3.371 \(\int \cos ^3(x) \sin ^3(x) \, dx\)

Optimal. Leaf size=17 \[ \frac {\sin ^4(x)}{4}-\frac {\sin ^6(x)}{6} \]

[Out]

1/4*sin(x)^4-1/6*sin(x)^6

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2564, 14} \[ \frac {\sin ^4(x)}{4}-\frac {\sin ^6(x)}{6} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^3*Sin[x]^3,x]

[Out]

Sin[x]^4/4 - Sin[x]^6/6

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rubi steps

\begin {align*} \int \cos ^3(x) \sin ^3(x) \, dx &=\operatorname {Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,\sin (x)\right )\\ &=\operatorname {Subst}\left (\int \left (x^3-x^5\right ) \, dx,x,\sin (x)\right )\\ &=\frac {\sin ^4(x)}{4}-\frac {\sin ^6(x)}{6}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 17, normalized size = 1.00 \[ \frac {1}{192} \cos (6 x)-\frac {3}{64} \cos (2 x) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^3*Sin[x]^3,x]

[Out]

(-3*Cos[2*x])/64 + Cos[6*x]/192

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fricas [A]  time = 0.45, size = 13, normalized size = 0.76 \[ \frac {1}{6} \, \cos \relax (x)^{6} - \frac {1}{4} \, \cos \relax (x)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3*sin(x)^3,x, algorithm="fricas")

[Out]

1/6*cos(x)^6 - 1/4*cos(x)^4

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giac [A]  time = 0.98, size = 13, normalized size = 0.76 \[ \frac {1}{6} \, \cos \relax (x)^{6} - \frac {1}{4} \, \cos \relax (x)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3*sin(x)^3,x, algorithm="giac")

[Out]

1/6*cos(x)^6 - 1/4*cos(x)^4

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maple [A]  time = 0.01, size = 18, normalized size = 1.06 \[ -\frac {\left (\cos ^{4}\relax (x )\right ) \left (\sin ^{2}\relax (x )\right )}{6}-\frac {\left (\cos ^{4}\relax (x )\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3*sin(x)^3,x)

[Out]

-1/6*cos(x)^4*sin(x)^2-1/12*cos(x)^4

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maxima [A]  time = 0.44, size = 13, normalized size = 0.76 \[ -\frac {1}{6} \, \sin \relax (x)^{6} + \frac {1}{4} \, \sin \relax (x)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3*sin(x)^3,x, algorithm="maxima")

[Out]

-1/6*sin(x)^6 + 1/4*sin(x)^4

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mupad [B]  time = 0.05, size = 14, normalized size = 0.82 \[ -\frac {{\sin \relax (x)}^4\,\left (2\,{\sin \relax (x)}^2-3\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3*sin(x)^3,x)

[Out]

-(sin(x)^4*(2*sin(x)^2 - 3))/12

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sympy [A]  time = 0.07, size = 12, normalized size = 0.71 \[ - \frac {\sin ^{6}{\relax (x )}}{6} + \frac {\sin ^{4}{\relax (x )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)**3*sin(x)**3,x)

[Out]

-sin(x)**6/6 + sin(x)**4/4

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