\(\int \cos ^2(x) \sin ^2(x) \, dx\) [262]

Optimal result

Integrand size = 9, antiderivative size = 24 \[ \int \cos ^2(x) \sin ^2(x) \, dx=\frac {x}{8}+\frac {1}{8} \cos (x) \sin (x)-\frac {1}{4} \cos ^3(x) \sin (x) \]

[Out]

1/8*x+1/8*cos(x)*sin(x)-1/4*cos(x)^3*sin(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2648, 2715, 8} \[ \int \cos ^2(x) \sin ^2(x) \, dx=\frac {x}{8}-\frac {1}{4} \sin (x) \cos ^3(x)+\frac {1}{8} \sin (x) \cos (x) \]

[In]

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

[Out]

x/8 + (Cos[x]*Sin[x])/8 - (Cos[x]^3*Sin[x])/4

Rule 8

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

Rule 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*n]

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{4} \cos ^3(x) \sin (x)+\frac {1}{4} \int \cos ^2(x) \, dx \\ & = \frac {1}{8} \cos (x) \sin (x)-\frac {1}{4} \cos ^3(x) \sin (x)+\frac {\int 1 \, dx}{8} \\ & = \frac {x}{8}+\frac {1}{8} \cos (x) \sin (x)-\frac {1}{4} \cos ^3(x) \sin (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \cos ^2(x) \sin ^2(x) \, dx=\frac {x}{8}-\frac {1}{32} \sin (4 x) \]

[In]

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

[Out]

x/8 - Sin[4*x]/32

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.46

method	result	size
risch	\(\frac {x}{8}-\frac {\sin \left (4 x \right )}{32}\)	\(11\)
parallelrisch	\(\frac {x}{8}-\frac {\sin \left (4 x \right )}{32}\)	\(11\)
default	\(\frac {x}{8}+\frac {\cos \left (x \right ) \sin \left (x \right )}{8}-\frac {\left (\cos ^{3}\left (x \right )\right ) \sin \left (x \right )}{4}\)	\(19\)
norman	\(\frac {\frac {x}{8}+\frac {7 \left (\tan ^{3}\left (\frac {x}{2}\right )\right )}{4}-\frac {7 \left (\tan ^{5}\left (\frac {x}{2}\right )\right )}{4}+\frac {\left (\tan ^{7}\left (\frac {x}{2}\right )\right )}{4}+\frac {x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}+\frac {3 x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{4}+\frac {x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{2}+\frac {x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )}{8}-\frac {\tan \left (\frac {x}{2}\right )}{4}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{4}}\)	\(82\)

[In]

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

[Out]

1/8*x-1/32*sin(4*x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \cos ^2(x) \sin ^2(x) \, dx=-\frac {1}{8} \, {\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + \frac {1}{8} \, x \]

[In]

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

[Out]

-1/8*(2*cos(x)^3 - cos(x))*sin(x) + 1/8*x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \cos ^2(x) \sin ^2(x) \, dx=\frac {x}{8} - \frac {\sin {\left (2 x \right )} \cos {\left (2 x \right )}}{16} \]

[In]

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

[Out]

x/8 - sin(2*x)*cos(2*x)/16

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.42 \[ \int \cos ^2(x) \sin ^2(x) \, dx=\frac {1}{8} \, x - \frac {1}{32} \, \sin \left (4 \, x\right ) \]

[In]

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

[Out]

1/8*x - 1/32*sin(4*x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.42 \[ \int \cos ^2(x) \sin ^2(x) \, dx=\frac {1}{8} \, x - \frac {1}{32} \, \sin \left (4 \, x\right ) \]

[In]

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

[Out]

1/8*x - 1/32*sin(4*x)

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \cos ^2(x) \sin ^2(x) \, dx=\frac {\cos \left (x\right )\,{\sin \left (x\right )}^3}{4}-\frac {\cos \left (x\right )\,\sin \left (x\right )}{8}+\frac {x}{8} \]

[In]

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

[Out]

x/8 - (cos(x)*sin(x))/8 + (cos(x)*sin(x)^3)/4