3.1.50 \(\int e^x x^2 \cos (x) \, dx\) [50]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 51, normalized size of antiderivative = 1.00, 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 = {4518, 4554, 14, 4517, 4553} \begin {gather*} \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) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(E^x*Cos[x]) + (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 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 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 4518

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 4553

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 4554

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^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*}

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Mathematica [A]
time = 0.04, size = 23, normalized size = 0.45 \begin {gather*} \frac {1}{2} e^x (-1+x) ((1+x) \cos (x)+(-1+x) \sin (x)) \end {gather*}

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 = 0.06, size = 28, normalized size = 0.55

method	result	size
default	\(\left (\frac {x^{2}}{2}-\frac {1}{2}\right ) {\mathrm e}^{x} \cos \left (x \right )-\left (-\frac {1}{2} x^{2}+x -\frac {1}{2}\right ) {\mathrm e}^{x} \sin \left (x \right )\)	\(28\)
risch	\(\left (\frac {1}{4}-\frac {i}{4}\right ) \left (x^{2}+i x -x -i\right ) {\mathrm e}^{\left (1+i\right ) x}+\left (\frac {1}{4}+\frac {i}{4}\right ) \left (x^{2}-i x -x +i\right ) {\mathrm e}^{\left (1-i\right ) x}\)	\(48\)
norman	\(\frac {{\mathrm e}^{x} \tan \left (\frac {x}{2}\right )+{\mathrm e}^{x} x^{2} \tan \left (\frac {x}{2}\right )+\frac {{\mathrm e}^{x} x^{2}}{2}+\frac {{\mathrm e}^{x} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}-2 \,{\mathrm e}^{x} x \tan \left (\frac {x}{2}\right )-\frac {{\mathrm e}^{x} x^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}-\frac {{\mathrm e}^{x}}{2}}{1+\tan ^{2}\left (\frac {x}{2}\right )}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.28, size = 26, normalized size = 0.51 \begin {gather*} \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 {gather*}

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 = 1.52, size = 26, normalized size = 0.51 \begin {gather*} \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 {gather*}

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 = 0.32, size = 48, normalized size = 0.94 \begin {gather*} \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 {gather*}

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 = 0.38, size = 24, normalized size = 0.47 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 2.36, size = 22, normalized size = 0.43 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (x-1\right )\,\left (\cos \left (x\right )-\sin \left (x\right )+x\,\cos \left (x\right )+x\,\sin \left (x\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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