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

Optimal. Leaf size=75 \[ -\frac{3}{10} e^{-3 x} x^2 \sin (x)-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{4}{25} e^{-3 x} x \sin (x)-\frac{9}{250} e^{-3 x} \sin (x)-\frac{3}{25} e^{-3 x} x \cos (x)-\frac{13}{250} e^{-3 x} \cos (x) \]

[Out]

(-13*Cos[x])/(250*E^(3*x)) - (3*x*Cos[x])/(25*E^(3*x)) - (x^2*Cos[x])/(10*E^(3*x)) - (9*Sin[x])/(250*E^(3*x))
- (4*x*Sin[x])/(25*E^(3*x)) - (3*x^2*Sin[x])/(10*E^(3*x))

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Rubi [A]  time = 0.140277, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4432, 4465, 14, 4433, 4466} \[ -\frac{3}{10} e^{-3 x} x^2 \sin (x)-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{4}{25} e^{-3 x} x \sin (x)-\frac{9}{250} e^{-3 x} \sin (x)-\frac{3}{25} e^{-3 x} x \cos (x)-\frac{13}{250} e^{-3 x} \cos (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-13*Cos[x])/(250*E^(3*x)) - (3*x*Cos[x])/(25*E^(3*x)) - (x^2*Cos[x])/(10*E^(3*x)) - (9*Sin[x])/(250*E^(3*x))
- (4*x*Sin[x])/(25*E^(3*x)) - (3*x^2*Sin[x])/(10*E^(3*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]

Rule 4465

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_.)*Sin[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Module[{u
= IntHide[F^(c*(a + b*x))*Sin[d + e*x]^n, x]}, Dist[(f*x)^m, u, x] - Dist[f*m, Int[(f*x)^(m - 1)*u, x], x]] /
; FreeQ[{F, a, b, c, d, e, f}, x] && IGtQ[n, 0] && GtQ[m, 0]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 4433

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[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 4466

Int[Cos[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_.), x_Symbol] :> Module[{u
= IntHide[F^(c*(a + b*x))*Cos[d + e*x]^n, x]}, Dist[(f*x)^m, u, x] - Dist[f*m, Int[(f*x)^(m - 1)*u, x], x]] /
; FreeQ[{F, a, b, c, d, e, f}, x] && IGtQ[n, 0] && GtQ[m, 0]

Rubi steps

\begin{align*} \int e^{-3 x} x^2 \sin (x) \, dx &=-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{3}{10} e^{-3 x} x^2 \sin (x)-2 \int x \left (-\frac{1}{10} e^{-3 x} \cos (x)-\frac{3}{10} e^{-3 x} \sin (x)\right ) \, dx\\ &=-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{3}{10} e^{-3 x} x^2 \sin (x)-2 \int \left (-\frac{1}{10} e^{-3 x} x \cos (x)-\frac{3}{10} e^{-3 x} x \sin (x)\right ) \, dx\\ &=-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{3}{10} e^{-3 x} x^2 \sin (x)+\frac{1}{5} \int e^{-3 x} x \cos (x) \, dx+\frac{3}{5} \int e^{-3 x} x \sin (x) \, dx\\ &=-\frac{3}{25} e^{-3 x} x \cos (x)-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{4}{25} e^{-3 x} x \sin (x)-\frac{3}{10} e^{-3 x} x^2 \sin (x)-\frac{1}{5} \int \left (-\frac{3}{10} e^{-3 x} \cos (x)+\frac{1}{10} e^{-3 x} \sin (x)\right ) \, dx-\frac{3}{5} \int \left (-\frac{1}{10} e^{-3 x} \cos (x)-\frac{3}{10} e^{-3 x} \sin (x)\right ) \, dx\\ &=-\frac{3}{25} e^{-3 x} x \cos (x)-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{4}{25} e^{-3 x} x \sin (x)-\frac{3}{10} e^{-3 x} x^2 \sin (x)-\frac{1}{50} \int e^{-3 x} \sin (x) \, dx+2 \left (\frac{3}{50} \int e^{-3 x} \cos (x) \, dx\right )+\frac{9}{50} \int e^{-3 x} \sin (x) \, dx\\ &=-\frac{2}{125} e^{-3 x} \cos (x)-\frac{3}{25} e^{-3 x} x \cos (x)-\frac{1}{10} e^{-3 x} x^2 \cos (x)-\frac{6}{125} e^{-3 x} \sin (x)-\frac{4}{25} e^{-3 x} x \sin (x)-\frac{3}{10} e^{-3 x} x^2 \sin (x)+2 \left (-\frac{9}{500} e^{-3 x} \cos (x)+\frac{3}{500} e^{-3 x} \sin (x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0347447, size = 38, normalized size = 0.51 \[ \frac{1}{250} e^{-3 x} \left (-\left (75 x^2+40 x+9\right ) \sin (x)-\left (25 x^2+30 x+13\right ) \cos (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((13 + 30*x + 25*x^2)*Cos[x]) - (9 + 40*x + 75*x^2)*Sin[x])/(250*E^(3*x))

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Maple [A]  time = 0.006, size = 36, normalized size = 0.5 \begin{align*} \left ( -{\frac{{x}^{2}}{10}}-{\frac{3\,x}{25}}-{\frac{13}{250}} \right ){{\rm e}^{-3\,x}}\cos \left ( x \right ) + \left ( -{\frac{3\,{x}^{2}}{10}}-{\frac{4\,x}{25}}-{\frac{9}{250}} \right ){{\rm e}^{-3\,x}}\sin \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/10*x^2-3/25*x-13/250)*exp(-3*x)*cos(x)+(-3/10*x^2-4/25*x-9/250)*exp(-3*x)*sin(x)

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Maxima [A]  time = 0.969134, size = 45, normalized size = 0.6 \begin{align*} -\frac{1}{250} \,{\left ({\left (25 \, x^{2} + 30 \, x + 13\right )} \cos \left (x\right ) +{\left (75 \, x^{2} + 40 \, x + 9\right )} \sin \left (x\right )\right )} e^{\left (-3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/250*((25*x^2 + 30*x + 13)*cos(x) + (75*x^2 + 40*x + 9)*sin(x))*e^(-3*x)

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Fricas [A]  time = 1.90528, size = 120, normalized size = 1.6 \begin{align*} -\frac{1}{250} \,{\left (25 \, x^{2} + 30 \, x + 13\right )} \cos \left (x\right ) e^{\left (-3 \, x\right )} - \frac{1}{250} \,{\left (75 \, x^{2} + 40 \, x + 9\right )} e^{\left (-3 \, x\right )} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/250*(25*x^2 + 30*x + 13)*cos(x)*e^(-3*x) - 1/250*(75*x^2 + 40*x + 9)*e^(-3*x)*sin(x)

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Sympy [A]  time = 2.22796, size = 80, normalized size = 1.07 \begin{align*} - \frac{3 x^{2} e^{- 3 x} \sin{\left (x \right )}}{10} - \frac{x^{2} e^{- 3 x} \cos{\left (x \right )}}{10} - \frac{4 x e^{- 3 x} \sin{\left (x \right )}}{25} - \frac{3 x e^{- 3 x} \cos{\left (x \right )}}{25} - \frac{9 e^{- 3 x} \sin{\left (x \right )}}{250} - \frac{13 e^{- 3 x} \cos{\left (x \right )}}{250} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x**2*exp(-3*x)*sin(x)/10 - x**2*exp(-3*x)*cos(x)/10 - 4*x*exp(-3*x)*sin(x)/25 - 3*x*exp(-3*x)*cos(x)/25 - 9
*exp(-3*x)*sin(x)/250 - 13*exp(-3*x)*cos(x)/250

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Giac [A]  time = 1.15373, size = 45, normalized size = 0.6 \begin{align*} -\frac{1}{250} \,{\left ({\left (25 \, x^{2} + 30 \, x + 13\right )} \cos \left (x\right ) +{\left (75 \, x^{2} + 40 \, x + 9\right )} \sin \left (x\right )\right )} e^{\left (-3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/250*((25*x^2 + 30*x + 13)*cos(x) + (75*x^2 + 40*x + 9)*sin(x))*e^(-3*x)