3.50 \(\int e^x x^2 \cos (x) \, dx\)

Optimal. Leaf size=51 \[ \frac{1}{2} e^x x^2 \sin (x)+\frac{1}{2} e^x x^2 \cos (x)-e^x x \sin (x)+\frac{1}{2} e^x \sin (x)-\frac{1}{2} e^x \cos (x) \]

[Out]

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

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Rubi [A]  time = 0.117484, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 5, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {4433, 4466, 14, 4432, 4465} \[ \frac{1}{2} e^x x^2 \sin (x)+\frac{1}{2} e^x x^2 \cos (x)-e^x x \sin (x)+\frac{1}{2} e^x \sin (x)-\frac{1}{2} e^x \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[E^x*x^2*Cos[x],x]

[Out]

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

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]

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 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]

Rubi steps

\begin{align*} \int e^x x^2 \cos (x) \, dx &=\frac{1}{2} e^x x^2 \cos (x)+\frac{1}{2} e^x x^2 \sin (x)-2 \int x \left (\frac{1}{2} e^x \cos (x)+\frac{1}{2} e^x \sin (x)\right ) \, dx\\ &=\frac{1}{2} e^x x^2 \cos (x)+\frac{1}{2} e^x x^2 \sin (x)-2 \int \left (\frac{1}{2} e^x x \cos (x)+\frac{1}{2} e^x x \sin (x)\right ) \, dx\\ &=\frac{1}{2} e^x x^2 \cos (x)+\frac{1}{2} e^x x^2 \sin (x)-\int e^x x \cos (x) \, dx-\int e^x x \sin (x) \, dx\\ &=\frac{1}{2} e^x x^2 \cos (x)-e^x x \sin (x)+\frac{1}{2} e^x x^2 \sin (x)+\int \left (-\frac{1}{2} e^x \cos (x)+\frac{1}{2} e^x \sin (x)\right ) \, dx+\int \left (\frac{1}{2} e^x \cos (x)+\frac{1}{2} e^x \sin (x)\right ) \, dx\\ &=\frac{1}{2} e^x x^2 \cos (x)-e^x x \sin (x)+\frac{1}{2} e^x x^2 \sin (x)+2 \left (\frac{1}{2} \int e^x \sin (x) \, dx\right )\\ &=\frac{1}{2} e^x x^2 \cos (x)-e^x x \sin (x)+\frac{1}{2} e^x x^2 \sin (x)+2 \left (-\frac{1}{4} e^x \cos (x)+\frac{1}{4} e^x \sin (x)\right )\\ \end{align*}

Mathematica [A]  time = 0.0307821, size = 23, normalized size = 0.45 \[ \frac{1}{2} e^x (x-1) ((x-1) \sin (x)+(x+1) \cos (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*x^2*Cos[x],x]

[Out]

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

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Maple [A]  time = 0.007, size = 28, normalized size = 0.6 \begin{align*} \left ({\frac{{x}^{2}}{2}}-{\frac{1}{2}} \right ){{\rm e}^{x}}\cos \left ( x \right ) - \left ( -{\frac{{x}^{2}}{2}}+x-{\frac{1}{2}} \right ){{\rm e}^{x}}\sin \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*x^2*cos(x),x)

[Out]

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

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Maxima [A]  time = 1.05925, size = 35, normalized size = 0.69 \begin{align*} \frac{1}{2} \,{\left (x^{2} - 1\right )} \cos \left (x\right ) e^{x} + \frac{1}{2} \,{\left (x^{2} - 2 \, x + 1\right )} e^{x} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.463794, size = 80, normalized size = 1.57 \begin{align*} \frac{1}{2} \,{\left (x^{2} - 1\right )} \cos \left (x\right ) e^{x} + \frac{1}{2} \,{\left (x^{2} - 2 \, x + 1\right )} e^{x} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.41969, size = 48, normalized size = 0.94 \begin{align*} \frac{x^{2} e^{x} \sin{\left (x \right )}}{2} + \frac{x^{2} e^{x} \cos{\left (x \right )}}{2} - x e^{x} \sin{\left (x \right )} + \frac{e^{x} \sin{\left (x \right )}}{2} - \frac{e^{x} \cos{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x**2*cos(x),x)

[Out]

x**2*exp(x)*sin(x)/2 + x**2*exp(x)*cos(x)/2 - x*exp(x)*sin(x) + exp(x)*sin(x)/2 - exp(x)*cos(x)/2

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Giac [A]  time = 1.0891, size = 32, normalized size = 0.63 \begin{align*} \frac{1}{2} \,{\left ({\left (x^{2} - 1\right )} \cos \left (x\right ) +{\left (x^{2} - 2 \, x + 1\right )} \sin \left (x\right )\right )} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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