\(\int x \cos (x) \sin (x) \, dx\) [27]

Optimal result

Integrand size = 6, antiderivative size = 23 \[ \int x \cos (x) \sin (x) \, dx=-\frac {x}{4}+\frac {1}{4} \cos (x) \sin (x)+\frac {1}{2} x \sin ^2(x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3524, 2715, 8} \[ \int x \cos (x) \sin (x) \, dx=-\frac {x}{4}+\frac {1}{2} x \sin ^2(x)+\frac {1}{4} \sin (x) \cos (x) \]

[In]

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

[Out]

-1/4*x + (Cos[x]*Sin[x])/4 + (x*Sin[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]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int x \cos (x) \sin (x) \, dx=-\frac {1}{4} x \cos (2 x)+\frac {1}{8} \sin (2 x) \]

[In]

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

[Out]

-1/4*(x*Cos[2*x]) + Sin[2*x]/8

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {x \cos \left (2 x \right )}{4}+\frac {\sin \left (2 x \right )}{8}\)	\(15\)
parallelrisch	\(-\frac {x \cos \left (2 x \right )}{4}+\frac {\sin \left (2 x \right )}{8}\)	\(15\)
default	\(-\frac {\left (\cos ^{2}\left (x \right )\right ) x}{2}+\frac {\cos \left (x \right ) \sin \left (x \right )}{4}+\frac {x}{4}\)	\(18\)
meijerg	\(\frac {\sqrt {\pi }\, \left (-\frac {x \cos \left (2 x \right )}{\sqrt {\pi }}+\frac {\sin \left (2 x \right )}{2 \sqrt {\pi }}\right )}{4}\)	\(26\)
norman	\(\frac {-\frac {x}{4}-\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{2}+\frac {3 x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}-\frac {x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}+\frac {\tan \left (\frac {x}{2}\right )}{2}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{2}}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int x \cos (x) \sin (x) \, dx=-\frac {1}{2} \, x \cos \left (x\right )^{2} + \frac {1}{4} \, \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{4} \, x \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int x \cos (x) \sin (x) \, dx=\frac {x \sin ^{2}{\left (x \right )}}{4} - \frac {x \cos ^{2}{\left (x \right )}}{4} + \frac {\sin {\left (x \right )} \cos {\left (x \right )}}{4} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int x \cos (x) \sin (x) \, dx=-\frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{8} \, \sin \left (2 \, x\right ) \]

[In]

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

[Out]

-1/4*x*cos(2*x) + 1/8*sin(2*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int x \cos (x) \sin (x) \, dx=-\frac {1}{4} \, x \cos \left (2 \, x\right ) + \frac {1}{8} \, \sin \left (2 \, x\right ) \]

[In]

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

[Out]

-1/4*x*cos(2*x) + 1/8*sin(2*x)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int x \cos (x) \sin (x) \, dx=\frac {\sin \left (2\,x\right )}{8}+\frac {x\,\left (2\,{\sin \left (x\right )}^2-1\right )}{4} \]

[In]

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

[Out]

sin(2*x)/8 + (x*(2*sin(x)^2 - 1))/4