3.130 \(\int \cos (a+b x) \sin ^6(2 a+2 b x) \, dx\)

Optimal. Leaf size=61 \[ -\frac{64 \sin ^{13}(a+b x)}{13 b}+\frac{192 \sin ^{11}(a+b x)}{11 b}-\frac{64 \sin ^9(a+b x)}{3 b}+\frac{64 \sin ^7(a+b x)}{7 b} \]

[Out]

(64*Sin[a + b*x]^7)/(7*b) - (64*Sin[a + b*x]^9)/(3*b) + (192*Sin[a + b*x]^11)/(11*b) - (64*Sin[a + b*x]^13)/(1
3*b)

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Rubi [A]  time = 0.0608379, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4287, 2564, 270} \[ -\frac{64 \sin ^{13}(a+b x)}{13 b}+\frac{192 \sin ^{11}(a+b x)}{11 b}-\frac{64 \sin ^9(a+b x)}{3 b}+\frac{64 \sin ^7(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^6,x]

[Out]

(64*Sin[a + b*x]^7)/(7*b) - (64*Sin[a + b*x]^9)/(3*b) + (192*Sin[a + b*x]^11)/(11*b) - (64*Sin[a + b*x]^13)/(1
3*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 (a+b x) \sin ^6(2 a+2 b x) \, dx &=64 \int \cos ^7(a+b x) \sin ^6(a+b x) \, dx\\ &=\frac{64 \operatorname{Subst}\left (\int x^6 \left (1-x^2\right )^3 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{64 \operatorname{Subst}\left (\int \left (x^6-3 x^8+3 x^{10}-x^{12}\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{64 \sin ^7(a+b x)}{7 b}-\frac{64 \sin ^9(a+b x)}{3 b}+\frac{192 \sin ^{11}(a+b x)}{11 b}-\frac{64 \sin ^{13}(a+b x)}{13 b}\\ \end{align*}

Mathematica [A]  time = 0.315134, size = 47, normalized size = 0.77 \[ \frac{2 \sin ^7(a+b x) (6377 \cos (2 (a+b x))+1890 \cos (4 (a+b x))+231 \cos (6 (a+b x))+5230)}{3003 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^6,x]

[Out]

(2*(5230 + 6377*Cos[2*(a + b*x)] + 1890*Cos[4*(a + b*x)] + 231*Cos[6*(a + b*x)])*Sin[a + b*x]^7)/(3003*b)

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Maple [A]  time = 0.036, size = 97, normalized size = 1.6 \begin{align*}{\frac{5\,\sin \left ( bx+a \right ) }{16\,b}}-{\frac{5\,\sin \left ( 3\,bx+3\,a \right ) }{64\,b}}-{\frac{3\,\sin \left ( 5\,bx+5\,a \right ) }{64\,b}}+{\frac{3\,\sin \left ( 7\,bx+7\,a \right ) }{224\,b}}+{\frac{\sin \left ( 9\,bx+9\,a \right ) }{96\,b}}-{\frac{\sin \left ( 11\,bx+11\,a \right ) }{704\,b}}-{\frac{\sin \left ( 13\,bx+13\,a \right ) }{832\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)*sin(2*b*x+2*a)^6,x)

[Out]

5/16*sin(b*x+a)/b-5/64*sin(3*b*x+3*a)/b-3/64/b*sin(5*b*x+5*a)+3/224/b*sin(7*b*x+7*a)+1/96/b*sin(9*b*x+9*a)-1/7
04/b*sin(11*b*x+11*a)-1/832/b*sin(13*b*x+13*a)

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Maxima [A]  time = 1.1206, size = 108, normalized size = 1.77 \begin{align*} -\frac{231 \, \sin \left (13 \, b x + 13 \, a\right ) + 273 \, \sin \left (11 \, b x + 11 \, a\right ) - 2002 \, \sin \left (9 \, b x + 9 \, a\right ) - 2574 \, \sin \left (7 \, b x + 7 \, a\right ) + 9009 \, \sin \left (5 \, b x + 5 \, a\right ) + 15015 \, \sin \left (3 \, b x + 3 \, a\right ) - 60060 \, \sin \left (b x + a\right )}{192192 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="maxima")

[Out]

-1/192192*(231*sin(13*b*x + 13*a) + 273*sin(11*b*x + 11*a) - 2002*sin(9*b*x + 9*a) - 2574*sin(7*b*x + 7*a) + 9
009*sin(5*b*x + 5*a) + 15015*sin(3*b*x + 3*a) - 60060*sin(b*x + a))/b

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Fricas [A]  time = 0.519273, size = 205, normalized size = 3.36 \begin{align*} -\frac{64 \,{\left (231 \, \cos \left (b x + a\right )^{12} - 567 \, \cos \left (b x + a\right )^{10} + 371 \, \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 )}{3003 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="fricas")

[Out]

-64/3003*(231*cos(b*x + a)^12 - 567*cos(b*x + a)^10 + 371*cos(b*x + a)^8 - 5*cos(b*x + a)^6 - 6*cos(b*x + a)^4
- 8*cos(b*x + a)^2 - 16)*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)*sin(2*b*x+2*a)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.52838, size = 130, normalized size = 2.13 \begin{align*} -\frac{\sin \left (13 \, b x + 13 \, a\right )}{832 \, b} - \frac{\sin \left (11 \, b x + 11 \, a\right )}{704 \, b} + \frac{\sin \left (9 \, b x + 9 \, a\right )}{96 \, b} + \frac{3 \, \sin \left (7 \, b x + 7 \, a\right )}{224 \, b} - \frac{3 \, \sin \left (5 \, b x + 5 \, a\right )}{64 \, b} - \frac{5 \, \sin \left (3 \, b x + 3 \, a\right )}{64 \, b} + \frac{5 \, \sin \left (b x + a\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)*sin(2*b*x+2*a)^6,x, algorithm="giac")

[Out]

-1/832*sin(13*b*x + 13*a)/b - 1/704*sin(11*b*x + 11*a)/b + 1/96*sin(9*b*x + 9*a)/b + 3/224*sin(7*b*x + 7*a)/b
- 3/64*sin(5*b*x + 5*a)/b - 5/64*sin(3*b*x + 3*a)/b + 5/16*sin(b*x + a)/b