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

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

[Out]

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

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Rubi [A]  time = 0.0580589, 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 = {4288, 2565, 270} \[ \frac{64 \cos ^{13}(a+b x)}{13 b}-\frac{192 \cos ^{11}(a+b x)}{11 b}+\frac{64 \cos ^9(a+b x)}{3 b}-\frac{64 \cos ^7(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4288

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

Rule 2565

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

Mathematica [A]  time = 0.325048, size = 47, normalized size = 0.77 \[ \frac{2 \cos ^7(a+b x) (6377 \cos (2 (a+b x))-1890 \cos (4 (a+b x))+231 \cos (6 (a+b x))-5230)}{3003 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.023, size = 97, normalized size = 1.6 \begin{align*} -{\frac{5\,\cos \left ( bx+a \right ) }{16\,b}}-{\frac{5\,\cos \left ( 3\,bx+3\,a \right ) }{64\,b}}+{\frac{3\,\cos \left ( 5\,bx+5\,a \right ) }{64\,b}}+{\frac{3\,\cos \left ( 7\,bx+7\,a \right ) }{224\,b}}-{\frac{\cos \left ( 9\,bx+9\,a \right ) }{96\,b}}-{\frac{\cos \left ( 11\,bx+11\,a \right ) }{704\,b}}+{\frac{\cos \left ( 13\,bx+13\,a \right ) }{832\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.17533, size = 108, normalized size = 1.77 \begin{align*} \frac{231 \, \cos \left (13 \, b x + 13 \, a\right ) - 273 \, \cos \left (11 \, b x + 11 \, a\right ) - 2002 \, \cos \left (9 \, b x + 9 \, a\right ) + 2574 \, \cos \left (7 \, b x + 7 \, a\right ) + 9009 \, \cos \left (5 \, b x + 5 \, a\right ) - 15015 \, \cos \left (3 \, b x + 3 \, a\right ) - 60060 \, \cos \left (b x + a\right )}{192192 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192192*(231*cos(13*b*x + 13*a) - 273*cos(11*b*x + 11*a) - 2002*cos(9*b*x + 9*a) + 2574*cos(7*b*x + 7*a) + 90
09*cos(5*b*x + 5*a) - 15015*cos(3*b*x + 3*a) - 60060*cos(b*x + a))/b

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Fricas [A]  time = 0.522097, size = 132, normalized size = 2.16 \begin{align*} \frac{64 \,{\left (231 \, \cos \left (b x + a\right )^{13} - 819 \, \cos \left (b x + a\right )^{11} + 1001 \, \cos \left (b x + a\right )^{9} - 429 \, \cos \left (b x + a\right )^{7}\right )}}{3003 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

64/3003*(231*cos(b*x + a)^13 - 819*cos(b*x + a)^11 + 1001*cos(b*x + a)^9 - 429*cos(b*x + a)^7)/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(sin(b*x+a)*sin(2*b*x+2*a)**6,x)

[Out]

Timed out

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Giac [A]  time = 1.43786, size = 130, normalized size = 2.13 \begin{align*} \frac{\cos \left (13 \, b x + 13 \, a\right )}{832 \, b} - \frac{\cos \left (11 \, b x + 11 \, a\right )}{704 \, b} - \frac{\cos \left (9 \, b x + 9 \, a\right )}{96 \, b} + \frac{3 \, \cos \left (7 \, b x + 7 \, a\right )}{224 \, b} + \frac{3 \, \cos \left (5 \, b x + 5 \, a\right )}{64 \, b} - \frac{5 \, \cos \left (3 \, b x + 3 \, a\right )}{64 \, b} - \frac{5 \, \cos \left (b x + a\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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