3.777 \(\int e^{-3 x^3} x^2 \sin (2 x^3) \, dx\)

Optimal. Leaf size=35 \[ -\frac{1}{13} e^{-3 x^3} \sin \left (2 x^3\right )-\frac{2}{39} e^{-3 x^3} \cos \left (2 x^3\right ) \]

[Out]

(-2*Cos[2*x^3])/(39*E^(3*x^3)) - Sin[2*x^3]/(13*E^(3*x^3))

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Rubi [A]  time = 0.157812, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {6715, 4432} \[ -\frac{1}{13} e^{-3 x^3} \sin \left (2 x^3\right )-\frac{2}{39} e^{-3 x^3} \cos \left (2 x^3\right ) \]

Antiderivative was successfully verified.

[In]

Int[(x^2*Sin[2*x^3])/E^(3*x^3),x]

[Out]

(-2*Cos[2*x^3])/(39*E^(3*x^3)) - Sin[2*x^3]/(13*E^(3*x^3))

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 4432

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*S
in[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] - Simp[(e*F^(c*(a + b*x))*Cos[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int e^{-3 x} \sin (2 x) \, dx,x,x^3\right )\\ &=-\frac{2}{39} e^{-3 x^3} \cos \left (2 x^3\right )-\frac{1}{13} e^{-3 x^3} \sin \left (2 x^3\right )\\ \end{align*}

Mathematica [A]  time = 0.0474003, size = 28, normalized size = 0.8 \[ -\frac{1}{39} e^{-3 x^3} \left (3 \sin \left (2 x^3\right )+2 \cos \left (2 x^3\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*Sin[2*x^3])/E^(3*x^3),x]

[Out]

-(2*Cos[2*x^3] + 3*Sin[2*x^3])/(39*E^(3*x^3))

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Maple [A]  time = 0.018, size = 36, normalized size = 1. \begin{align*}{\frac{1}{ \left ( 1+ \left ( \tan \left ({x}^{3} \right ) \right ) ^{2} \right ){{\rm e}^{3\,{x}^{3}}}} \left ( -{\frac{2}{39}}+{\frac{2\, \left ( \tan \left ({x}^{3} \right ) \right ) ^{2}}{39}}-{\frac{2\,\tan \left ({x}^{3} \right ) }{13}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*sin(2*x^3)/exp(3*x^3),x)

[Out]

(-2/39+2/39*tan(x^3)^2-2/13*tan(x^3))/(1+tan(x^3)^2)/exp(3*x^3)

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Maxima [A]  time = 0.961542, size = 34, normalized size = 0.97 \begin{align*} -\frac{1}{39} \,{\left (2 \, \cos \left (2 \, x^{3}\right ) + 3 \, \sin \left (2 \, x^{3}\right )\right )} e^{\left (-3 \, x^{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(2*x^3)/exp(3*x^3),x, algorithm="maxima")

[Out]

-1/39*(2*cos(2*x^3) + 3*sin(2*x^3))*e^(-3*x^3)

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Fricas [A]  time = 2.03139, size = 78, normalized size = 2.23 \begin{align*} -\frac{2}{39} \, \cos \left (2 \, x^{3}\right ) e^{\left (-3 \, x^{3}\right )} - \frac{1}{13} \, e^{\left (-3 \, x^{3}\right )} \sin \left (2 \, x^{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(2*x^3)/exp(3*x^3),x, algorithm="fricas")

[Out]

-2/39*cos(2*x^3)*e^(-3*x^3) - 1/13*e^(-3*x^3)*sin(2*x^3)

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Sympy [A]  time = 2.31183, size = 32, normalized size = 0.91 \begin{align*} - \frac{e^{- 3 x^{3}} \sin{\left (2 x^{3} \right )}}{13} - \frac{2 e^{- 3 x^{3}} \cos{\left (2 x^{3} \right )}}{39} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*sin(2*x**3)/exp(3*x**3),x)

[Out]

-exp(-3*x**3)*sin(2*x**3)/13 - 2*exp(-3*x**3)*cos(2*x**3)/39

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Giac [A]  time = 1.09278, size = 34, normalized size = 0.97 \begin{align*} -\frac{1}{39} \,{\left (2 \, \cos \left (2 \, x^{3}\right ) + 3 \, \sin \left (2 \, x^{3}\right )\right )} e^{\left (-3 \, x^{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*sin(2*x^3)/exp(3*x^3),x, algorithm="giac")

[Out]

-1/39*(2*cos(2*x^3) + 3*sin(2*x^3))*e^(-3*x^3)