ln \left (x \right )-x \right ) {\mathrm e}^{-x \,{\mathrm e}^{-3}}\)	\(18\)
risch	\(\ln \left (x \right ) {\mathrm e}^{-x \,{\mathrm e}^{-3}}-\left (-2+x \right ) {\mathrm e}^{-x \,{\mathrm e}^{-3}}\)	\(22\)
parallelrisch	\(-{\mathrm e}^{-3} \left (x \,{\mathrm e}^{3}-\ln \left (x \right ) {\mathrm e}^{3}-2 \,{\mathrm e}^{3}\right ) {\mathrm e}^{-x \,{\mathrm e}^{-3}}\)	\(31\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (15) = 30\).

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 2.19 \[ \int \frac {e^{-3-\frac {x}{e^3}} \left (e^3 (1-x)-2 x+x^2-x \log (x)\right )}{x} \, dx=-{\left (x - 2\right )} e^{\left (-{\left (x + 3 \, e^{3}\right )} e^{\left (-3\right )} + 3\right )} + e^{\left (-{\left (x + 3 \, e^{3}\right )} e^{\left (-3\right )} + 3\right )} \log \left (x\right ) \]

[In]

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

[Out]

-(x - 2)*e^(-(x + 3*e^3)*e^(-3) + 3) + e^(-(x + 3*e^3)*e^(-3) + 3)*log(x)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-3-\frac {x}{e^3}} \left (e^3 (1-x)-2 x+x^2-x \log (x)\right )}{x} \, dx=\left (- x + \log {\left (x \right )} + 2\right ) e^{- \frac {x}{e^{3}}} \]

[In]

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

[Out]

(-x + log(x) + 2)*exp(-x*exp(-3))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (15) = 30\).

Time = 0.20 (sec) , antiderivative size = 38, normalized size of antiderivative = 2.38 \[ \int \frac {e^{-3-\frac {x}{e^3}} \left (e^3 (1-x)-2 x+x^2-x \log (x)\right )}{x} \, dx=-{\left (x + e^{3}\right )} e^{\left (-x e^{\left (-3\right )}\right )} + e^{\left (-x e^{\left (-3\right )}\right )} \log \left (x\right ) + e^{\left (-x e^{\left (-3\right )} + 3\right )} + 2 \, e^{\left (-x e^{\left (-3\right )}\right )} \]

[In]

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

[Out]

-(x + e^3)*e^(-x*e^(-3)) + e^(-x*e^(-3))*log(x) + e^(-x*e^(-3) + 3) + 2*e^(-x*e^(-3))

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.85 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-3-\frac {x}{e^3}} \left (e^3 (1-x)-2 x+x^2-x \log (x)\right )}{x} \, dx={\mathrm {e}}^{-x\,{\mathrm {e}}^{-3}}\,\left (\ln \left (x\right )-x+2\right ) \]

[In]

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

[Out]

exp(-x*exp(-3))*(log(x) - x + 2)