3.83.6 \(\int \frac {1}{6} (x+e^{-2 x+x^3} (12-18 x^2)) \, dx\)

Optimal. Leaf size=20 \[ 1-e^{x \left (-2+x^2\right )}+\frac {x^2}{12} \]

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Rubi [A]  time = 0.04, antiderivative size = 19, normalized size of antiderivative = 0.95, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {12, 6706} \begin {gather*} \frac {x^2}{12}-e^{x^3-2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + E^(-2*x + x^3)*(12 - 18*x^2))/6,x]

[Out]

-E^(-2*x + x^3) + x^2/12

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}{6} \int \left (x+e^{-2 x+x^3} \left (12-18 x^2\right )\right ) \, dx\\ &=\frac {x^2}{12}+\frac {1}{6} \int e^{-2 x+x^3} \left (12-18 x^2\right ) \, dx\\ &=-e^{-2 x+x^3}+\frac {x^2}{12}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 23, normalized size = 1.15 \begin {gather*} \frac {1}{6} \left (-6 e^{-2 x+x^3}+\frac {x^2}{2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + E^(-2*x + x^3)*(12 - 18*x^2))/6,x]

[Out]

(-6*E^(-2*x + x^3) + x^2/2)/6

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fricas [A]  time = 0.76, size = 16, normalized size = 0.80 \begin {gather*} \frac {1}{12} \, x^{2} - e^{\left (x^{3} - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-18*x^2+12)*exp(x^3-2*x)+1/6*x,x, algorithm="fricas")

[Out]

1/12*x^2 - e^(x^3 - 2*x)

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giac [A]  time = 0.16, size = 16, normalized size = 0.80 \begin {gather*} \frac {1}{12} \, x^{2} - e^{\left (x^{3} - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-18*x^2+12)*exp(x^3-2*x)+1/6*x,x, algorithm="giac")

[Out]

1/12*x^2 - e^(x^3 - 2*x)

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maple [A]  time = 0.03, size = 17, normalized size = 0.85

method	result	size
default	\(\frac {x^{2}}{12}-{\mathrm e}^{x^{3}-2 x}\)	\(17\)
norman	\(\frac {x^{2}}{12}-{\mathrm e}^{x^{3}-2 x}\)	\(17\)
risch	\(\frac {x^{2}}{12}-{\mathrm e}^{\left (x^{2}-2\right ) x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(-18*x^2+12)*exp(x^3-2*x)+1/6*x,x,method=_RETURNVERBOSE)

[Out]

1/12*x^2-exp(x^3-2*x)

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maxima [A]  time = 0.35, size = 16, normalized size = 0.80 \begin {gather*} \frac {1}{12} \, x^{2} - e^{\left (x^{3} - 2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-18*x^2+12)*exp(x^3-2*x)+1/6*x,x, algorithm="maxima")

[Out]

1/12*x^2 - e^(x^3 - 2*x)

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mupad [B]  time = 0.09, size = 16, normalized size = 0.80 \begin {gather*} \frac {x^2}{12}-{\mathrm {e}}^{x^3-2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/6 - (exp(x^3 - 2*x)*(18*x^2 - 12))/6,x)

[Out]

x^2/12 - exp(x^3 - 2*x)

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sympy [A]  time = 0.09, size = 12, normalized size = 0.60 \begin {gather*} \frac {x^{2}}{12} - e^{x^{3} - 2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(-18*x**2+12)*exp(x**3-2*x)+1/6*x,x)

[Out]

x**2/12 - exp(x**3 - 2*x)

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