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]

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

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 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(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 = 0.01, 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

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 26, normalized size = 0.84 \[ \frac {2 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4}}{12 \, b} \]

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

________________________________________________________________________________________

giac [A]  time = 0.20, size = 26, normalized size = 0.84 \[ -\frac {2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}}{12 \, b} \]

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

________________________________________________________________________________________

maple [A]  time = 0.02, size = 34, normalized size = 1.10 \[ \frac {-\frac {\left (\cos ^{4}\left (b x +a \right )\right ) \left (\sin ^{2}\left (b x +a \right )\right )}{6}-\frac {\left (\cos ^{4}\left (b x +a \right )\right )}{12}}{b} \]

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)

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 26, normalized size = 0.84 \[ -\frac {2 \, \sin \left (b x + a\right )^{6} - 3 \, \sin \left (b x + a\right )^{4}}{12 \, b} \]

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

________________________________________________________________________________________

mupad [B]  time = 0.51, size = 37, normalized size = 1.19 \[ \frac {{\cos \left (a+b\,x\right )}^4\,\left ({\cos \left (a+b\,x\right )}^2-1\right )}{4\,b}-\frac {{\cos \left (a+b\,x\right )}^6}{12\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3*sin(a + b*x)^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 4.33, size = 44, normalized size = 1.42 \[ \begin {cases} - \frac {\sin ^{2}{\left (a + b x \right )} \cos ^{4}{\left (a + b x \right )}}{4 b} - \frac {\cos ^{6}{\left (a + b x \right )}}{12 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\relax (a )} \cos ^{3}{\relax (a )} & \text {otherwise} \end {cases} \]

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

________________________________________________________________________________________