\(\int \cos (x) \sin ^2(6 x) \, dx\) [131]

Optimal result

Integrand size = 9, antiderivative size = 23 \[ \int \cos (x) \sin ^2(6 x) \, dx=\frac {\sin (x)}{2}-\frac {1}{44} \sin (11 x)-\frac {1}{52} \sin (13 x) \]

[Out]

1/2*sin(x)-1/44*sin(11*x)-1/52*sin(13*x)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {4439, 2717} \[ \int \cos (x) \sin ^2(6 x) \, dx=\frac {\sin (x)}{2}-\frac {1}{44} \sin (11 x)-\frac {1}{52} \sin (13 x) \]

[In]

Int[Cos[x]*Sin[6*x]^2,x]

[Out]

Sin[x]/2 - Sin[11*x]/44 - Sin[13*x]/52

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 {\cos (x)}{2}-\frac {1}{4} \cos (11 x)-\frac {1}{4} \cos (13 x)\right ) \, dx \\ & = -\left (\frac {1}{4} \int \cos (11 x) \, dx\right )-\frac {1}{4} \int \cos (13 x) \, dx+\frac {1}{2} \int \cos (x) \, dx \\ & = \frac {\sin (x)}{2}-\frac {1}{44} \sin (11 x)-\frac {1}{52} \sin (13 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \cos (x) \sin ^2(6 x) \, dx=\frac {\sin (x)}{2}-\frac {1}{44} \sin (11 x)-\frac {1}{52} \sin (13 x) \]

[In]

Integrate[Cos[x]*Sin[6*x]^2,x]

[Out]

Sin[x]/2 - Sin[11*x]/44 - Sin[13*x]/52

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method	result	size
default	\(\frac {\sin \left (x \right )}{2}-\frac {\sin \left (11 x \right )}{44}-\frac {\sin \left (13 x \right )}{52}\)	\(18\)
risch	\(\frac {\sin \left (x \right )}{2}-\frac {\sin \left (11 x \right )}{44}-\frac {\sin \left (13 x \right )}{52}\)	\(18\)
parallelrisch	\(\frac {\sin \left (x \right )}{2}-\frac {\sin \left (11 x \right )}{44}-\frac {\sin \left (13 x \right )}{52}\)	\(18\)
norman	\(\frac {\frac {24 \tan \left (3 x \right )^{3}}{143}+\frac {24 \tan \left (3 x \right ) \tan \left (\frac {x}{2}\right )^{2}}{143}+\frac {280 \tan \left (3 x \right )^{2} \tan \left (\frac {x}{2}\right )}{143}-\frac {24 \tan \left (3 x \right )^{3} \tan \left (\frac {x}{2}\right )^{2}}{143}+\frac {144 \tan \left (3 x \right )^{4} \tan \left (\frac {x}{2}\right )}{143}-\frac {24 \tan \left (3 x \right )}{143}+\frac {144 \tan \left (\frac {x}{2}\right )}{143}}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right ) \left (1+\tan \left (3 x \right )^{2}\right )^{2}}\)	\(93\)

[In]

int(cos(x)*sin(6*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/2*sin(x)-1/44*sin(11*x)-1/52*sin(13*x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \cos (x) \sin ^2(6 x) \, dx=-\frac {4}{143} \, {\left (2816 \, \cos \left (x\right )^{12} - 6912 \, \cos \left (x\right )^{10} + 6048 \, \cos \left (x\right )^{8} - 2240 \, \cos \left (x\right )^{6} + 315 \, \cos \left (x\right )^{4} - 9 \, \cos \left (x\right )^{2} - 18\right )} \sin \left (x\right ) \]

[In]

integrate(cos(x)*sin(6*x)^2,x, algorithm="fricas")

[Out]

-4/143*(2816*cos(x)^12 - 6912*cos(x)^10 + 6048*cos(x)^8 - 2240*cos(x)^6 + 315*cos(x)^4 - 9*cos(x)^2 - 18)*sin(
x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).

Time = 0.27 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \cos (x) \sin ^2(6 x) \, dx=\frac {71 \sin {\left (x \right )} \sin ^{2}{\left (6 x \right )}}{143} + \frac {72 \sin {\left (x \right )} \cos ^{2}{\left (6 x \right )}}{143} - \frac {12 \sin {\left (6 x \right )} \cos {\left (x \right )} \cos {\left (6 x \right )}}{143} \]

[In]

integrate(cos(x)*sin(6*x)**2,x)

[Out]

71*sin(x)*sin(6*x)**2/143 + 72*sin(x)*cos(6*x)**2/143 - 12*sin(6*x)*cos(x)*cos(6*x)/143

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \cos (x) \sin ^2(6 x) \, dx=-\frac {1}{52} \, \sin \left (13 \, x\right ) - \frac {1}{44} \, \sin \left (11 \, x\right ) + \frac {1}{2} \, \sin \left (x\right ) \]

[In]

integrate(cos(x)*sin(6*x)^2,x, algorithm="maxima")

[Out]

-1/52*sin(13*x) - 1/44*sin(11*x) + 1/2*sin(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \cos (x) \sin ^2(6 x) \, dx=-\frac {1}{52} \, \sin \left (13 \, x\right ) - \frac {1}{44} \, \sin \left (11 \, x\right ) + \frac {1}{2} \, \sin \left (x\right ) \]

[In]

integrate(cos(x)*sin(6*x)^2,x, algorithm="giac")

[Out]

-1/52*sin(13*x) - 1/44*sin(11*x) + 1/2*sin(x)

Mupad [B] (verification not implemented)

Time = 26.96 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \cos (x) \sin ^2(6 x) \, dx=\frac {\sin \left (x\right )}{2}-\frac {\sin \left (13\,x\right )}{52}-\frac {\sin \left (11\,x\right )}{44} \]

[In]

int(sin(6*x)^2*cos(x),x)

[Out]

sin(x)/2 - sin(13*x)/52 - sin(11*x)/44