Integrand size = 12, antiderivative size = 57 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{4} x \cos (2 x)+\frac {1}{64} x \cos (8 x)-\frac {1}{8} \sin (2 x)+\frac {1}{4} x^2 \sin (2 x)-\frac {1}{512} \sin (8 x)+\frac {1}{16} x^2 \sin (8 x) \] Output:
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)
Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{512} \left (128 x \cos (2 x)+8 x \cos (8 x)-64 \sin (2 x)+128 x^2 \sin (2 x)-\sin (8 x)+32 x^2 \sin (8 x)\right ) \] Input:
Integrate[x^2*Cos[3*x]*Cos[5*x],x]
Output:
(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
Time = 0.27 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4929, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \cos (3 x) \cos (5 x) \, dx\) |
\(\Big \downarrow \) 4929 |
\(\displaystyle \int \left (\frac {1}{2} x^2 \cos (2 x)+\frac {1}{2} x^2 \cos (8 x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} x^2 \sin (2 x)+\frac {1}{16} x^2 \sin (8 x)-\frac {1}{8} \sin (2 x)-\frac {1}{512} \sin (8 x)+\frac {1}{4} x \cos (2 x)+\frac {1}{64} x \cos (8 x)\) |
Input:
Int[x^2*Cos[3*x]*Cos[5*x],x]
Output:
(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
Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*Cos[(c_.) + (d_.)*(x_)]^(q_.)*((e_.) + (f _.)*(x_))^(m_.), x_Symbol] :> Int[ExpandTrigReduce[(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]
Time = 0.68 (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 | \(\frac {\frac {17 x}{64}-\frac {17 x \tan \left (\frac {5 x}{2}\right )^{2}}{64}-\frac {3 x^{2} \tan \left (\frac {3 x}{2}\right )}{8}+\frac {5 x^{2} \tan \left (\frac {5 x}{2}\right )}{8}-\frac {63 \tan \left (\frac {3 x}{2}\right ) \tan \left (\frac {5 x}{2}\right )^{2}}{256}-\frac {17 \tan \left (\frac {3 x}{2}\right )^{2} x}{64}+\frac {65 \tan \left (\frac {3 x}{2}\right )^{2} \tan \left (\frac {5 x}{2}\right )}{256}+\frac {15 x \tan \left (\frac {3 x}{2}\right ) \tan \left (\frac {5 x}{2}\right )}{16}+\frac {3 x^{2} \tan \left (\frac {3 x}{2}\right ) \tan \left (\frac {5 x}{2}\right )^{2}}{8}-\frac {5 x^{2} \tan \left (\frac {3 x}{2}\right )^{2} \tan \left (\frac {5 x}{2}\right )}{8}+\frac {17 \tan \left (\frac {3 x}{2}\right )^{2} x \tan \left (\frac {5 x}{2}\right )^{2}}{64}+\frac {63 \tan \left (\frac {3 x}{2}\right )}{256}-\frac {65 \tan \left (\frac {5 x}{2}\right )}{256}}{\left (1+\tan \left (\frac {3 x}{2}\right )^{2}\right ) \left (1+\tan \left (\frac {5 x}{2}\right )^{2}\right )}\) | \(154\) |
orering | \(\frac {\left (1088 x^{4}-129 x^{2}-51\right ) \cos \left (3 x \right ) \cos \left (5 x \right )}{1024 x^{3}}-\frac {\left (1088 x^{4}-130 x^{2}-153\right ) \left (2 x \cos \left (3 x \right ) \cos \left (5 x \right )-3 x^{2} \sin \left (3 x \right ) \cos \left (5 x \right )-5 x^{2} \cos \left (3 x \right ) \sin \left (5 x \right )\right )}{4096 x^{4}}+\frac {\left (128 x^{2}-51\right ) \left (2 \cos \left (3 x \right ) \cos \left (5 x \right )-12 \cos \left (5 x \right ) \sin \left (3 x \right ) x -20 \cos \left (3 x \right ) \sin \left (5 x \right ) x -34 x^{2} \cos \left (3 x \right ) \cos \left (5 x \right )+30 x^{2} \sin \left (3 x \right ) \sin \left (5 x \right )\right )}{4096 x^{3}}-\frac {\left (32 x^{2}-17\right ) \left (-18 \sin \left (3 x \right ) \cos \left (5 x \right )-30 \sin \left (5 x \right ) \cos \left (3 x \right )+180 \sin \left (5 x \right ) \sin \left (3 x \right ) x -204 x \cos \left (3 x \right ) \cos \left (5 x \right )+252 x^{2} \sin \left (3 x \right ) \cos \left (5 x \right )+260 x^{2} \cos \left (3 x \right ) \sin \left (5 x \right )\right )}{8192 x^{2}}\) | \(234\) |
Input:
int(x^2*cos(3*x)*cos(5*x),x,method=_RETURNVERBOSE)
Output:
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)
Time = 0.08 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.28 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=2 \, x \cos \left (x\right )^{8} - 4 \, x \cos \left (x\right )^{6} + \frac {5}{2} \, x \cos \left (x\right )^{4} + \frac {1}{64} \, {\left (16 \, {\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{7} - 24 \, {\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{5} + 10 \, {\left (32 \, x^{2} - 1\right )} \cos \left (x\right )^{3} - 15 \, \cos \left (x\right )\right )} \sin \left (x\right ) - \frac {15}{64} \, x \] Input:
integrate(x^2*cos(3*x)*cos(5*x),x, algorithm="fricas")
Output:
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
Time = 0.59 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.58 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=- \frac {3 x^{2} \sin {\left (3 x \right )} \cos {\left (5 x \right )}}{16} + \frac {5 x^{2} \sin {\left (5 x \right )} \cos {\left (3 x \right )}}{16} + \frac {15 x \sin {\left (3 x \right )} \sin {\left (5 x \right )}}{64} + \frac {17 x \cos {\left (3 x \right )} \cos {\left (5 x \right )}}{64} + \frac {63 \sin {\left (3 x \right )} \cos {\left (5 x \right )}}{512} - \frac {65 \sin {\left (5 x \right )} \cos {\left (3 x \right )}}{512} \] Input:
integrate(x**2*cos(3*x)*cos(5*x),x)
Output:
-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*s in(5*x)*cos(3*x)/512
Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{64} \, x \cos \left (8 \, x\right ) + \frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{512} \, {\left (32 \, x^{2} - 1\right )} \sin \left (8 \, x\right ) + \frac {1}{8} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \] Input:
integrate(x^2*cos(3*x)*cos(5*x),x, algorithm="maxima")
Output:
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)
Time = 0.14 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.72 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {1}{64} \, x \cos \left (8 \, x\right ) + \frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{512} \, {\left (32 \, x^{2} - 1\right )} \sin \left (8 \, x\right ) + \frac {1}{8} \, {\left (2 \, x^{2} - 1\right )} \sin \left (2 \, x\right ) \] Input:
integrate(x^2*cos(3*x)*cos(5*x),x, algorithm="giac")
Output:
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)
Time = 16.22 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {x\,\cos \left (2\,x\right )}{4}-\frac {\sin \left (8\,x\right )}{512}-\frac {\sin \left (2\,x\right )}{8}+\frac {x\,\cos \left (8\,x\right )}{64}+\frac {x^2\,\sin \left (2\,x\right )}{4}+\frac {x^2\,\sin \left (8\,x\right )}{16} \] Input:
int(x^2*cos(3*x)*cos(5*x),x)
Output:
(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
Time = 0.17 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.21 \[ \int x^2 \cos (3 x) \cos (5 x) \, dx=\frac {17 \cos \left (5 x \right ) \cos \left (3 x \right ) x}{64}-\frac {3 \cos \left (5 x \right ) \sin \left (3 x \right ) x^{2}}{16}+\frac {63 \cos \left (5 x \right ) \sin \left (3 x \right )}{512}+\frac {5 \cos \left (3 x \right ) \sin \left (5 x \right ) x^{2}}{16}-\frac {65 \cos \left (3 x \right ) \sin \left (5 x \right )}{512}+\frac {15 \sin \left (5 x \right ) \sin \left (3 x \right ) x}{64} \] Input:
int(x^2*cos(3*x)*cos(5*x),x)
Output:
(136*cos(5*x)*cos(3*x)*x - 96*cos(5*x)*sin(3*x)*x**2 + 63*cos(5*x)*sin(3*x ) + 160*cos(3*x)*sin(5*x)*x**2 - 65*cos(3*x)*sin(5*x) + 120*sin(5*x)*sin(3 *x)*x)/512