3.152 \(\int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx\)

Optimal. Leaf size=61 \[ -\frac{16 \sin ^{11}(a+b x)}{11 b}+\frac{16 \sin ^9(a+b x)}{3 b}-\frac{48 \sin ^7(a+b x)}{7 b}+\frac{16 \sin ^5(a+b x)}{5 b} \]

[Out]

(16*Sin[a + b*x]^5)/(5*b) - (48*Sin[a + b*x]^7)/(7*b) + (16*Sin[a + b*x]^9)/(3*b) - (16*Sin[a + b*x]^11)/(11*b
)

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Rubi [A]  time = 0.064816, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4287, 2564, 270} \[ -\frac{16 \sin ^{11}(a+b x)}{11 b}+\frac{16 \sin ^9(a+b x)}{3 b}-\frac{48 \sin ^7(a+b x)}{7 b}+\frac{16 \sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

(16*Sin[a + b*x]^5)/(5*b) - (48*Sin[a + b*x]^7)/(7*b) + (16*Sin[a + b*x]^9)/(3*b) - (16*Sin[a + b*x]^11)/(11*b
)

Rule 4287

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]

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \cos ^3(a+b x) \sin ^4(2 a+2 b x) \, dx &=16 \int \cos ^7(a+b x) \sin ^4(a+b x) \, dx\\ &=\frac{16 \operatorname{Subst}\left (\int x^4 \left (1-x^2\right )^3 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{16 \operatorname{Subst}\left (\int \left (x^4-3 x^6+3 x^8-x^{10}\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{16 \sin ^5(a+b x)}{5 b}-\frac{48 \sin ^7(a+b x)}{7 b}+\frac{16 \sin ^9(a+b x)}{3 b}-\frac{16 \sin ^{11}(a+b x)}{11 b}\\ \end{align*}

Mathematica [A]  time = 0.212373, size = 47, normalized size = 0.77 \[ \frac{\sin ^5(a+b x) (3335 \cos (2 (a+b x))+910 \cos (4 (a+b x))+105 \cos (6 (a+b x))+3042)}{2310 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

((3042 + 3335*Cos[2*(a + b*x)] + 910*Cos[4*(a + b*x)] + 105*Cos[6*(a + b*x)])*Sin[a + b*x]^5)/(2310*b)

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Maple [A]  time = 0.03, size = 83, normalized size = 1.4 \begin{align*}{\frac{7\,\sin \left ( bx+a \right ) }{32\,b}}-{\frac{\sin \left ( 3\,bx+3\,a \right ) }{32\,b}}-{\frac{11\,\sin \left ( 5\,bx+5\,a \right ) }{320\,b}}-{\frac{\sin \left ( 7\,bx+7\,a \right ) }{448\,b}}+{\frac{\sin \left ( 9\,bx+9\,a \right ) }{192\,b}}+{\frac{\sin \left ( 11\,bx+11\,a \right ) }{704\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x)

[Out]

7/32*sin(b*x+a)/b-1/32*sin(3*b*x+3*a)/b-11/320/b*sin(5*b*x+5*a)-1/448/b*sin(7*b*x+7*a)+1/192/b*sin(9*b*x+9*a)+
1/704/b*sin(11*b*x+11*a)

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Maxima [A]  time = 1.16585, size = 93, normalized size = 1.52 \begin{align*} \frac{105 \, \sin \left (11 \, b x + 11 \, a\right ) + 385 \, \sin \left (9 \, b x + 9 \, a\right ) - 165 \, \sin \left (7 \, b x + 7 \, a\right ) - 2541 \, \sin \left (5 \, b x + 5 \, a\right ) - 2310 \, \sin \left (3 \, b x + 3 \, a\right ) + 16170 \, \sin \left (b x + a\right )}{73920 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

1/73920*(105*sin(11*b*x + 11*a) + 385*sin(9*b*x + 9*a) - 165*sin(7*b*x + 7*a) - 2541*sin(5*b*x + 5*a) - 2310*s
in(3*b*x + 3*a) + 16170*sin(b*x + a))/b

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Fricas [A]  time = 0.510513, size = 174, normalized size = 2.85 \begin{align*} \frac{16 \,{\left (105 \, \cos \left (b x + a\right )^{10} - 140 \, \cos \left (b x + a\right )^{8} + 5 \, \cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4} + 8 \, \cos \left (b x + a\right )^{2} + 16\right )} \sin \left (b x + a\right )}{1155 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

16/1155*(105*cos(b*x + a)^10 - 140*cos(b*x + a)^8 + 5*cos(b*x + a)^6 + 6*cos(b*x + a)^4 + 8*cos(b*x + a)^2 + 1
6)*sin(b*x + a)/b

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**3*sin(2*b*x+2*a)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.21293, size = 111, normalized size = 1.82 \begin{align*} \frac{\sin \left (11 \, b x + 11 \, a\right )}{704 \, b} + \frac{\sin \left (9 \, b x + 9 \, a\right )}{192 \, b} - \frac{\sin \left (7 \, b x + 7 \, a\right )}{448 \, b} - \frac{11 \, \sin \left (5 \, b x + 5 \, a\right )}{320 \, b} - \frac{\sin \left (3 \, b x + 3 \, a\right )}{32 \, b} + \frac{7 \, \sin \left (b x + a\right )}{32 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

1/704*sin(11*b*x + 11*a)/b + 1/192*sin(9*b*x + 9*a)/b - 1/448*sin(7*b*x + 7*a)/b - 11/320*sin(5*b*x + 5*a)/b -
1/32*sin(3*b*x + 3*a)/b + 7/32*sin(b*x + a)/b