\(\int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx\) [652]

Optimal result

Integrand size = 11, antiderivative size = 28 \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=\frac {\cos (x)}{4}-\frac {1}{4} \cos (\cos (x)) \sin (\cos (x))-\frac {1}{2} \cos (x) \sin ^2(\cos (x)) \]

[Out]

1/4*cos(x)-1/4*cos(cos(x))*sin(cos(x))-1/2*cos(x)*sin(cos(x))^2

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4420, 3524, 2715, 8} \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=\frac {\cos (x)}{4}-\frac {1}{2} \cos (x) \sin ^2(\cos (x))-\frac {1}{4} \cos (\cos (x)) \sin (\cos (x)) \]

[In]

Int[Cos[x]*Cos[Cos[x]]*Sin[x]*Sin[Cos[x]],x]

[Out]

Cos[x]/4 - (Cos[Cos[x]]*Sin[Cos[x]])/4 - (Cos[x]*Sin[Cos[x]]^2)/2

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3524

Int[Cos[(a_.) + (b_.)*(x_)^(n_.)]*(x_)^(m_.)*Sin[(a_.) + (b_.)*(x_)^(n_.)]^(p_.), x_Symbol] :> Simp[x^(m - n +
1)*(Sin[a + b*x^n]^(p + 1)/(b*n*(p + 1))), x] - Dist[(m - n + 1)/(b*n*(p + 1)), Int[x^(m - n)*Sin[a + b*x^n]^
(p + 1), x], x] /; FreeQ[{a, b, p}, x] && LtQ[0, n, m + 1] && NeQ[p, -1]

Rule 4420

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, Dist[-d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}(\int x \cos (x) \sin (x) \, dx,x,\cos (x)) \\ & = -\frac {1}{2} \cos (x) \sin ^2(\cos (x))+\frac {1}{2} \text {Subst}\left (\int \sin ^2(x) \, dx,x,\cos (x)\right ) \\ & = -\frac {1}{4} \cos (\cos (x)) \sin (\cos (x))-\frac {1}{2} \cos (x) \sin ^2(\cos (x))+\frac {1}{4} \text {Subst}(\int 1 \, dx,x,\cos (x)) \\ & = \frac {\cos (x)}{4}-\frac {1}{4} \cos (\cos (x)) \sin (\cos (x))-\frac {1}{2} \cos (x) \sin ^2(\cos (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 2.05 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=\frac {1}{4} \cos (x) \cos (2 \cos (x))-\frac {1}{8} \sin (2 \cos (x)) \]

[In]

Integrate[Cos[x]*Cos[Cos[x]]*Sin[x]*Sin[Cos[x]],x]

[Out]

(Cos[x]*Cos[2*Cos[x]])/4 - Sin[2*Cos[x]]/8

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {\cos \left (x \right ) \cos \left (\cos \left (x \right )\right )^{2}}{2}-\frac {\cos \left (\cos \left (x \right )\right ) \sin \left (\cos \left (x \right )\right )}{4}-\frac {\cos \left (x \right )}{4}\)	\(23\)
default	\(\frac {\cos \left (x \right ) \cos \left (\cos \left (x \right )\right )^{2}}{2}-\frac {\cos \left (\cos \left (x \right )\right ) \sin \left (\cos \left (x \right )\right )}{4}-\frac {\cos \left (x \right )}{4}\)	\(23\)
risch	\(\frac {\cos \left (-2 \cos \left (x \right )+x \right )}{8}+\frac {\cos \left (2 \cos \left (x \right )+x \right )}{8}-\frac {\sin \left (2 \cos \left (x \right )\right )}{8}\)	\(27\)
parallelrisch	\(-\frac {1}{4}+\frac {\cos \left (-2 \cos \left (x \right )+x \right )}{8}+\frac {\cos \left (2 \cos \left (x \right )+x \right )}{8}-\frac {\sin \left (2 \cos \left (x \right )\right )}{8}\)	\(28\)

[In]

int(cos(x)*cos(cos(x))*sin(x)*sin(cos(x)),x,method=_RETURNVERBOSE)

[Out]

1/2*cos(x)*cos(cos(x))^2-1/4*cos(cos(x))*sin(cos(x))-1/4*cos(x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (22) = 44\).

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 2.61 \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=\frac {1}{2} \, \cos \left (x\right ) \cos \left (\frac {\tan \left (\frac {1}{2} \, x\right )^{2} - 1}{\tan \left (\frac {1}{2} \, x\right )^{2} + 1}\right )^{2} + \frac {1}{4} \, \cos \left (\frac {\tan \left (\frac {1}{2} \, x\right )^{2} - 1}{\tan \left (\frac {1}{2} \, x\right )^{2} + 1}\right ) \sin \left (\frac {\tan \left (\frac {1}{2} \, x\right )^{2} - 1}{\tan \left (\frac {1}{2} \, x\right )^{2} + 1}\right ) - \frac {1}{4} \, \cos \left (x\right ) \]

[In]

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

[Out]

1/2*cos(x)*cos((tan(1/2*x)^2 - 1)/(tan(1/2*x)^2 + 1))^2 + 1/4*cos((tan(1/2*x)^2 - 1)/(tan(1/2*x)^2 + 1))*sin((
tan(1/2*x)^2 - 1)/(tan(1/2*x)^2 + 1)) - 1/4*cos(x)

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=- \frac {\sin ^{2}{\left (\cos {\left (x \right )} \right )} \cos {\left (x \right )}}{4} - \frac {\sin {\left (\cos {\left (x \right )} \right )} \cos {\left (\cos {\left (x \right )} \right )}}{4} + \frac {\cos {\left (x \right )} \cos ^{2}{\left (\cos {\left (x \right )} \right )}}{4} \]

[In]

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

[Out]

-sin(cos(x))**2*cos(x)/4 - sin(cos(x))*cos(cos(x))/4 + cos(x)*cos(cos(x))**2/4

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=\frac {1}{4} \, \cos \left (x\right ) \cos \left (2 \, \cos \left (x\right )\right ) - \frac {1}{8} \, \sin \left (2 \, \cos \left (x\right )\right ) \]

[In]

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

[Out]

1/4*cos(x)*cos(2*cos(x)) - 1/8*sin(2*cos(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.61 \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=\frac {1}{4} \, \cos \left (x\right ) \cos \left (2 \, \cos \left (x\right )\right ) - \frac {1}{8} \, \sin \left (2 \, \cos \left (x\right )\right ) \]

[In]

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

[Out]

1/4*cos(x)*cos(2*cos(x)) - 1/8*sin(2*cos(x))

Mupad [B] (verification not implemented)

Time = 26.84 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \cos (x) \cos (\cos (x)) \sin (x) \sin (\cos (x)) \, dx=\frac {\cos \left (x\right )\,{\cos \left (\cos \left (x\right )\right )}^2}{2}-\frac {\sin \left (\cos \left (x\right )\right )\,\cos \left (\cos \left (x\right )\right )}{4}-\frac {\cos \left (x\right )}{4} \]

[In]

int(cos(cos(x))*sin(cos(x))*cos(x)*sin(x),x)

[Out]

(cos(cos(x))^2*cos(x))/2 - cos(x)/4 - (cos(cos(x))*sin(cos(x)))/4