3.911 \(\int x^2 \cos (3 x) \cos (5 x) \, dx\)

Optimal. Leaf size=57 \[ \frac{1}{4} x^2 \sin (2 x)+\frac{1}{16} x^2 \sin (8 x)-\frac{1}{8} \sin (2 x)-\frac{1}{512} \sin (8 x)+\frac{1}{4} x \cos (2 x)+\frac{1}{64} x \cos (8 x) \]

[Out]

(x*Cos[2*x])/4 + (x*Cos[8*x])/64 - Sin[2*x]/8 + (x^2*Sin[2*x])/4 - Sin[8*x]/512 + (x^2*Sin[8*x])/16

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Rubi [A]  time = 0.0734621, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4429, 3296, 2637} \[ \frac{1}{4} x^2 \sin (2 x)+\frac{1}{16} x^2 \sin (8 x)-\frac{1}{8} \sin (2 x)-\frac{1}{512} \sin (8 x)+\frac{1}{4} x \cos (2 x)+\frac{1}{64} x \cos (8 x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cos[3*x]*Cos[5*x],x]

[Out]

(x*Cos[2*x])/4 + (x*Cos[8*x])/64 - Sin[2*x]/8 + (x^2*Sin[2*x])/4 - Sin[8*x]/512 + (x^2*Sin[8*x])/16

Rule 4429

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*Cos[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Int[E
xpandTrigReduce[(e + f*x)^m, Cos[a + b*x]^p*Cos[c + d*x]^q, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p,
0] && IGtQ[q, 0] && IntegerQ[m]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int x^2 \cos (3 x) \cos (5 x) \, dx &=\int \left (\frac{1}{2} x^2 \cos (2 x)+\frac{1}{2} x^2 \cos (8 x)\right ) \, dx\\ &=\frac{1}{2} \int x^2 \cos (2 x) \, dx+\frac{1}{2} \int x^2 \cos (8 x) \, dx\\ &=\frac{1}{4} x^2 \sin (2 x)+\frac{1}{16} x^2 \sin (8 x)-\frac{1}{8} \int x \sin (8 x) \, dx-\frac{1}{2} \int x \sin (2 x) \, dx\\ &=\frac{1}{4} x \cos (2 x)+\frac{1}{64} x \cos (8 x)+\frac{1}{4} x^2 \sin (2 x)+\frac{1}{16} x^2 \sin (8 x)-\frac{1}{64} \int \cos (8 x) \, dx-\frac{1}{4} \int \cos (2 x) \, dx\\ &=\frac{1}{4} x \cos (2 x)+\frac{1}{64} x \cos (8 x)-\frac{1}{8} \sin (2 x)+\frac{1}{4} x^2 \sin (2 x)-\frac{1}{512} \sin (8 x)+\frac{1}{16} x^2 \sin (8 x)\\ \end{align*}

Mathematica [A]  time = 0.0914518, size = 49, normalized size = 0.86 \[ \frac{1}{512} \left (128 x^2 \sin (2 x)+32 x^2 \sin (8 x)-64 \sin (2 x)-\sin (8 x)+128 x \cos (2 x)+8 x \cos (8 x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cos[3*x]*Cos[5*x],x]

[Out]

(128*x*Cos[2*x] + 8*x*Cos[8*x] - 64*Sin[2*x] + 128*x^2*Sin[2*x] - Sin[8*x] + 32*x^2*Sin[8*x])/512

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Maple [A]  time = 0.046, size = 46, normalized size = 0.8 \begin{align*}{\frac{x\cos \left ( 2\,x \right ) }{4}}+{\frac{x\cos \left ( 8\,x \right ) }{64}}-{\frac{\sin \left ( 2\,x \right ) }{8}}+{\frac{{x}^{2}\sin \left ( 2\,x \right ) }{4}}-{\frac{\sin \left ( 8\,x \right ) }{512}}+{\frac{{x}^{2}\sin \left ( 8\,x \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cos(3*x)*cos(5*x),x)

[Out]

1/4*x*cos(2*x)+1/64*x*cos(8*x)-1/8*sin(2*x)+1/4*x^2*sin(2*x)-1/512*sin(8*x)+1/16*x^2*sin(8*x)

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Maxima [A]  time = 0.985726, size = 55, normalized size = 0.96 \begin{align*} \frac{1}{64} \, x \cos \left (8 \, x\right ) + \frac{1}{4} \, x \cos \left (2 \, x\right ) + \frac{1}{512} \,{\left (32 \, x^{2} - 1\right )} \sin \left (8 \, x\right ) + \frac{1}{8} \,{\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(3*x)*cos(5*x),x, algorithm="maxima")

[Out]

1/64*x*cos(8*x) + 1/4*x*cos(2*x) + 1/512*(32*x^2 - 1)*sin(8*x) + 1/8*(2*x^2 - 1)*sin(2*x)

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Fricas [A]  time = 2.25457, size = 220, normalized size = 3.86 \begin{align*} 2 \, x \cos \left (x\right )^{8} - 4 \, x \cos \left (x\right )^{6} + \frac{5}{2} \, x \cos \left (x\right )^{4} + \frac{1}{64} \,{\left (16 \,{\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{7} - 24 \,{\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{5} + 10 \,{\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{3} - 15 \, \cos \left (x\right )\right )} \sin \left (x\right ) - \frac{15}{64} \, x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(3*x)*cos(5*x),x, algorithm="fricas")

[Out]

2*x*cos(x)^8 - 4*x*cos(x)^6 + 5/2*x*cos(x)^4 + 1/64*(16*(32*x^2 - 1)*cos(x)^7 - 24*(32*x^2 - 1)*cos(x)^5 + 10*
(32*x^2 - 1)*cos(x)^3 - 15*cos(x))*sin(x) - 15/64*x

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Sympy [A]  time = 7.72247, size = 90, normalized size = 1.58 \begin{align*} - \frac{3 x^{2} \sin{\left (3 x \right )} \cos{\left (5 x \right )}}{16} + \frac{5 x^{2} \sin{\left (5 x \right )} \cos{\left (3 x \right )}}{16} + \frac{15 x \sin{\left (3 x \right )} \sin{\left (5 x \right )}}{64} + \frac{17 x \cos{\left (3 x \right )} \cos{\left (5 x \right )}}{64} + \frac{63 \sin{\left (3 x \right )} \cos{\left (5 x \right )}}{512} - \frac{65 \sin{\left (5 x \right )} \cos{\left (3 x \right )}}{512} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cos(3*x)*cos(5*x),x)

[Out]

-3*x**2*sin(3*x)*cos(5*x)/16 + 5*x**2*sin(5*x)*cos(3*x)/16 + 15*x*sin(3*x)*sin(5*x)/64 + 17*x*cos(3*x)*cos(5*x
)/64 + 63*sin(3*x)*cos(5*x)/512 - 65*sin(5*x)*cos(3*x)/512

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Giac [A]  time = 1.09981, size = 55, normalized size = 0.96 \begin{align*} \frac{1}{64} \, x \cos \left (8 \, x\right ) + \frac{1}{4} \, x \cos \left (2 \, x\right ) + \frac{1}{512} \,{\left (32 \, x^{2} - 1\right )} \sin \left (8 \, x\right ) + \frac{1}{8} \,{\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cos(3*x)*cos(5*x),x, algorithm="giac")

[Out]

1/64*x*cos(8*x) + 1/4*x*cos(2*x) + 1/512*(32*x^2 - 1)*sin(8*x) + 1/8*(2*x^2 - 1)*sin(2*x)