\(\int e^{-3 x^3} x^2 \sin (2 x^3) \, dx\) [777]

Optimal result

Integrand size = 17, antiderivative size = 35 \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=-\frac {2}{39} e^{-3 x^3} \cos \left (2 x^3\right )-\frac {1}{13} e^{-3 x^3} \sin \left (2 x^3\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6847, 4517} \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=-\frac {1}{13} e^{-3 x^3} \sin \left (2 x^3\right )-\frac {2}{39} e^{-3 x^3} \cos \left (2 x^3\right ) \]

[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 4517

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]

Rule 6847

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {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] (verified)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=-\frac {1}{39} e^{-3 x^3} \left (2 \cos \left (2 x^3\right )+3 \sin \left (2 x^3\right )\right ) \]

[In]

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

[Out]

-1/39*(2*Cos[2*x^3] + 3*Sin[2*x^3])/E^(3*x^3)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.74

method	result	size
parallelrisch	\(-\frac {{\mathrm e}^{-3 x^{3}} \left (2 \cos \left (2 x^{3}\right )+3 \sin \left (2 x^{3}\right )\right )}{39}\)	\(26\)
norman	\(\frac {\left (-\frac {2}{39}+\frac {2 \tan \left (x^{3}\right )^{2}}{39}-\frac {2 \tan \left (x^{3}\right )}{13}\right ) {\mathrm e}^{-3 x^{3}}}{1+\tan \left (x^{3}\right )^{2}}\)	\(36\)
risch	\(-\frac {{\mathrm e}^{\left (-3+2 i\right ) x^{3}}}{39}+\frac {i {\mathrm e}^{\left (-3+2 i\right ) x^{3}}}{26}-\frac {{\mathrm e}^{\left (-3-2 i\right ) x^{3}}}{39}-\frac {i {\mathrm e}^{\left (-3-2 i\right ) x^{3}}}{26}\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=-\frac {2}{39} \, \cos \left (2 \, x^{3}\right ) e^{\left (-3 \, x^{3}\right )} - \frac {1}{13} \, e^{\left (-3 \, x^{3}\right )} \sin \left (2 \, x^{3}\right ) \]

[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)

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=- \frac {e^{- 3 x^{3}} \sin {\left (2 x^{3} \right )}}{13} - \frac {2 e^{- 3 x^{3}} \cos {\left (2 x^{3} \right )}}{39} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=-\frac {1}{39} \, {\left (2 \, \cos \left (2 \, x^{3}\right ) + 3 \, \sin \left (2 \, x^{3}\right )\right )} e^{\left (-3 \, x^{3}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=-\frac {1}{39} \, {\left (2 \, \cos \left (2 \, x^{3}\right ) + 3 \, \sin \left (2 \, x^{3}\right )\right )} e^{\left (-3 \, x^{3}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 26.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int e^{-3 x^3} x^2 \sin \left (2 x^3\right ) \, dx=-\frac {{\mathrm {e}}^{-3\,x^3}\,\left (2\,\cos \left (2\,x^3\right )+3\,\sin \left (2\,x^3\right )\right )}{39} \]

[In]

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

[Out]

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