\(\int \cos ^2(3 x) \sin ^3(2 x) \, dx\) [134]

Optimal result

Integrand size = 13, antiderivative size = 41 \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=-\frac {3}{16} \cos (2 x)+\frac {3}{64} \cos (4 x)+\frac {1}{48} \cos (6 x)-\frac {3}{128} \cos (8 x)+\frac {1}{192} \cos (12 x) \]

[Out]

-3/16*cos(2*x)+3/64*cos(4*x)+1/48*cos(6*x)-3/128*cos(8*x)+1/192*cos(12*x)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4439, 2718} \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=-\frac {3}{16} \cos (2 x)+\frac {3}{64} \cos (4 x)+\frac {1}{48} \cos (6 x)-\frac {3}{128} \cos (8 x)+\frac {1}{192} \cos (12 x) \]

[In]

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

[Out]

(-3*Cos[2*x])/16 + (3*Cos[4*x])/64 + Cos[6*x]/48 - (3*Cos[8*x])/128 + Cos[12*x]/192

Rule 2718

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

Rule 4439

Int[(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + (d_.)*(x_)]^(q_.), x_Symbol] :> Int[ExpandTrigReduce[ActivateT
rig[F[a + b*x]^p*G[c + d*x]^q], x], x] /; FreeQ[{a, b, c, d}, x] && (EqQ[F, sin] || EqQ[F, cos]) && (EqQ[G, si
n] || EqQ[G, cos]) && IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{8} \sin (2 x)-\frac {3}{16} \sin (4 x)-\frac {1}{8} \sin (6 x)+\frac {3}{16} \sin (8 x)-\frac {1}{16} \sin (12 x)\right ) \, dx \\ & = -\left (\frac {1}{16} \int \sin (12 x) \, dx\right )-\frac {1}{8} \int \sin (6 x) \, dx-\frac {3}{16} \int \sin (4 x) \, dx+\frac {3}{16} \int \sin (8 x) \, dx+\frac {3}{8} \int \sin (2 x) \, dx \\ & = -\frac {3}{16} \cos (2 x)+\frac {3}{64} \cos (4 x)+\frac {1}{48} \cos (6 x)-\frac {3}{128} \cos (8 x)+\frac {1}{192} \cos (12 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=-\frac {3}{16} \cos (2 x)+\frac {3}{64} \cos (4 x)+\frac {1}{48} \cos (6 x)-\frac {3}{128} \cos (8 x)+\frac {1}{192} \cos (12 x) \]

[In]

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

[Out]

(-3*Cos[2*x])/16 + (3*Cos[4*x])/64 + Cos[6*x]/48 - (3*Cos[8*x])/128 + Cos[12*x]/192

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {3 \cos \left (2 x \right )}{16}+\frac {3 \cos \left (4 x \right )}{64}+\frac {\cos \left (6 x \right )}{48}-\frac {3 \cos \left (8 x \right )}{128}+\frac {\cos \left (12 x \right )}{192}\)	\(32\)
risch	\(-\frac {3 \cos \left (2 x \right )}{16}+\frac {3 \cos \left (4 x \right )}{64}+\frac {\cos \left (6 x \right )}{48}-\frac {3 \cos \left (8 x \right )}{128}+\frac {\cos \left (12 x \right )}{192}\)	\(32\)
parallelrisch	\(\frac {377}{1920}-\frac {3 \cos \left (8 x \right )}{128}+\frac {\cos \left (12 x \right )}{192}-\frac {3 \cos \left (2 x \right )}{16}+\frac {\cos \left (6 x \right )}{48}+\frac {3 \cos \left (4 x \right )}{64}\)	\(33\)

[In]

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

[Out]

-3/16*cos(2*x)+3/64*cos(4*x)+1/48*cos(6*x)-3/128*cos(8*x)+1/192*cos(12*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=\frac {32}{3} \, \cos \left (x\right )^{12} - 32 \, \cos \left (x\right )^{10} + 33 \, \cos \left (x\right )^{8} - 12 \, \cos \left (x\right )^{6} \]

[In]

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

[Out]

32/3*cos(x)^12 - 32*cos(x)^10 + 33*cos(x)^8 - 12*cos(x)^6

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 226 vs. \(2 (37) = 74\).

Time = 1.51 (sec) , antiderivative size = 226, normalized size of antiderivative = 5.51 \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=- \frac {x \sin ^{3}{\left (2 x \right )} \sin ^{2}{\left (3 x \right )}}{16} + \frac {x \sin ^{3}{\left (2 x \right )} \cos ^{2}{\left (3 x \right )}}{16} - \frac {3 x \sin ^{2}{\left (2 x \right )} \sin {\left (3 x \right )} \cos {\left (2 x \right )} \cos {\left (3 x \right )}}{8} + \frac {3 x \sin {\left (2 x \right )} \sin ^{2}{\left (3 x \right )} \cos ^{2}{\left (2 x \right )}}{16} - \frac {3 x \sin {\left (2 x \right )} \cos ^{2}{\left (2 x \right )} \cos ^{2}{\left (3 x \right )}}{16} + \frac {x \sin {\left (3 x \right )} \cos ^{3}{\left (2 x \right )} \cos {\left (3 x \right )}}{8} - \frac {\sin ^{3}{\left (2 x \right )} \sin {\left (3 x \right )} \cos {\left (3 x \right )}}{48} - \frac {\sin ^{2}{\left (2 x \right )} \cos {\left (2 x \right )} \cos ^{2}{\left (3 x \right )}}{2} + \frac {5 \sin {\left (2 x \right )} \sin {\left (3 x \right )} \cos ^{2}{\left (2 x \right )} \cos {\left (3 x \right )}}{8} - \frac {9 \sin ^{2}{\left (3 x \right )} \cos ^{3}{\left (2 x \right )}}{32} - \frac {5 \cos ^{3}{\left (2 x \right )} \cos ^{2}{\left (3 x \right )}}{96} \]

[In]

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

[Out]

-x*sin(2*x)**3*sin(3*x)**2/16 + x*sin(2*x)**3*cos(3*x)**2/16 - 3*x*sin(2*x)**2*sin(3*x)*cos(2*x)*cos(3*x)/8 +
3*x*sin(2*x)*sin(3*x)**2*cos(2*x)**2/16 - 3*x*sin(2*x)*cos(2*x)**2*cos(3*x)**2/16 + x*sin(3*x)*cos(2*x)**3*cos
(3*x)/8 - sin(2*x)**3*sin(3*x)*cos(3*x)/48 - sin(2*x)**2*cos(2*x)*cos(3*x)**2/2 + 5*sin(2*x)*sin(3*x)*cos(2*x)
**2*cos(3*x)/8 - 9*sin(3*x)**2*cos(2*x)**3/32 - 5*cos(2*x)**3*cos(3*x)**2/96

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=\frac {1}{192} \, \cos \left (12 \, x\right ) - \frac {3}{128} \, \cos \left (8 \, x\right ) + \frac {1}{48} \, \cos \left (6 \, x\right ) + \frac {3}{64} \, \cos \left (4 \, x\right ) - \frac {3}{16} \, \cos \left (2 \, x\right ) \]

[In]

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

[Out]

1/192*cos(12*x) - 3/128*cos(8*x) + 1/48*cos(6*x) + 3/64*cos(4*x) - 3/16*cos(2*x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=\frac {1}{192} \, \cos \left (12 \, x\right ) - \frac {3}{128} \, \cos \left (8 \, x\right ) + \frac {1}{48} \, \cos \left (6 \, x\right ) + \frac {3}{64} \, \cos \left (4 \, x\right ) - \frac {3}{16} \, \cos \left (2 \, x\right ) \]

[In]

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

[Out]

1/192*cos(12*x) - 3/128*cos(8*x) + 1/48*cos(6*x) + 3/64*cos(4*x) - 3/16*cos(2*x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.61 \[ \int \cos ^2(3 x) \sin ^3(2 x) \, dx=\frac {32\,{\cos \left (x\right )}^{12}}{3}-32\,{\cos \left (x\right )}^{10}+33\,{\cos \left (x\right )}^8-12\,{\cos \left (x\right )}^6 \]

[In]

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

[Out]

33*cos(x)^8 - 12*cos(x)^6 - 32*cos(x)^10 + (32*cos(x)^12)/3