3.2.53 \(\int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx\) [153]

Optimal. Leaf size=31 \[ -\frac {8 \cos ^7(a+b x)}{7 b}+\frac {8 \cos ^9(a+b x)}{9 b} \]

[Out]

-8/7*cos(b*x+a)^7/b+8/9*cos(b*x+a)^9/b

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Rubi [A]
time = 0.04, antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4372, 2645, 14} \begin {gather*} \frac {8 \cos ^9(a+b x)}{9 b}-\frac {8 \cos ^7(a+b x)}{7 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^3,x]

[Out]

(-8*Cos[a + b*x]^7)/(7*b) + (8*Cos[a + b*x]^9)/(9*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 2645

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

Rule 4372

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rubi steps

\begin {align*} \int \cos ^3(a+b x) \sin ^3(2 a+2 b x) \, dx &=8 \int \cos ^6(a+b x) \sin ^3(a+b x) \, dx\\ &=-\frac {8 \text {Subst}\left (\int x^6 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {8 \text {Subst}\left (\int \left (x^6-x^8\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {8 \cos ^7(a+b x)}{7 b}+\frac {8 \cos ^9(a+b x)}{9 b}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 27, normalized size = 0.87 \begin {gather*} \frac {4 \cos ^7(a+b x) (-11+7 \cos (2 (a+b x)))}{63 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^3,x]

[Out]

(4*Cos[a + b*x]^7*(-11 + 7*Cos[2*(a + b*x)]))/(63*b)

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Maple [A]
time = 0.18, size = 55, normalized size = 1.77

method	result	size
default	\(-\frac {3 \cos \left (x b +a \right )}{16 b}-\frac {\cos \left (3 x b +3 a \right )}{12 b}+\frac {3 \cos \left (7 x b +7 a \right )}{224 b}+\frac {\cos \left (9 x b +9 a \right )}{288 b}\)	\(55\)
risch	\(-\frac {3 \cos \left (x b +a \right )}{16 b}-\frac {\cos \left (3 x b +3 a \right )}{12 b}+\frac {3 \cos \left (7 x b +7 a \right )}{224 b}+\frac {\cos \left (9 x b +9 a \right )}{288 b}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x,method=_RETURNVERBOSE)

[Out]

-3/16*cos(b*x+a)/b-1/12*cos(3*b*x+3*a)/b+3/224*cos(7*b*x+7*a)/b+1/288*cos(9*b*x+9*a)/b

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Maxima [A]
time = 0.28, size = 47, normalized size = 1.52 \begin {gather*} \frac {7 \, \cos \left (9 \, b x + 9 \, a\right ) + 27 \, \cos \left (7 \, b x + 7 \, a\right ) - 168 \, \cos \left (3 \, b x + 3 \, a\right ) - 378 \, \cos \left (b x + a\right )}{2016 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="maxima")

[Out]

1/2016*(7*cos(9*b*x + 9*a) + 27*cos(7*b*x + 7*a) - 168*cos(3*b*x + 3*a) - 378*cos(b*x + a))/b

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Fricas [A]
time = 1.91, size = 26, normalized size = 0.84 \begin {gather*} \frac {8 \, {\left (7 \, \cos \left (b x + a\right )^{9} - 9 \, \cos \left (b x + a\right )^{7}\right )}}{63 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="fricas")

[Out]

8/63*(7*cos(b*x + a)^9 - 9*cos(b*x + a)^7)/b

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (26) = 52\).
time = 6.37, size = 284, normalized size = 9.16 \begin {gather*} \begin {cases} - \frac {94 \sin ^{3}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{315 b} - \frac {32 \sin ^{3}{\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} - \frac {4 \sin ^{2}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{7 b} - \frac {64 \sin ^{2}{\left (a + b x \right )} \cos {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{105 b} + \frac {13 \sin {\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{105 b} + \frac {8 \sin {\left (a + b x \right )} \sin {\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac {46 \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos {\left (2 a + 2 b x \right )}}{105 b} - \frac {16 \cos ^{3}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{63 b} & \text {for}\: b \neq 0 \\x \sin ^{3}{\left (2 a \right )} \cos ^{3}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**3,x)

[Out]

Piecewise((-94*sin(a + b*x)**3*sin(2*a + 2*b*x)**3/(315*b) - 32*sin(a + b*x)**3*sin(2*a + 2*b*x)*cos(2*a + 2*b
*x)**2/(105*b) - 4*sin(a + b*x)**2*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)/(7*b) - 64*sin(a + b*x)**
2*cos(a + b*x)*cos(2*a + 2*b*x)**3/(105*b) + 13*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(a + b*x)**2/(105*b) + 8*s
in(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/(35*b) - 46*sin(2*a + 2*b*x)**2*cos(a + b*x)*
*3*cos(2*a + 2*b*x)/(105*b) - 16*cos(a + b*x)**3*cos(2*a + 2*b*x)**3/(63*b), Ne(b, 0)), (x*sin(2*a)**3*cos(a)*
*3, True))

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Giac [A]
time = 0.55, size = 26, normalized size = 0.84 \begin {gather*} \frac {8 \, {\left (7 \, \cos \left (b x + a\right )^{9} - 9 \, \cos \left (b x + a\right )^{7}\right )}}{63 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^3,x, algorithm="giac")

[Out]

8/63*(7*cos(b*x + a)^9 - 9*cos(b*x + a)^7)/b

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Mupad [B]
time = 0.15, size = 26, normalized size = 0.84 \begin {gather*} -\frac {8\,\left (9\,{\cos \left (a+b\,x\right )}^7-7\,{\cos \left (a+b\,x\right )}^9\right )}{63\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^3*sin(2*a + 2*b*x)^3,x)

[Out]

-(8*(9*cos(a + b*x)^7 - 7*cos(a + b*x)^9))/(63*b)

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