3.71.12 \(\int (2+e^{12 e^{\frac {4-x}{x}}} (-192 e^{\frac {4-x}{x}}+8 x)) \, dx\)

Optimal. Leaf size=22 \[ x \left (2+4 e^{12 e^{\frac {4-x}{x}}} x\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 36, normalized size of antiderivative = 1.64, number of steps used = 2, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2288} \begin {gather*} \frac {16 e^{12 e^{\frac {4-x}{x}}}}{\frac {4-x}{x^2}+\frac {1}{x}}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2 + E^(12*E^((4 - x)/x))*(-192*E^((4 - x)/x) + 8*x),x]

[Out]

(16*E^(12*E^((4 - x)/x)))/((4 - x)/x^2 + x^(-1)) + 2*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} &=2 x+\int e^{12 e^{\frac {4-x}{x}}} \left (-192 e^{\frac {4-x}{x}}+8 x\right ) \, dx\\ &=\frac {16 e^{12 e^{\frac {4-x}{x}}}}{\frac {4-x}{x^2}+\frac {1}{x}}+2 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 22, normalized size = 1.00 \begin {gather*} 2 x+4 e^{12 e^{-1+\frac {4}{x}}} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2 + E^(12*E^((4 - x)/x))*(-192*E^((4 - x)/x) + 8*x),x]

[Out]

2*x + 4*E^(12*E^(-1 + 4/x))*x^2

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fricas [A]  time = 0.57, size = 21, normalized size = 0.95 \begin {gather*} 4 \, x^{2} e^{\left (12 \, e^{\left (-\frac {x - 4}{x}\right )}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-192*exp((-x+4)/x)+8*x)*exp(12*exp((-x+4)/x))+2,x, algorithm="fricas")

[Out]

4*x^2*e^(12*e^(-(x - 4)/x)) + 2*x

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 8 \, {\left (x - 24 \, e^{\left (-\frac {x - 4}{x}\right )}\right )} e^{\left (12 \, e^{\left (-\frac {x - 4}{x}\right )}\right )} + 2\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-192*exp((-x+4)/x)+8*x)*exp(12*exp((-x+4)/x))+2,x, algorithm="giac")

[Out]

integrate(8*(x - 24*e^(-(x - 4)/x))*e^(12*e^(-(x - 4)/x)) + 2, x)

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maple [A]  time = 0.05, size = 22, normalized size = 1.00

method	result	size
risch	\(2 x +4 x^{2} {\mathrm e}^{12 \,{\mathrm e}^{-\frac {x -4}{x}}}\)	\(22\)
default	\(2 x +4 x^{2} {\mathrm e}^{12 \,{\mathrm e}^{\frac {-x +4}{x}}}\)	\(23\)
norman	\(2 x +4 x^{2} {\mathrm e}^{12 \,{\mathrm e}^{\frac {-x +4}{x}}}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-192*exp((-x+4)/x)+8*x)*exp(12*exp((-x+4)/x))+2,x,method=_RETURNVERBOSE)

[Out]

2*x+4*x^2*exp(12*exp(-(x-4)/x))

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maxima [A]  time = 0.43, size = 20, normalized size = 0.91 \begin {gather*} 4 \, x^{2} e^{\left (12 \, e^{\left (\frac {4}{x} - 1\right )}\right )} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-192*exp((-x+4)/x)+8*x)*exp(12*exp((-x+4)/x))+2,x, algorithm="maxima")

[Out]

4*x^2*e^(12*e^(4/x - 1)) + 2*x

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mupad [B]  time = 4.13, size = 20, normalized size = 0.91 \begin {gather*} 2\,x+4\,x^2\,{\mathrm {e}}^{12\,{\mathrm {e}}^{-1}\,{\mathrm {e}}^{4/x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(12*exp(-(x - 4)/x))*(8*x - 192*exp(-(x - 4)/x)) + 2,x)

[Out]

2*x + 4*x^2*exp(12*exp(-1)*exp(4/x))

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sympy [A]  time = 0.85, size = 17, normalized size = 0.77 \begin {gather*} 4 x^{2} e^{12 e^{\frac {4 - x}{x}}} + 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-192*exp((-x+4)/x)+8*x)*exp(12*exp((-x+4)/x))+2,x)

[Out]

4*x**2*exp(12*exp((4 - x)/x)) + 2*x

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