3.101.84 \(\int \frac {1}{2} (4+e^{-e^{-e^{\frac {1}{2} (-2 x^2-x^3+(-2 x-x^2) \log (3))}+x}+2 x} (-4+e^{-e^{\frac {1}{2} (-2 x^2-x^3+(-2 x-x^2) \log (3))}+x} (2+e^{\frac {1}{2} (-2 x^2-x^3+(-2 x-x^2) \log (3))} (4 x+3 x^2+(2+2 x) \log (3))))) \, dx\)

Optimal. Leaf size=36 \[ -e^{-e^{-e^{\frac {1}{2} (-2-x) x (x+\log (3))}+x}+2 x}+2 x \]

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Rubi [A]  time = 1.60, antiderivative size = 52, normalized size of antiderivative = 1.44, number of steps used = 3, number of rules used = 2, integrand size = 137, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {12, 6706} \begin {gather*} 2 x-\exp \left (2 x-\exp \left (x-3^{\frac {1}{2} \left (-x^2-2 x\right )} e^{\frac {1}{2} \left (-x^3-2 x^2\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-E^(-E^(-(3^((-2*x - x^2)/2)*E^((-2*x^2 - x^3)/2)) + x) + 2*x) + 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 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \left (4+\exp \left (-\exp \left (-\exp \left (\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )\right )+x\right )+2 x\right ) \left (-4+\exp \left (-\exp \left (\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )\right )+x\right ) \left (2+\exp \left (\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )\right ) \left (4 x+3 x^2+(2+2 x) \log (3)\right )\right )\right )\right ) \, dx\\ &=2 x+\frac {1}{2} \int \exp \left (-\exp \left (-\exp \left (\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )\right )+x\right )+2 x\right ) \left (-4+\exp \left (-\exp \left (\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )\right )+x\right ) \left (2+\exp \left (\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )\right ) \left (4 x+3 x^2+(2+2 x) \log (3)\right )\right )\right ) \, dx\\ &=-\exp \left (-\exp \left (-3^{\frac {1}{2} \left (-2 x-x^2\right )} e^{\frac {1}{2} \left (-2 x^2-x^3\right )}+x\right )+2 x\right )+2 x\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 7.93, size = 139, normalized size = 3.86 \begin {gather*} \frac {1}{2} \int \left (4+e^{-e^{-e^{\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )}+x}+2 x} \left (-4+e^{-e^{\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )}+x} \left (2+e^{\frac {1}{2} \left (-2 x^2-x^3+\left (-2 x-x^2\right ) \log (3)\right )} \left (4 x+3 x^2+(2+2 x) \log (3)\right )\right )\right )\right ) \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.73, size = 41, normalized size = 1.14 \begin {gather*} 2 \, x - e^{\left (2 \, x - e^{\left (x - e^{\left (-\frac {1}{2} \, x^{3} - x^{2} - \frac {1}{2} \, {\left (x^{2} + 2 \, x\right )} \log \relax (3)\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((2*x+2)*log(3)+3*x^2+4*x)*exp(1/2*(-x^2-2*x)*log(3)-1/2*x^3-x^2)+2)*exp(-exp(1/2*(-x^2-2*x)*l
og(3)-1/2*x^3-x^2)+x)-4)*exp(-exp(-exp(1/2*(-x^2-2*x)*log(3)-1/2*x^3-x^2)+x)+2*x)+2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.89, size = 42, normalized size = 1.17 \begin {gather*} 2 \, x - e^{\left (2 \, x - e^{\left (x - e^{\left (-\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} \log \relax (3) - x^{2} - x \log \relax (3)\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((2*x+2)*log(3)+3*x^2+4*x)*exp(1/2*(-x^2-2*x)*log(3)-1/2*x^3-x^2)+2)*exp(-exp(1/2*(-x^2-2*x)*l
og(3)-1/2*x^3-x^2)+x)-4)*exp(-exp(-exp(1/2*(-x^2-2*x)*log(3)-1/2*x^3-x^2)+x)+2*x)+2,x, algorithm="giac")

[Out]

2*x - e^(2*x - e^(x - e^(-1/2*x^3 - 1/2*x^2*log(3) - x^2 - x*log(3))))

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maple [A]  time = 0.28, size = 36, normalized size = 1.00

method	result	size
risch	\(2 x -{\mathrm e}^{-{\mathrm e}^{-3^{-\frac {x \left (2+x \right )}{2}} {\mathrm e}^{-\frac {x^{2} \left (2+x \right )}{2}}+x}+2 x}\)	\(36\)
default	\(2 x -{\mathrm e}^{-{\mathrm e}^{-{\mathrm e}^{\frac {\left (-x^{2}-2 x \right ) \ln \relax (3)}{2}-\frac {x^{3}}{2}-x^{2}}+x}+2 x}\)	\(44\)
norman	\(2 x -{\mathrm e}^{-{\mathrm e}^{-{\mathrm e}^{\frac {\left (-x^{2}-2 x \right ) \ln \relax (3)}{2}-\frac {x^{3}}{2}-x^{2}}+x}+2 x}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((((2*x+2)*ln(3)+3*x^2+4*x)*exp(1/2*(-x^2-2*x)*ln(3)-1/2*x^3-x^2)+2)*exp(-exp(1/2*(-x^2-2*x)*ln(3)-1/2
*x^3-x^2)+x)-4)*exp(-exp(-exp(1/2*(-x^2-2*x)*ln(3)-1/2*x^3-x^2)+x)+2*x)+2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.65, size = 42, normalized size = 1.17 \begin {gather*} 2 \, x - e^{\left (2 \, x - e^{\left (x - e^{\left (-\frac {1}{2} \, x^{3} - \frac {1}{2} \, x^{2} \log \relax (3) - x^{2} - x \log \relax (3)\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((2*x+2)*log(3)+3*x^2+4*x)*exp(1/2*(-x^2-2*x)*log(3)-1/2*x^3-x^2)+2)*exp(-exp(1/2*(-x^2-2*x)*l
og(3)-1/2*x^3-x^2)+x)-4)*exp(-exp(-exp(1/2*(-x^2-2*x)*log(3)-1/2*x^3-x^2)+x)+2*x)+2,x, algorithm="maxima")

[Out]

2*x - e^(2*x - e^(x - e^(-1/2*x^3 - 1/2*x^2*log(3) - x^2 - x*log(3))))

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mupad [B]  time = 7.96, size = 42, normalized size = 1.17 \begin {gather*} 2\,x-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-{\mathrm {e}}^{-\frac {{\mathrm {e}}^{-x^2}}{3^x\,\sqrt {{\mathrm {e}}^{x^3}}\,\sqrt {3^{x^2}}}}\,{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 3.04, size = 32, normalized size = 0.89 \begin {gather*} 2 x - e^{2 x - e^{x - e^{- \frac {x^{3}}{2} - x^{2} + \left (- \frac {x^{2}}{2} - x\right ) \log {\relax (3 )}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((((2*x+2)*ln(3)+3*x**2+4*x)*exp(1/2*(-x**2-2*x)*ln(3)-1/2*x**3-x**2)+2)*exp(-exp(1/2*(-x**2-2*x
)*ln(3)-1/2*x**3-x**2)+x)-4)*exp(-exp(-exp(1/2*(-x**2-2*x)*ln(3)-1/2*x**3-x**2)+x)+2*x)+2,x)

[Out]

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

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