3.39.19 \(\int \frac {1}{3} e^{\frac {1}{3} (3+9 x-6 x^2-4 x^2 \log ^2(x))} (9-12 x-8 x \log (x)-8 x \log ^2(x)) \, dx\) [3819]

Optimal. Leaf size=22 \[ e^{1+x-x \left (-2+2 x+\frac {4}{3} x \log ^2(x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 25, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {12, 6838} \begin {gather*} e^{\frac {1}{3} \left (-6 x^2-4 x^2 \log ^2(x)+9 x+3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((3 + 9*x - 6*x^2 - 4*x^2*Log[x]^2)/3)*(9 - 12*x - 8*x*Log[x] - 8*x*Log[x]^2))/3,x]

[Out]

E^((3 + 9*x - 6*x^2 - 4*x^2*Log[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 6838

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 e^{\frac {1}{3} \left (3+9 x-6 x^2-4 x^2 \log ^2(x)\right )} \left (9-12 x-8 x \log (x)-8 x \log ^2(x)\right ) \, dx\\ &=e^{\frac {1}{3} \left (3+9 x-6 x^2-4 x^2 \log ^2(x)\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.07, size = 23, normalized size = 1.05 \begin {gather*} e^{1+3 x-2 x^2-\frac {4}{3} x^2 \log ^2(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((3 + 9*x - 6*x^2 - 4*x^2*Log[x]^2)/3)*(9 - 12*x - 8*x*Log[x] - 8*x*Log[x]^2))/3,x]

[Out]

E^(1 + 3*x - 2*x^2 - (4*x^2*Log[x]^2)/3)

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Maple [A]
time = 0.13, size = 21, normalized size = 0.95

method	result	size
norman	\({\mathrm e}^{-\frac {4 x^{2} \ln \left (x \right )^{2}}{3}-2 x^{2}+3 x +1}\)	\(21\)
risch	\({\mathrm e}^{-\frac {4 x^{2} \ln \left (x \right )^{2}}{3}-2 x^{2}+3 x +1}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(-8*x*ln(x)^2-8*x*ln(x)-12*x+9)*exp(-4/3*x^2*ln(x)^2-2*x^2+3*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 20, normalized size = 0.91 \begin {gather*} e^{\left (-\frac {4}{3} \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} + 3 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-8*x*log(x)^2-8*x*log(x)-12*x+9)*exp(-4/3*x^2*log(x)^2-2*x^2+3*x+1),x, algorithm="maxima")

[Out]

e^(-4/3*x^2*log(x)^2 - 2*x^2 + 3*x + 1)

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Fricas [A]
time = 0.34, size = 20, normalized size = 0.91 \begin {gather*} e^{\left (-\frac {4}{3} \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} + 3 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-8*x*log(x)^2-8*x*log(x)-12*x+9)*exp(-4/3*x^2*log(x)^2-2*x^2+3*x+1),x, algorithm="fricas")

[Out]

e^(-4/3*x^2*log(x)^2 - 2*x^2 + 3*x + 1)

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Sympy [A]
time = 0.10, size = 22, normalized size = 1.00 \begin {gather*} e^{- \frac {4 x^{2} \log {\left (x \right )}^{2}}{3} - 2 x^{2} + 3 x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-8*x*ln(x)**2-8*x*ln(x)-12*x+9)*exp(-4/3*x**2*ln(x)**2-2*x**2+3*x+1),x)

[Out]

exp(-4*x**2*log(x)**2/3 - 2*x**2 + 3*x + 1)

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Giac [A]
time = 0.41, size = 20, normalized size = 0.91 \begin {gather*} e^{\left (-\frac {4}{3} \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} + 3 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(-8*x*log(x)^2-8*x*log(x)-12*x+9)*exp(-4/3*x^2*log(x)^2-2*x^2+3*x+1),x, algorithm="giac")

[Out]

e^(-4/3*x^2*log(x)^2 - 2*x^2 + 3*x + 1)

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Mupad [B]
time = 2.39, size = 23, normalized size = 1.05 \begin {gather*} {\mathrm {e}}^{3\,x}\,\mathrm {e}\,{\mathrm {e}}^{-2\,x^2}\,{\mathrm {e}}^{-\frac {4\,x^2\,{\ln \left (x\right )}^2}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3*x - (4*x^2*log(x)^2)/3 - 2*x^2 + 1)*(12*x + 8*x*log(x)^2 + 8*x*log(x) - 9))/3,x)

[Out]

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

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