3.568 \(\int e^{2 x} x^2 \sin (4 x) \, dx\)

Optimal. Leaf size=87 \[ \frac{1}{10} e^{2 x} x^2 \sin (4 x)-\frac{1}{5} e^{2 x} x^2 \cos (4 x)+\frac{3}{50} e^{2 x} x \sin (4 x)-\frac{11}{500} e^{2 x} \sin (4 x)+\frac{2}{25} e^{2 x} x \cos (4 x)+\frac{1}{250} e^{2 x} \cos (4 x) \]

[Out]

(E^(2*x)*Cos[4*x])/250 + (2*E^(2*x)*x*Cos[4*x])/25 - (E^(2*x)*x^2*Cos[4*x])/5 - (11*E^(2*x)*Sin[4*x])/500 + (3
*E^(2*x)*x*Sin[4*x])/50 + (E^(2*x)*x^2*Sin[4*x])/10

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Rubi [A]  time = 0.156602, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {4432, 4465, 14, 4433, 4466} \[ \frac{1}{10} e^{2 x} x^2 \sin (4 x)-\frac{1}{5} e^{2 x} x^2 \cos (4 x)+\frac{3}{50} e^{2 x} x \sin (4 x)-\frac{11}{500} e^{2 x} \sin (4 x)+\frac{2}{25} e^{2 x} x \cos (4 x)+\frac{1}{250} e^{2 x} \cos (4 x) \]

Antiderivative was successfully verified.

[In]

Int[E^(2*x)*x^2*Sin[4*x],x]

[Out]

(E^(2*x)*Cos[4*x])/250 + (2*E^(2*x)*x*Cos[4*x])/25 - (E^(2*x)*x^2*Cos[4*x])/5 - (11*E^(2*x)*Sin[4*x])/500 + (3
*E^(2*x)*x*Sin[4*x])/50 + (E^(2*x)*x^2*Sin[4*x])/10

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^{2 x} x^2 \sin (4 x) \, dx &=-\frac{1}{5} e^{2 x} x^2 \cos (4 x)+\frac{1}{10} e^{2 x} x^2 \sin (4 x)-2 \int x \left (-\frac{1}{5} e^{2 x} \cos (4 x)+\frac{1}{10} e^{2 x} \sin (4 x)\right ) \, dx\\ &=-\frac{1}{5} e^{2 x} x^2 \cos (4 x)+\frac{1}{10} e^{2 x} x^2 \sin (4 x)-2 \int \left (-\frac{1}{5} e^{2 x} x \cos (4 x)+\frac{1}{10} e^{2 x} x \sin (4 x)\right ) \, dx\\ &=-\frac{1}{5} e^{2 x} x^2 \cos (4 x)+\frac{1}{10} e^{2 x} x^2 \sin (4 x)-\frac{1}{5} \int e^{2 x} x \sin (4 x) \, dx+\frac{2}{5} \int e^{2 x} x \cos (4 x) \, dx\\ &=\frac{2}{25} e^{2 x} x \cos (4 x)-\frac{1}{5} e^{2 x} x^2 \cos (4 x)+\frac{3}{50} e^{2 x} x \sin (4 x)+\frac{1}{10} e^{2 x} x^2 \sin (4 x)+\frac{1}{5} \int \left (-\frac{1}{5} e^{2 x} \cos (4 x)+\frac{1}{10} e^{2 x} \sin (4 x)\right ) \, dx-\frac{2}{5} \int \left (\frac{1}{10} e^{2 x} \cos (4 x)+\frac{1}{5} e^{2 x} \sin (4 x)\right ) \, dx\\ &=\frac{2}{25} e^{2 x} x \cos (4 x)-\frac{1}{5} e^{2 x} x^2 \cos (4 x)+\frac{3}{50} e^{2 x} x \sin (4 x)+\frac{1}{10} e^{2 x} x^2 \sin (4 x)+\frac{1}{50} \int e^{2 x} \sin (4 x) \, dx-2 \left (\frac{1}{25} \int e^{2 x} \cos (4 x) \, dx\right )-\frac{2}{25} \int e^{2 x} \sin (4 x) \, dx\\ &=\frac{3}{250} e^{2 x} \cos (4 x)+\frac{2}{25} e^{2 x} x \cos (4 x)-\frac{1}{5} e^{2 x} x^2 \cos (4 x)-\frac{3}{500} e^{2 x} \sin (4 x)+\frac{3}{50} e^{2 x} x \sin (4 x)+\frac{1}{10} e^{2 x} x^2 \sin (4 x)-2 \left (\frac{1}{250} e^{2 x} \cos (4 x)+\frac{1}{125} e^{2 x} \sin (4 x)\right )\\ \end{align*}

Mathematica [A]  time = 0.077555, size = 40, normalized size = 0.46 \[ \frac{1}{500} e^{2 x} \left (\left (50 x^2+30 x-11\right ) \sin (4 x)+\left (-100 x^2+40 x+2\right ) \cos (4 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*x)*x^2*Sin[4*x],x]

[Out]

(E^(2*x)*((2 + 40*x - 100*x^2)*Cos[4*x] + (-11 + 30*x + 50*x^2)*Sin[4*x]))/500

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Maple [A]  time = 0.004, size = 40, normalized size = 0.5 \begin{align*} \left ( -{\frac{{x}^{2}}{5}}+{\frac{2\,x}{25}}+{\frac{1}{250}} \right ){{\rm e}^{2\,x}}\cos \left ( 4\,x \right ) + \left ({\frac{{x}^{2}}{10}}+{\frac{3\,x}{50}}-{\frac{11}{500}} \right ){{\rm e}^{2\,x}}\sin \left ( 4\,x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*x^2*sin(4*x),x)

[Out]

(-1/5*x^2+2/25*x+1/250)*exp(2*x)*cos(4*x)+(1/10*x^2+3/50*x-11/500)*exp(2*x)*sin(4*x)

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Maxima [A]  time = 0.956134, size = 55, normalized size = 0.63 \begin{align*} -\frac{1}{250} \,{\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac{1}{500} \,{\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*x^2*sin(4*x),x, algorithm="maxima")

[Out]

-1/250*(50*x^2 - 20*x - 1)*cos(4*x)*e^(2*x) + 1/500*(50*x^2 + 30*x - 11)*e^(2*x)*sin(4*x)

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Fricas [A]  time = 1.75816, size = 123, normalized size = 1.41 \begin{align*} -\frac{1}{250} \,{\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac{1}{500} \,{\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*x^2*sin(4*x),x, algorithm="fricas")

[Out]

-1/250*(50*x^2 - 20*x - 1)*cos(4*x)*e^(2*x) + 1/500*(50*x^2 + 30*x - 11)*e^(2*x)*sin(4*x)

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Sympy [A]  time = 2.20705, size = 85, normalized size = 0.98 \begin{align*} \frac{x^{2} e^{2 x} \sin{\left (4 x \right )}}{10} - \frac{x^{2} e^{2 x} \cos{\left (4 x \right )}}{5} + \frac{3 x e^{2 x} \sin{\left (4 x \right )}}{50} + \frac{2 x e^{2 x} \cos{\left (4 x \right )}}{25} - \frac{11 e^{2 x} \sin{\left (4 x \right )}}{500} + \frac{e^{2 x} \cos{\left (4 x \right )}}{250} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*x**2*sin(4*x),x)

[Out]

x**2*exp(2*x)*sin(4*x)/10 - x**2*exp(2*x)*cos(4*x)/5 + 3*x*exp(2*x)*sin(4*x)/50 + 2*x*exp(2*x)*cos(4*x)/25 - 1
1*exp(2*x)*sin(4*x)/500 + exp(2*x)*cos(4*x)/250

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Giac [A]  time = 1.12231, size = 53, normalized size = 0.61 \begin{align*} -\frac{1}{500} \,{\left (2 \,{\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) -{\left (50 \, x^{2} + 30 \, x - 11\right )} \sin \left (4 \, x\right )\right )} e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*x)*x^2*sin(4*x),x, algorithm="giac")

[Out]

-1/500*(2*(50*x^2 - 20*x - 1)*cos(4*x) - (50*x^2 + 30*x - 11)*sin(4*x))*e^(2*x)