3.72.10 \(\int e^{-e^{-2 e+5 x}} x^{-2-\frac {1}{x^2}} (-3+3 x^2-15 e^{-2 e+5 x} x^3+6 \log (x)) \, dx\)

Optimal. Leaf size=28 \[ 3 e^{-e^{x+2 (-e+2 x)}} x^{1-\frac {1}{x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 24, normalized size of antiderivative = 0.86, number of steps used = 1, number of rules used = 1, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {2288} \begin {gather*} 3 e^{-e^{5 x-2 e}} x^{1-\frac {1}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(-2 - x^(-2))*(-3 + 3*x^2 - 15*E^(-2*E + 5*x)*x^3 + 6*Log[x]))/E^E^(-2*E + 5*x),x]

[Out]

(3*x^(1 - x^(-2)))/E^E^(-2*E + 5*x)

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} &=3 e^{-e^{-2 e+5 x}} x^{1-\frac {1}{x^2}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 24, normalized size = 0.86 \begin {gather*} 3 e^{-e^{-2 e+5 x}} x^{1-\frac {1}{x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(-2 - x^(-2))*(-3 + 3*x^2 - 15*E^(-2*E + 5*x)*x^3 + 6*Log[x]))/E^E^(-2*E + 5*x),x]

[Out]

(3*x^(1 - x^(-2)))/E^E^(-2*E + 5*x)

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fricas [A]  time = 0.92, size = 22, normalized size = 0.79 \begin {gather*} \frac {3 \, x e^{\left (-e^{\left (5 \, x - 2 \, e\right )}\right )}}{x^{\left (\frac {1}{x^{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(x)-15*x^3*exp(-2*exp(1)+5*x)+3*x^2-3)/x^2/exp(exp(-2*exp(1)+5*x))/exp(log(x)/x^2),x, algorith
m="fricas")

[Out]

3*x*e^(-e^(5*x - 2*e))/x^(x^(-2))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {3 \, {\left (5 \, x^{3} e^{\left (5 \, x - 2 \, e\right )} - x^{2} - 2 \, \log \relax (x) + 1\right )} e^{\left (-e^{\left (5 \, x - 2 \, e\right )}\right )}}{x^{2} x^{\left (\frac {1}{x^{2}}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(x)-15*x^3*exp(-2*exp(1)+5*x)+3*x^2-3)/x^2/exp(exp(-2*exp(1)+5*x))/exp(log(x)/x^2),x, algorith
m="giac")

[Out]

integrate(-3*(5*x^3*e^(5*x - 2*e) - x^2 - 2*log(x) + 1)*e^(-e^(5*x - 2*e))/(x^2*x^(x^(-2))), x)

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maple [A]  time = 0.03, size = 23, normalized size = 0.82

method	result	size
risch	\(3 x \,x^{-\frac {1}{x^{2}}} {\mathrm e}^{-{\mathrm e}^{-2 \,{\mathrm e}+5 x}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*ln(x)-15*x^3*exp(-2*exp(1)+5*x)+3*x^2-3)/x^2/exp(exp(-2*exp(1)+5*x))/exp(ln(x)/x^2),x,method=_RETURNVER
BOSE)

[Out]

3*x/(x^(1/x^2))*exp(-exp(-2*exp(1)+5*x))

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maxima [A]  time = 0.60, size = 23, normalized size = 0.82 \begin {gather*} 3 \, x e^{\left (-\frac {\log \relax (x)}{x^{2}} - e^{\left (5 \, x - 2 \, e\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*log(x)-15*x^3*exp(-2*exp(1)+5*x)+3*x^2-3)/x^2/exp(exp(-2*exp(1)+5*x))/exp(log(x)/x^2),x, algorith
m="maxima")

[Out]

3*x*e^(-log(x)/x^2 - e^(5*x - 2*e))

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mupad [B]  time = 4.39, size = 23, normalized size = 0.82 \begin {gather*} 3\,x^{1-\frac {1}{x^2}}\,{\mathrm {e}}^{-{\mathrm {e}}^{-2\,\mathrm {e}}\,{\mathrm {e}}^{5\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-exp(5*x - 2*exp(1)))*exp(-log(x)/x^2)*(6*log(x) - 15*x^3*exp(5*x - 2*exp(1)) + 3*x^2 - 3))/x^2,x)

[Out]

3*x^(1 - 1/x^2)*exp(-exp(-2*exp(1))*exp(5*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((6*ln(x)-15*x**3*exp(-2*exp(1)+5*x)+3*x**2-3)/x**2/exp(exp(-2*exp(1)+5*x))/exp(ln(x)/x**2),x)

[Out]

Timed out

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