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

Optimal. Leaf size=22 \[ e^{\frac {-2-x-x^2+\log (x)}{x}} x^2 \]

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Rubi [B]  time = 0.04, antiderivative size = 63, normalized size of antiderivative = 2.86, number of steps used = 1, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2288} \begin {gather*} -\frac {e^{-\frac {x^2+x+2}{x}} x^{\frac {1}{x}} \left (-x^2-\log (x)+3\right )}{\frac {2 x-\frac {1}{x}+1}{x}-\frac {x^2+x-\log (x)+2}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2288

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

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Mathematica [A]  time = 0.36, size = 20, normalized size = 0.91 \begin {gather*} e^{-1-\frac {2}{x}-x} x^{2+\frac {1}{x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-1 - 2/x - x)*x^(2 + x^(-1))

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fricas [A]  time = 1.10, size = 20, normalized size = 0.91 \begin {gather*} x^{2} e^{\left (-\frac {x^{2} + x - \log \relax (x) + 2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(-(x^2 + x - log(x) + 2)/x)

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giac [A]  time = 0.22, size = 20, normalized size = 0.91 \begin {gather*} x^{2} e^{\left (-\frac {x^{2} + x - \log \relax (x) + 2}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(-(x^2 + x - log(x) + 2)/x)

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maple [A]  time = 0.02, size = 22, normalized size = 1.00

method	result	size
norman	\({\mathrm e}^{\frac {\ln \relax (x )-x^{2}-x -2}{x}} x^{2}\)	\(22\)
risch	\({\mathrm e}^{\frac {\ln \relax (x )-x^{2}-x -2}{x}} x^{2}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp((ln(x)-x^2-x-2)/x)*x^2

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maxima [A]  time = 0.43, size = 21, normalized size = 0.95 \begin {gather*} x^{2} e^{\left (-x + \frac {\log \relax (x)}{x} - \frac {2}{x} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^2*e^(-x + log(x)/x - 2/x - 1)

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mupad [B]  time = 4.35, size = 20, normalized size = 0.91 \begin {gather*} x^{\frac {1}{x}+2}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-\frac {2}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1/x + 2)*exp(-x)*exp(-1)*exp(-2/x)

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sympy [A]  time = 0.32, size = 15, normalized size = 0.68 \begin {gather*} x^{2} e^{\frac {- x^{2} - x + \log {\relax (x )} - 2}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*exp((-x**2 - x + log(x) - 2)/x)

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