3.4.71 \(\int \cos ^3(x) \sin ^3(x) \, dx\) [371]

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 = 0.01, 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 = {2644, 14} \begin {gather*} \frac {\sin ^4(x)}{4}-\frac {\sin ^6(x)}{6} \end {gather*}

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 2644

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 &=\text {Subst}\left (\int x^3 \left (1-x^2\right ) \, dx,x,\sin (x)\right )\\ &=\text {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 \begin {gather*} -\frac {3}{64} \cos (2 x)+\frac {1}{192} \cos (6 x) \end {gather*}

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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Mathics [A]
time = 1.78, size = 12, normalized size = 0.71 \begin {gather*} \frac {\left (2+\text {Cos}\left [2 x\right ]\right ) \text {Sin}\left [x\right ]^4}{12} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[Cos[x]^3*Sin[x]^3,x]')

[Out]

(2 + Cos[2 x]) Sin[x] ^ 4 / 12

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Maple [A]
time = 0.03, size = 18, normalized size = 1.06

method	result	size
risch	\(\frac {\cos \left (6 x \right )}{192}-\frac {3 \cos \left (2 x \right )}{64}\)	\(14\)
default	\(-\frac {\left (\cos ^{4}\left (x \right )\right ) \left (\sin ^{2}\left (x \right )\right )}{6}-\frac {\left (\cos ^{4}\left (x \right )\right )}{12}\)	\(18\)
norman	\(\frac {6 \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+6 \left (\tan ^{8}\left (\frac {x}{2}\right )\right )+\frac {2 \left (\tan ^{12}\left (\frac {x}{2}\right )\right )}{15}+\frac {4 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{5}+\frac {4 \left (\tan ^{10}\left (\frac {x}{2}\right )\right )}{5}+\frac {2}{15}}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{6}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3*sin(x)^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.27, size = 13, normalized size = 0.76 \begin {gather*} -\frac {1}{6} \, \sin \left (x\right )^{6} + \frac {1}{4} \, \sin \left (x\right )^{4} \end {gather*}

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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Fricas [A]
time = 0.37, size = 13, normalized size = 0.76 \begin {gather*} \frac {1}{6} \, \cos \left (x\right )^{6} - \frac {1}{4} \, \cos \left (x\right )^{4} \end {gather*}

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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Sympy [A]
time = 0.03, size = 12, normalized size = 0.71 \begin {gather*} - \frac {\sin ^{6}{\left (x \right )}}{6} + \frac {\sin ^{4}{\left (x \right )}}{4} \end {gather*}

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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Giac [A]
time = 0.00, size = 19, normalized size = 1.12 \begin {gather*} \frac {\frac {\cos ^{6}x}{3}-\frac {\cos ^{4}x}{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*cos(x)^6 - 1/4*cos(x)^4

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Mupad [B]
time = 0.05, size = 14, normalized size = 0.82 \begin {gather*} -\frac {{\sin \left (x\right )}^4\,\left (2\,{\sin \left (x\right )}^2-3\right )}{12} \end {gather*}

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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