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

Optimal. Leaf size=19 \[ e^{2-\frac {1+\frac {\log (x)}{x}}{x^2}} x \]

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Rubi [B]  time = 0.10, antiderivative size = 79, normalized size of antiderivative = 4.16, number of steps used = 1, number of rules used = 1, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2288} \begin {gather*} -\frac {e^{\frac {2 x-\frac {x+\log (x)}{x^2}}{x}} (-2 x-3 \log (x)+1)}{x^3 \left (\frac {\frac {2 (x+\log (x))}{x^3}-\frac {\frac {1}{x}+1}{x^2}+2}{x}-\frac {2 x-\frac {x+\log (x)}{x^2}}{x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((E^((2*x - (x + Log[x])/x^2)/x)*(1 - 2*x - 3*Log[x]))/(x^3*((2 - (1 + x^(-1))/x^2 + (2*(x + Log[x]))/x^3)/x
- (2*x - (x + Log[x])/x^2)/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-\frac {x+\log (x)}{x^2}}{x}} (1-2 x-3 \log (x))}{x^3 \left (\frac {2-\frac {1+\frac {1}{x}}{x^2}+\frac {2 (x+\log (x))}{x^3}}{x}-\frac {2 x-\frac {x+\log (x)}{x^2}}{x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 19, normalized size = 1.00 \begin {gather*} e^{2-\frac {1}{x^2}} x^{1-\frac {1}{x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(2 - x^(-2))*x^(1 - x^(-3))

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fricas [A]  time = 0.78, size = 20, normalized size = 1.05 \begin {gather*} x e^{\left (\frac {2 \, x^{3} - x - \log \relax (x)}{x^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^((2*x^3 - x - log(x))/x^3)

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giac [A]  time = 0.16, size = 20, normalized size = 1.05 \begin {gather*} x e^{\left (\frac {2 \, x^{3} - x - \log \relax (x)}{x^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^((2*x^3 - x - log(x))/x^3)

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maple [A]  time = 0.03, size = 18, normalized size = 0.95

method	result	size
risch	\(x \,{\mathrm e}^{-\frac {-2 x^{3}+\ln \relax (x )+x}{x^{3}}}\)	\(18\)
norman	\(x \,{\mathrm e}^{\frac {\frac {-x -\ln \relax (x )}{x^{2}}+2 x}{x}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 17, normalized size = 0.89 \begin {gather*} x e^{\left (-\frac {1}{x^{2}} - \frac {\log \relax (x)}{x^{3}} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*e^(-1/x^2 - log(x)/x^3 + 2)

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mupad [B]  time = 7.43, size = 18, normalized size = 0.95 \begin {gather*} x^{1-\frac {1}{x^3}}\,{\mathrm {e}}^2\,{\mathrm {e}}^{-\frac {1}{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x^(1 - 1/x^3)*exp(2)*exp(-1/x^2)

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sympy [A]  time = 0.33, size = 17, normalized size = 0.89 \begin {gather*} x e^{\frac {2 x + \frac {- x - \log {\relax (x )}}{x^{2}}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp((2*x + (-x - log(x))/x**2)/x)

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