\(\int x^2 \cos (3 x) \cos (5 x) \, dx\) [911]

Optimal result

Integrand size = 12, antiderivative size = 57 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{4} x \cos (2 x)+\frac {1}{64} x \cos (8 x)-\frac {1}{8} \sin (2 x)+\frac {1}{4} x^2 \sin (2 x)-\frac {1}{512} \sin (8 x)+\frac {1}{16} x^2 \sin (8 x) \]

[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)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4514, 3377, 2717} \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{4} x^2 \sin (2 x)+\frac {1}{16} x^2 \sin (8 x)-\frac {1}{8} \sin (2 x)-\frac {1}{512} \sin (8 x)+\frac {1}{4} x \cos (2 x)+\frac {1}{64} x \cos (8 x) \]

[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 2717

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

Rule 3377

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 4514

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{512} \left (128 x \cos (2 x)+8 x \cos (8 x)-64 \sin (2 x)+128 x^2 \sin (2 x)-\sin (8 x)+32 x^2 \sin (8 x)\right ) \]

[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

Maple [A] (verified)

Time = 0.90 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.74

method	result	size
risch	\(\frac {x \cos \left (8 x \right )}{64}+\frac {\left (32 x^{2}-1\right ) \sin \left (8 x \right )}{512}+\frac {x \cos \left (2 x \right )}{4}+\frac {\left (2 x^{2}-1\right ) \sin \left (2 x \right )}{8}\)	\(42\)
default	\(\frac {x \cos \left (2 x \right )}{4}+\frac {x \cos \left (8 x \right )}{64}-\frac {\sin \left (2 x \right )}{8}+\frac {x^{2} \sin \left (2 x \right )}{4}-\frac {\sin \left (8 x \right )}{512}+\frac {x^{2} \sin \left (8 x \right )}{16}\)	\(46\)
parallelrisch	\(\frac {x \cos \left (2 x \right )}{4}+\frac {x \cos \left (8 x \right )}{64}-\frac {\sin \left (2 x \right )}{8}+\frac {x^{2} \sin \left (2 x \right )}{4}-\frac {\sin \left (8 x \right )}{512}+\frac {x^{2} \sin \left (8 x \right )}{16}\)	\(46\)
norman	\(\frac {\frac {17 x}{64}-\frac {17 x \tan \left (\frac {5 x}{2}\right )^{2}}{64}-\frac {3 x^{2} \tan \left (\frac {3 x}{2}\right )}{8}+\frac {5 x^{2} \tan \left (\frac {5 x}{2}\right )}{8}-\frac {63 \tan \left (\frac {3 x}{2}\right ) \tan \left (\frac {5 x}{2}\right )^{2}}{256}-\frac {17 \tan \left (\frac {3 x}{2}\right )^{2} x}{64}+\frac {65 \tan \left (\frac {3 x}{2}\right )^{2} \tan \left (\frac {5 x}{2}\right )}{256}+\frac {3 x^{2} \tan \left (\frac {3 x}{2}\right ) \tan \left (\frac {5 x}{2}\right )^{2}}{8}-\frac {5 x^{2} \tan \left (\frac {3 x}{2}\right )^{2} \tan \left (\frac {5 x}{2}\right )}{8}+\frac {15 \tan \left (\frac {3 x}{2}\right ) x \tan \left (\frac {5 x}{2}\right )}{16}+\frac {17 \tan \left (\frac {3 x}{2}\right )^{2} x \tan \left (\frac {5 x}{2}\right )^{2}}{64}+\frac {63 \tan \left (\frac {3 x}{2}\right )}{256}-\frac {65 \tan \left (\frac {5 x}{2}\right )}{256}}{\left (1+\tan \left (\frac {3 x}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {5 x}{2}\right )^{2}\right )}\)	\(154\)

[In]

int(x^2*cos(3*x)*cos(5*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=2 \, x \cos \left (x\right )^{8} - 4 \, x \cos \left (x\right )^{6} + \frac {5}{2} \, x \cos \left (x\right )^{4} + \frac {1}{64} \, {\left (16 \, {\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{7} - 24 \, {\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{5} + 10 \, {\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{3} - 15 \, \cos \left (x\right )\right )} \sin \left (x\right ) - \frac {15}{64} \, x \]

[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

Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.58 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=- \frac {3 x^{2} \sin {\left (3 x \right )} \cos {\left (5 x \right )}}{16} + \frac {5 x^{2} \sin {\left (5 x \right )} \cos {\left (3 x \right )}}{16} + \frac {15 x \sin {\left (3 x \right )} \sin {\left (5 x \right )}}{64} + \frac {17 x \cos {\left (3 x \right )} \cos {\left (5 x \right )}}{64} + \frac {63 \sin {\left (3 x \right )} \cos {\left (5 x \right )}}{512} - \frac {65 \sin {\left (5 x \right )} \cos {\left (3 x \right )}}{512} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{64} \, x \cos \left (8 \, x\right ) + \frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{512} \, {\left (32 \, x^{2} - 1\right )} \sin \left (8 \, x\right ) + \frac {1}{8} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{64} \, x \cos \left (8 \, x\right ) + \frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{512} \, {\left (32 \, x^{2} - 1\right )} \sin \left (8 \, x\right ) + \frac {1}{8} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \]

[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)

Mupad [B] (verification not implemented)

Time = 26.36 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {x\,\cos \left (2\,x\right )}{4}-\frac {\sin \left (8\,x\right )}{512}-\frac {\sin \left (2\,x\right )}{8}+\frac {x\,\cos \left (8\,x\right )}{64}+\frac {x^2\,\sin \left (2\,x\right )}{4}+\frac {x^2\,\sin \left (8\,x\right )}{16} \]

[In]

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

[Out]

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