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

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

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Rubi [A]  time = 0.16, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.08, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{2 x} x^2 \sin (4 x) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [A]  time = 1.27, size = 41, normalized size = 0.47 \[ -\frac {1}{250} \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac {1}{500} \, {\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \]

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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giac [A]  time = 0.57, size = 39, normalized size = 0.45 \[ -\frac {1}{500} \, {\left (2 \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) - {\left (50 \, x^{2} + 30 \, x - 11\right )} \sin \left (4 \, x\right )\right )} e^{\left (2 \, x\right )} \]

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)

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maple [A]  time = 0.07, size = 40, normalized size = 0.46

method	result	size
default	\(\left (-\frac {1}{5} x^{2}+\frac {2}{25} x +\frac {1}{250}\right ) {\mathrm e}^{2 x} \cos \left (4 x \right )+\left (\frac {1}{10} x^{2}+\frac {3}{50} x -\frac {11}{500}\right ) {\mathrm e}^{2 x} \sin \left (4 x \right )\)	\(40\)
risch	\(\left (-\frac {1}{500}-\frac {i}{1000}\right ) \left (50 x^{2}+20 i x -10 x -4 i-3\right ) {\mathrm e}^{\left (2+4 i\right ) x}+\left (-\frac {1}{500}+\frac {i}{1000}\right ) \left (50 x^{2}-20 i x -10 x +4 i-3\right ) {\mathrm e}^{\left (2-4 i\right ) x}\)	\(54\)
norman	\(\frac {\frac {2 x \,{\mathrm e}^{2 x}}{25}-\frac {{\mathrm e}^{2 x} x^{2}}{5}-\frac {11 \,{\mathrm e}^{2 x} \tan \left (2 x \right )}{250}-\frac {{\mathrm e}^{2 x} \left (\tan ^{2}\left (2 x \right )\right )}{250}+\frac {3 x \,{\mathrm e}^{2 x} \tan \left (2 x \right )}{25}-\frac {2 x \,{\mathrm e}^{2 x} \left (\tan ^{2}\left (2 x \right )\right )}{25}+\frac {{\mathrm e}^{2 x} x^{2} \tan \left (2 x \right )}{5}+\frac {{\mathrm e}^{2 x} x^{2} \left (\tan ^{2}\left (2 x \right )\right )}{5}+\frac {{\mathrm e}^{2 x}}{250}}{1+\tan ^{2}\left (2 x \right )}\)	\(109\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*x)*x^2*sin(4*x),x,method=_RETURNVERBOSE)

[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 = 0.45, size = 41, normalized size = 0.47 \[ -\frac {1}{250} \, {\left (50 \, x^{2} - 20 \, x - 1\right )} \cos \left (4 \, x\right ) e^{\left (2 \, x\right )} + \frac {1}{500} \, {\left (50 \, x^{2} + 30 \, x - 11\right )} e^{\left (2 \, x\right )} \sin \left (4 \, x\right ) \]

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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mupad [B]  time = 0.35, size = 51, normalized size = 0.59 \[ \frac {{\mathrm {e}}^{2\,x}\,\left (2\,\cos \left (4\,x\right )-11\,\sin \left (4\,x\right )+40\,x\,\cos \left (4\,x\right )+30\,x\,\sin \left (4\,x\right )-100\,x^2\,\cos \left (4\,x\right )+50\,x^2\,\sin \left (4\,x\right )\right )}{500} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(2*x)*(2*cos(4*x) - 11*sin(4*x) + 40*x*cos(4*x) + 30*x*sin(4*x) - 100*x^2*cos(4*x) + 50*x^2*sin(4*x)))/500

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sympy [A]  time = 2.11, size = 85, normalized size = 0.98 \[ \frac {x^{2} e^{2 x} \sin {\left (4 x \right )}}{10} - \frac {x^{2} e^{2 x} \cos {\left (4 x \right )}}{5} + \frac {3 x e^{2 x} \sin {\left (4 x \right )}}{50} + \frac {2 x e^{2 x} \cos {\left (4 x \right )}}{25} - \frac {11 e^{2 x} \sin {\left (4 x \right )}}{500} + \frac {e^{2 x} \cos {\left (4 x \right )}}{250} \]

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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