3.98.5 \(\int \frac {2^{-2 x} e^{e^{\frac {2-12 x-2 x^2}{3 x}}} (-3 x+e^{\frac {2-12 x-2 x^2}{3 x}} (2+2 x^2)+6 x^2 \log (2))}{3 x} \, dx\)

Optimal. Leaf size=29 \[ 4-2^{-2 x} e^{e^{-4+\frac {1}{3} \left (\frac {2}{x}-2 x\right )}} x \]

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Rubi [A]  time = 0.31, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {12, 2288} \begin {gather*} -2^{-2 x} e^{e^{\frac {2 \left (-x^2-6 x+1\right )}{3 x}}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^((2 - 12*x - 2*x^2)/(3*x))*(-3*x + E^((2 - 12*x - 2*x^2)/(3*x))*(2 + 2*x^2) + 6*x^2*Log[2]))/(3*2^(2*
x)*x),x]

[Out]

-((E^E^((2*(1 - 6*x - x^2))/(3*x))*x)/2^(2*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{3} \int \frac {2^{-2 x} e^{e^{\frac {2-12 x-2 x^2}{3 x}}} \left (-3 x+e^{\frac {2-12 x-2 x^2}{3 x}} \left (2+2 x^2\right )+6 x^2 \log (2)\right )}{x} \, dx\\ &=-2^{-2 x} e^{e^{\frac {2 \left (1-6 x-x^2\right )}{3 x}}} x\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.45, size = 74, normalized size = 2.55 \begin {gather*} \frac {1}{3} \int \frac {2^{-2 x} e^{e^{\frac {2-12 x-2 x^2}{3 x}}} \left (-3 x+e^{\frac {2-12 x-2 x^2}{3 x}} \left (2+2 x^2\right )+6 x^2 \log (2)\right )}{x} \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^((2 - 12*x - 2*x^2)/(3*x))*(-3*x + E^((2 - 12*x - 2*x^2)/(3*x))*(2 + 2*x^2) + 6*x^2*Log[2]))/(3
*2^(2*x)*x),x]

[Out]

Integrate[(E^E^((2 - 12*x - 2*x^2)/(3*x))*(-3*x + E^((2 - 12*x - 2*x^2)/(3*x))*(2 + 2*x^2) + 6*x^2*Log[2]))/(2
^(2*x)*x), x]/3

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fricas [A]  time = 0.97, size = 25, normalized size = 0.86 \begin {gather*} -\frac {x e^{\left (e^{\left (-\frac {2 \, {\left (x^{2} + 6 \, x - 1\right )}}{3 \, x}\right )}\right )}}{2^{2 \, x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*x^2+2)*exp(1/3*(-2*x^2-12*x+2)/x)+6*x^2*log(2)-3*x)*exp(exp(1/3*(-2*x^2-12*x+2)/x))/x/exp(x*
log(2))^2,x, algorithm="fricas")

[Out]

-x*e^(e^(-2/3*(x^2 + 6*x - 1)/x))/2^(2*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (6 \, x^{2} \log \relax (2) + 2 \, {\left (x^{2} + 1\right )} e^{\left (-\frac {2 \, {\left (x^{2} + 6 \, x - 1\right )}}{3 \, x}\right )} - 3 \, x\right )} e^{\left (e^{\left (-\frac {2 \, {\left (x^{2} + 6 \, x - 1\right )}}{3 \, x}\right )}\right )}}{3 \cdot 2^{2 \, x} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*x^2+2)*exp(1/3*(-2*x^2-12*x+2)/x)+6*x^2*log(2)-3*x)*exp(exp(1/3*(-2*x^2-12*x+2)/x))/x/exp(x*
log(2))^2,x, algorithm="giac")

[Out]

integrate(1/3*(6*x^2*log(2) + 2*(x^2 + 1)*e^(-2/3*(x^2 + 6*x - 1)/x) - 3*x)*e^(e^(-2/3*(x^2 + 6*x - 1)/x))/(2^
(2*x)*x), x)

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maple [A]  time = 0.07, size = 24, normalized size = 0.83

method	result	size
risch	\(-x 2^{-2 x} {\mathrm e}^{{\mathrm e}^{-\frac {2 \left (x^{2}+6 x -1\right )}{3 x}}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((2*x^2+2)*exp(1/3*(-2*x^2-12*x+2)/x)+6*x^2*ln(2)-3*x)*exp(exp(1/3*(-2*x^2-12*x+2)/x))/x/exp(x*ln(2))^
2,x,method=_RETURNVERBOSE)

[Out]

-x/(2^x)^2*exp(exp(-2/3*(x^2+6*x-1)/x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{3} \, \int \frac {{\left (6 \, x^{2} \log \relax (2) + 2 \, {\left (x^{2} + 1\right )} e^{\left (-\frac {2 \, {\left (x^{2} + 6 \, x - 1\right )}}{3 \, x}\right )} - 3 \, x\right )} e^{\left (e^{\left (-\frac {2 \, {\left (x^{2} + 6 \, x - 1\right )}}{3 \, x}\right )}\right )}}{2^{2 \, x} x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*x^2+2)*exp(1/3*(-2*x^2-12*x+2)/x)+6*x^2*log(2)-3*x)*exp(exp(1/3*(-2*x^2-12*x+2)/x))/x/exp(x*
log(2))^2,x, algorithm="maxima")

[Out]

1/3*integrate((6*x^2*log(2) + 2*(x^2 + 1)*e^(-2/3*(x^2 + 6*x - 1)/x) - 3*x)*e^(e^(-2/3*(x^2 + 6*x - 1)/x))/(2^
(2*x)*x), x)

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mupad [B]  time = 5.96, size = 20, normalized size = 0.69 \begin {gather*} -{\left (\frac {1}{4}\right )}^x\,x\,{\mathrm {e}}^{{\mathrm {e}}^{-\frac {2\,x}{3}}\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^{\frac {2}{3\,x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(-(4*x + (2*x^2)/3 - 2/3)/x))*exp(-2*x*log(2))*(6*x^2*log(2) - 3*x + exp(-(4*x + (2*x^2)/3 - 2/3)/
x)*(2*x^2 + 2)))/(3*x),x)

[Out]

-(1/4)^x*x*exp(exp(-(2*x)/3)*exp(-4)*exp(2/(3*x)))

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sympy [A]  time = 115.41, size = 29, normalized size = 1.00 \begin {gather*} - x e^{- 2 x \log {\relax (2 )}} e^{e^{\frac {- \frac {2 x^{2}}{3} - 4 x + \frac {2}{3}}{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((2*x**2+2)*exp(1/3*(-2*x**2-12*x+2)/x)+6*x**2*ln(2)-3*x)*exp(exp(1/3*(-2*x**2-12*x+2)/x))/x/exp
(x*ln(2))**2,x)

[Out]

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

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