3.16.76 \(\int \frac {e^{-x+\frac {2-x-x^2+x^3-\log (x)}{x^2}} (-20+4 x+8 x^2+8 \log (x))}{x} \, dx\)

Optimal. Leaf size=20 \[ 4 e^{-\frac {-2+x+x^2+\log (x)}{x^2}} x^2 \]

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Rubi [B]  time = 0.16, antiderivative size = 84, normalized size of antiderivative = 4.20, number of steps used = 1, number of rules used = 1, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {2288} \begin {gather*} \frac {4 e^{\frac {x^3-x^2-x-\log (x)+2}{x^2}-x} (-x-2 \log (x)+5)}{x \left (\frac {-3 x^2+2 x+\frac {1}{x}+1}{x^2}+\frac {2 \left (x^3-x^2-x-\log (x)+2\right )}{x^3}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-x + (2 - x - x^2 + x^3 - Log[x])/x^2)*(-20 + 4*x + 8*x^2 + 8*Log[x]))/x,x]

[Out]

(4*E^(-x + (2 - x - x^2 + x^3 - Log[x])/x^2)*(5 - x - 2*Log[x]))/(x*(1 + (1 + x^(-1) + 2*x - 3*x^2)/x^2 + (2*(
2 - x - x^2 + x^3 - Log[x]))/x^3))

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 {4 e^{-x+\frac {2-x-x^2+x^3-\log (x)}{x^2}} (5-x-2 \log (x))}{x \left (1+\frac {1+\frac {1}{x}+2 x-3 x^2}{x^2}+\frac {2 \left (2-x-x^2+x^3-\log (x)\right )}{x^3}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.44, size = 25, normalized size = 1.25 \begin {gather*} 4 e^{-1+\frac {2}{x^2}-\frac {1}{x}} x^{2-\frac {1}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-x + (2 - x - x^2 + x^3 - Log[x])/x^2)*(-20 + 4*x + 8*x^2 + 8*Log[x]))/x,x]

[Out]

4*E^(-1 + 2/x^2 - x^(-1))*x^(2 - x^(-2))

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fricas [A]  time = 0.84, size = 19, normalized size = 0.95 \begin {gather*} 4 \, x^{2} e^{\left (-\frac {x^{2} + x + \log \relax (x) - 2}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(x)+8*x^2+4*x-20)*exp((-log(x)+x^3-x^2-x+2)/x^2)/exp(x)/x,x, algorithm="fricas")

[Out]

4*x^2*e^(-(x^2 + x + log(x) - 2)/x^2)

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giac [A]  time = 0.28, size = 19, normalized size = 0.95 \begin {gather*} 4 \, x^{2} e^{\left (-\frac {x^{2} + x + \log \relax (x) - 2}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(x)+8*x^2+4*x-20)*exp((-log(x)+x^3-x^2-x+2)/x^2)/exp(x)/x,x, algorithm="giac")

[Out]

4*x^2*e^(-(x^2 + x + log(x) - 2)/x^2)

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maple [A]  time = 0.02, size = 20, normalized size = 1.00




method result size



risch \(4 x^{2} {\mathrm e}^{-\frac {\ln \relax (x )+x^{2}+x -2}{x^{2}}}\) \(20\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*ln(x)+8*x^2+4*x-20)*exp((-ln(x)+x^3-x^2-x+2)/x^2)/exp(x)/x,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.63, size = 25, normalized size = 1.25 \begin {gather*} 4 \, x^{2} e^{\left (-\frac {1}{x} - \frac {\log \relax (x)}{x^{2}} + \frac {2}{x^{2}} - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*log(x)+8*x^2+4*x-20)*exp((-log(x)+x^3-x^2-x+2)/x^2)/exp(x)/x,x, algorithm="maxima")

[Out]

4*x^2*e^(-1/x - log(x)/x^2 + 2/x^2 - 1)

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mupad [B]  time = 1.08, size = 25, normalized size = 1.25 \begin {gather*} 4\,x^{2-\frac {1}{x^2}}\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{-\frac {1}{x}}\,{\mathrm {e}}^{\frac {2}{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*exp(-(x + log(x) + x^2 - x^3 - 2)/x^2)*(4*x + 8*log(x) + 8*x^2 - 20))/x,x)

[Out]

4*x^(2 - 1/x^2)*exp(-1)*exp(-1/x)*exp(2/x^2)

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sympy [A]  time = 67.50, size = 26, normalized size = 1.30 \begin {gather*} 4 x^{2} e^{- x} e^{\frac {x^{3} - x^{2} - x - \log {\relax (x )} + 2}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*ln(x)+8*x**2+4*x-20)*exp((-ln(x)+x**3-x**2-x+2)/x**2)/exp(x)/x,x)

[Out]

4*x**2*exp(-x)*exp((x**3 - x**2 - x - log(x) + 2)/x**2)

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