3.68 \(\int \cos ^3(a+b x) \sin ^3(a+b x) \, dx\)

Optimal. Leaf size=31 \[ \frac{\sin ^4(a+b x)}{4 b}-\frac{\sin ^6(a+b x)}{6 b} \]

[Out]

Sin[a + b*x]^4/(4*b) - Sin[a + b*x]^6/(6*b)

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Rubi [A]  time = 0.0333805, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2564, 14} \[ \frac{\sin ^4(a+b x)}{4 b}-\frac{\sin ^6(a+b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Sin[a + b*x]^3,x]

[Out]

Sin[a + b*x]^4/(4*b) - Sin[a + b*x]^6/(6*b)

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])

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]]

Rubi steps

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

Mathematica [A]  time = 0.0145893, size = 35, normalized size = 1.13 \[ \frac{1}{8} \left (\frac{\cos (6 (a+b x))}{24 b}-\frac{3 \cos (2 (a+b x))}{8 b}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Sin[a + b*x]^3,x]

[Out]

((-3*Cos[2*(a + b*x)])/(8*b) + Cos[6*(a + b*x)]/(24*b))/8

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Maple [A]  time = 0.01, size = 34, normalized size = 1.1 \begin{align*}{\frac{1}{b} \left ( -{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4} \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{6}}-{\frac{ \left ( \cos \left ( bx+a \right ) \right ) ^{4}}{12}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*sin(b*x+a)^3,x)

[Out]

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

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Maxima [A]  time = 0.98458, size = 35, normalized size = 1.13 \begin{align*} -\frac{2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*sin(b*x + a)^6 - 3*sin(b*x + a)^4)/b

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Fricas [A]  time = 1.61483, size = 62, normalized size = 2. \begin{align*} \frac{2 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*cos(b*x + a)^6 - 3*cos(b*x + a)^4)/b

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Sympy [A]  time = 3.582, size = 42, normalized size = 1.35 \begin{align*} \begin{cases} \frac{\sin ^{6}{\left (a + b x \right )}}{12 b} + \frac{\sin ^{4}{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )}}{4 b} & \text{for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \cos ^{3}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3*sin(b*x+a)**3,x)

[Out]

Piecewise((sin(a + b*x)**6/(12*b) + sin(a + b*x)**4*cos(a + b*x)**2/(4*b), Ne(b, 0)), (x*sin(a)**3*cos(a)**3,
True))

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Giac [A]  time = 1.23579, size = 35, normalized size = 1.13 \begin{align*} -\frac{2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*sin(b*x + a)^6 - 3*sin(b*x + a)^4)/b