### 3.65 $$\int \cos ^2(x) \sin ^2(x) \, dx$$

Optimal. Leaf size=24 $\frac{x}{8}-\frac{1}{4} \sin (x) \cos ^3(x)+\frac{1}{8} \sin (x) \cos (x)$

[Out]

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

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Rubi [A]  time = 0.0248675, antiderivative size = 24, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {2568, 2635, 8} $\frac{x}{8}-\frac{1}{4} \sin (x) \cos ^3(x)+\frac{1}{8} \sin (x) \cos (x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

Rule 2568

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 2635

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] && Integer
Q[2*n]

Rule 8

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

Rubi steps

\begin{align*} \int \cos ^2(x) \sin ^2(x) \, dx &=-\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]  time = 0.0037463, size = 14, normalized size = 0.58 $\frac{x}{8}-\frac{1}{32} \sin (4 x)$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

x/8 - Sin[4*x]/32

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Maple [A]  time = 0., size = 19, normalized size = 0.8 \begin{align*}{\frac{x}{8}}+{\frac{\cos \left ( x \right ) \sin \left ( x \right ) }{8}}-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{3}\sin \left ( x \right ) }{4}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.925287, size = 14, normalized size = 0.58 \begin{align*} \frac{1}{8} \, x - \frac{1}{32} \, \sin \left (4 \, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x - 1/32*sin(4*x)

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Fricas [A]  time = 2.03853, size = 58, normalized size = 2.42 \begin{align*} -\frac{1}{8} \,{\left (2 \, \cos \left (x\right )^{3} - \cos \left (x\right )\right )} \sin \left (x\right ) + \frac{1}{8} \, x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.065533, size = 14, normalized size = 0.58 \begin{align*} \frac{x}{8} - \frac{\sin{\left (2 x \right )} \cos{\left (2 x \right )}}{16} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/8 - sin(2*x)*cos(2*x)/16

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Giac [A]  time = 1.05108, size = 14, normalized size = 0.58 \begin{align*} \frac{1}{8} \, x - \frac{1}{32} \, \sin \left (4 \, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x - 1/32*sin(4*x)