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

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

[Out]

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

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Rubi [A]  time = 0.0685345, 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 = {4288, 2565, 270} \[ \frac{16 \cos ^{11}(a+b x)}{11 b}-\frac{16 \cos ^9(a+b x)}{3 b}+\frac{48 \cos ^7(a+b x)}{7 b}-\frac{16 \cos ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.225861, size = 47, normalized size = 0.77 \[ \frac{\cos ^5(a+b x) (3335 \cos (2 (a+b x))-910 \cos (4 (a+b x))+105 \cos (6 (a+b x))-3042)}{2310 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.011, size = 83, normalized size = 1.4 \begin{align*} -{\frac{7\,\cos \left ( bx+a \right ) }{32\,b}}-{\frac{\cos \left ( 3\,bx+3\,a \right ) }{32\,b}}+{\frac{11\,\cos \left ( 5\,bx+5\,a \right ) }{320\,b}}-{\frac{\cos \left ( 7\,bx+7\,a \right ) }{448\,b}}-{\frac{\cos \left ( 9\,bx+9\,a \right ) }{192\,b}}+{\frac{\cos \left ( 11\,bx+11\,a \right ) }{704\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.21816, size = 93, normalized size = 1.52 \begin{align*} \frac{105 \, \cos \left (11 \, b x + 11 \, a\right ) - 385 \, \cos \left (9 \, b x + 9 \, a\right ) - 165 \, \cos \left (7 \, b x + 7 \, a\right ) + 2541 \, \cos \left (5 \, b x + 5 \, a\right ) - 2310 \, \cos \left (3 \, b x + 3 \, a\right ) - 16170 \, \cos \left (b x + a\right )}{73920 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/73920*(105*cos(11*b*x + 11*a) - 385*cos(9*b*x + 9*a) - 165*cos(7*b*x + 7*a) + 2541*cos(5*b*x + 5*a) - 2310*c
os(3*b*x + 3*a) - 16170*cos(b*x + a))/b

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Fricas [A]  time = 0.503269, size = 130, normalized size = 2.13 \begin{align*} \frac{16 \,{\left (105 \, \cos \left (b x + a\right )^{11} - 385 \, \cos \left (b x + a\right )^{9} + 495 \, \cos \left (b x + a\right )^{7} - 231 \, \cos \left (b x + a\right )^{5}\right )}}{1155 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/1155*(105*cos(b*x + a)^11 - 385*cos(b*x + a)^9 + 495*cos(b*x + a)^7 - 231*cos(b*x + a)^5)/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)**3*sin(2*b*x+2*a)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.42921, size = 111, normalized size = 1.82 \begin{align*} \frac{\cos \left (11 \, b x + 11 \, a\right )}{704 \, b} - \frac{\cos \left (9 \, b x + 9 \, a\right )}{192 \, b} - \frac{\cos \left (7 \, b x + 7 \, a\right )}{448 \, b} + \frac{11 \, \cos \left (5 \, b x + 5 \, a\right )}{320 \, b} - \frac{\cos \left (3 \, b x + 3 \, a\right )}{32 \, b} - \frac{7 \, \cos \left (b x + a\right )}{32 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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