\(\int \cos (\cos (x)) \sin (x) \, dx\) [651]

Optimal result

Integrand size = 6, antiderivative size = 5 \[ \int \cos (\cos (x)) \sin (x) \, dx=-\sin (\cos (x)) \]

[Out]

-sin(cos(x))

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4420, 2717} \[ \int \cos (\cos (x)) \sin (x) \, dx=-\sin (\cos (x)) \]

[In]

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

[Out]

-Sin[Cos[x]]

Rule 2717

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

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 \cos (x) \, dx,x,\cos (x)) \\ & = -\sin (\cos (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 4.39 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \cos (\cos (x)) \sin (x) \, dx=-\sin (\cos (x)) \]

[In]

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

[Out]

-Sin[Cos[x]]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 6, normalized size of antiderivative = 1.20

method	result	size
derivativedivides	\(-\sin \left (\cos \left (x \right )\right )\)	\(6\)
default	\(-\sin \left (\cos \left (x \right )\right )\)	\(6\)
risch	\(-\sin \left (\cos \left (x \right )\right )\)	\(6\)
parallelrisch	\(-\sin \left (\cos \left (x \right )\right )\)	\(6\)
norman	\(\frac {-2 \tan \left (\frac {x}{2}\right )^{2} \tan \left (\frac {1-\tan \left (\frac {x}{2}\right )^{2}}{2+2 \tan \left (\frac {x}{2}\right )^{2}}\right )-2 \tan \left (\frac {1-\tan \left (\frac {x}{2}\right )^{2}}{2+2 \tan \left (\frac {x}{2}\right )^{2}}\right )}{\left (1+\tan \left (\frac {1-\tan \left (\frac {x}{2}\right )^{2}}{2+2 \tan \left (\frac {x}{2}\right )^{2}}\right )^{2}\right ) \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )}\)	\(98\)

[In]

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

[Out]

-sin(cos(x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 20 vs. \(2 (5) = 10\).

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 4.00 \[ \int \cos (\cos (x)) \sin (x) \, dx=\sin \left (\frac {\tan \left (\frac {1}{2} \, x\right )^{2} - 1}{\tan \left (\frac {1}{2} \, x\right )^{2} + 1}\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \cos (\cos (x)) \sin (x) \, dx=- \sin {\left (\cos {\left (x \right )} \right )} \]

[In]

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

[Out]

-sin(cos(x))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \cos (\cos (x)) \sin (x) \, dx=-\sin \left (\cos \left (x\right )\right ) \]

[In]

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

[Out]

-sin(cos(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \cos (\cos (x)) \sin (x) \, dx=-\sin \left (\cos \left (x\right )\right ) \]

[In]

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

[Out]

-sin(cos(x))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 5, normalized size of antiderivative = 1.00 \[ \int \cos (\cos (x)) \sin (x) \, dx=-\sin \left (\cos \left (x\right )\right ) \]

[In]

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

[Out]

-sin(cos(x))