3.74.14 \(\int \frac {1}{4} e^{\frac {1}{12} (-11-9 x-900 x^2-300 x^3)} (-3-600 x-300 x^2) \, dx\)

Optimal. Leaf size=20 \[ e^{\frac {4}{3}-\frac {1}{4} (3+x) \left (3+100 x^2\right )} \]

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Rubi [A]  time = 0.06, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {12, 6706} \begin {gather*} e^{\frac {1}{12} \left (-300 x^3-900 x^2-9 x-11\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((-11 - 9*x - 900*x^2 - 300*x^3)/12)*(-3 - 600*x - 300*x^2))/4,x]

[Out]

E^((-11 - 9*x - 900*x^2 - 300*x^3)/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}{4} \int e^{\frac {1}{12} \left (-11-9 x-900 x^2-300 x^3\right )} \left (-3-600 x-300 x^2\right ) \, dx\\ &=e^{\frac {1}{12} \left (-11-9 x-900 x^2-300 x^3\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 21, normalized size = 1.05 \begin {gather*} e^{-\frac {11}{12}-\frac {3 x}{4}-75 x^2-25 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-11 - 9*x - 900*x^2 - 300*x^3)/12)*(-3 - 600*x - 300*x^2))/4,x]

[Out]

E^(-11/12 - (3*x)/4 - 75*x^2 - 25*x^3)

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fricas [A]  time = 0.59, size = 16, normalized size = 0.80 \begin {gather*} e^{\left (-25 \, x^{3} - 75 \, x^{2} - \frac {3}{4} \, x - \frac {11}{12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-300*x^2-600*x-3)*exp(-25*x^3-75*x^2-3/4*x-11/12),x, algorithm="fricas")

[Out]

e^(-25*x^3 - 75*x^2 - 3/4*x - 11/12)

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giac [A]  time = 0.14, size = 16, normalized size = 0.80 \begin {gather*} e^{\left (-25 \, x^{3} - 75 \, x^{2} - \frac {3}{4} \, x - \frac {11}{12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-300*x^2-600*x-3)*exp(-25*x^3-75*x^2-3/4*x-11/12),x, algorithm="giac")

[Out]

e^(-25*x^3 - 75*x^2 - 3/4*x - 11/12)

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maple [A]  time = 0.04, size = 17, normalized size = 0.85

method	result	size
gosper	\({\mathrm e}^{-25 x^{3}-75 x^{2}-\frac {3}{4} x -\frac {11}{12}}\)	\(17\)
norman	\({\mathrm e}^{-25 x^{3}-75 x^{2}-\frac {3}{4} x -\frac {11}{12}}\)	\(17\)
risch	\({\mathrm e}^{-25 x^{3}-75 x^{2}-\frac {3}{4} x -\frac {11}{12}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(-300*x^2-600*x-3)*exp(-25*x^3-75*x^2-3/4*x-11/12),x,method=_RETURNVERBOSE)

[Out]

exp(-25*x^3-75*x^2-3/4*x-11/12)

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maxima [A]  time = 0.35, size = 16, normalized size = 0.80 \begin {gather*} e^{\left (-25 \, x^{3} - 75 \, x^{2} - \frac {3}{4} \, x - \frac {11}{12}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-300*x^2-600*x-3)*exp(-25*x^3-75*x^2-3/4*x-11/12),x, algorithm="maxima")

[Out]

e^(-25*x^3 - 75*x^2 - 3/4*x - 11/12)

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mupad [B]  time = 4.38, size = 19, normalized size = 0.95 \begin {gather*} {\mathrm {e}}^{-\frac {3\,x}{4}}\,{\mathrm {e}}^{-\frac {11}{12}}\,{\mathrm {e}}^{-25\,x^3}\,{\mathrm {e}}^{-75\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- (3*x)/4 - 75*x^2 - 25*x^3 - 11/12)*(600*x + 300*x^2 + 3))/4,x)

[Out]

exp(-(3*x)/4)*exp(-11/12)*exp(-25*x^3)*exp(-75*x^2)

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sympy [A]  time = 0.13, size = 20, normalized size = 1.00 \begin {gather*} e^{- 25 x^{3} - 75 x^{2} - \frac {3 x}{4} - \frac {11}{12}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(-300*x**2-600*x-3)*exp(-25*x**3-75*x**2-3/4*x-11/12),x)

[Out]

exp(-25*x**3 - 75*x**2 - 3*x/4 - 11/12)

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