3.14.64 \(\int \frac {1}{3} e^{\frac {1}{3} (-3 x+24 x^2-4 e^{e^x} x^2+3 \log (\frac {5}{4}))} (-12+192 x+e^{e^x} (-32 x-16 e^x x^2)) \, dx\)

Optimal. Leaf size=24 \[ 5 e^{-x-4 \left (-2+\frac {e^{e^x}}{3}\right ) x^2} \]

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Rubi [A]  time = 0.32, antiderivative size = 27, normalized size of antiderivative = 1.12, number of steps used = 4, number of rules used = 3, integrand size = 58, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {12, 2274, 6706} \begin {gather*} 5 e^{\frac {1}{3} \left (-4 e^{e^x} x^2+24 x^2-3 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-3*x + 24*x^2 - 4*E^E^x*x^2 + 3*Log[5/4])/3)*(-12 + 192*x + E^E^x*(-32*x - 16*E^x*x^2)))/3,x]

[Out]

5*E^((-3*x + 24*x^2 - 4*E^E^x*x^2)/3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
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}{3} \int \exp \left (\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2+3 \log \left (\frac {5}{4}\right )\right )\right ) \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx\\ &=\frac {1}{3} \int \frac {5}{4} e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx\\ &=\frac {5}{12} \int e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2\right )} \left (-12+192 x+e^{e^x} \left (-32 x-16 e^x x^2\right )\right ) \, dx\\ &=5 e^{\frac {1}{3} \left (-3 x+24 x^2-4 e^{e^x} x^2\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.62, size = 21, normalized size = 0.88 \begin {gather*} 5 e^{-\frac {1}{3} x \left (3+4 \left (-6+e^{e^x}\right ) x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-3*x + 24*x^2 - 4*E^E^x*x^2 + 3*Log[5/4])/3)*(-12 + 192*x + E^E^x*(-32*x - 16*E^x*x^2)))/3,x]

[Out]

5/E^((x*(3 + 4*(-6 + E^E^x)*x))/3)

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fricas [A]  time = 0.92, size = 22, normalized size = 0.92 \begin {gather*} 4 \, e^{\left (-\frac {4}{3} \, x^{2} e^{\left (e^{x}\right )} + 8 \, x^{2} - x + \log \left (\frac {5}{4}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-16*exp(x)*x^2-32*x)*exp(exp(x))+192*x-12)*exp(-4/3*exp(exp(x))*x^2+log(5/4)+8*x^2-x),x, algor
ithm="fricas")

[Out]

4*e^(-4/3*x^2*e^(e^x) + 8*x^2 - x + log(5/4))

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giac [A]  time = 0.46, size = 22, normalized size = 0.92 \begin {gather*} 4 \, e^{\left (-\frac {4}{3} \, x^{2} e^{\left (e^{x}\right )} + 8 \, x^{2} - x + \log \left (\frac {5}{4}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-16*exp(x)*x^2-32*x)*exp(exp(x))+192*x-12)*exp(-4/3*exp(exp(x))*x^2+log(5/4)+8*x^2-x),x, algor
ithm="giac")

[Out]

4*e^(-4/3*x^2*e^(e^x) + 8*x^2 - x + log(5/4))

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maple [A]  time = 0.06, size = 18, normalized size = 0.75

method	result	size
risch	\(5 \,{\mathrm e}^{-\frac {x \left (4 x \,{\mathrm e}^{{\mathrm e}^{x}}-24 x +3\right )}{3}}\)	\(18\)
norman	\(4 \,{\mathrm e}^{-\frac {4 \,{\mathrm e}^{{\mathrm e}^{x}} x^{2}}{3}+\ln \left (\frac {5}{4}\right )+8 x^{2}-x}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*((-16*exp(x)*x^2-32*x)*exp(exp(x))+192*x-12)*exp(-4/3*exp(exp(x))*x^2+ln(5/4)+8*x^2-x),x,method=_RETUR
NVERBOSE)

[Out]

5*exp(-1/3*x*(4*x*exp(exp(x))-24*x+3))

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maxima [A]  time = 0.51, size = 20, normalized size = 0.83 \begin {gather*} 5 \, e^{\left (-\frac {4}{3} \, x^{2} e^{\left (e^{x}\right )} + 8 \, x^{2} - x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-16*exp(x)*x^2-32*x)*exp(exp(x))+192*x-12)*exp(-4/3*exp(exp(x))*x^2+log(5/4)+8*x^2-x),x, algor
ithm="maxima")

[Out]

5*e^(-4/3*x^2*e^(e^x) + 8*x^2 - x)

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mupad [B]  time = 1.04, size = 21, normalized size = 0.88 \begin {gather*} 5\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{8\,x^2}\,{\mathrm {e}}^{-\frac {4\,x^2\,{\mathrm {e}}^{{\mathrm {e}}^x}}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(log(5/4) - x - (4*x^2*exp(exp(x)))/3 + 8*x^2)*(exp(exp(x))*(32*x + 16*x^2*exp(x)) - 192*x + 12))/3,x
)

[Out]

5*exp(-x)*exp(8*x^2)*exp(-(4*x^2*exp(exp(x)))/3)

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sympy [A]  time = 0.34, size = 20, normalized size = 0.83 \begin {gather*} 5 e^{- \frac {4 x^{2} e^{e^{x}}}{3} + 8 x^{2} - x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*((-16*exp(x)*x**2-32*x)*exp(exp(x))+192*x-12)*exp(-4/3*exp(exp(x))*x**2+ln(5/4)+8*x**2-x),x)

[Out]

5*exp(-4*x**2*exp(exp(x))/3 + 8*x**2 - x)

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