3.3 \(\int \sin (a+b x) \sin ^5(2 a+2 b x) \, dx\)

Optimal. Leaf size=46 \[ \frac{32 \sin ^{11}(a+b x)}{11 b}-\frac{64 \sin ^9(a+b x)}{9 b}+\frac{32 \sin ^7(a+b x)}{7 b} \]

[Out]

(32*Sin[a + b*x]^7)/(7*b) - (64*Sin[a + b*x]^9)/(9*b) + (32*Sin[a + b*x]^11)/(11*b)

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Rubi [A]  time = 0.0546081, antiderivative size = 46, 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, 2564, 270} \[ \frac{32 \sin ^{11}(a+b x)}{11 b}-\frac{64 \sin ^9(a+b x)}{9 b}+\frac{32 \sin ^7(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^5,x]

[Out]

(32*Sin[a + b*x]^7)/(7*b) - (64*Sin[a + b*x]^9)/(9*b) + (32*Sin[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 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 \sin (a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^6(a+b x) \, dx\\ &=\frac{32 \operatorname{Subst}\left (\int x^6 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{32 \operatorname{Subst}\left (\int \left (x^6-2 x^8+x^{10}\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{32 \sin ^7(a+b x)}{7 b}-\frac{64 \sin ^9(a+b x)}{9 b}+\frac{32 \sin ^{11}(a+b x)}{11 b}\\ \end{align*}

Mathematica [A]  time = 0.251399, size = 37, normalized size = 0.8 \[ \frac{4 \sin ^7(a+b x) (364 \cos (2 (a+b x))+63 \cos (4 (a+b x))+365)}{693 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*(365 + 364*Cos[2*(a + b*x)] + 63*Cos[4*(a + b*x)])*Sin[a + b*x]^7)/(693*b)

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Maple [B]  time = 0.012, size = 83, normalized size = 1.8 \begin{align*}{\frac{5\,\sin \left ( bx+a \right ) }{16\,b}}-{\frac{5\,\sin \left ( 3\,bx+3\,a \right ) }{48\,b}}-{\frac{\sin \left ( 5\,bx+5\,a \right ) }{32\,b}}+{\frac{5\,\sin \left ( 7\,bx+7\,a \right ) }{224\,b}}+{\frac{\sin \left ( 9\,bx+9\,a \right ) }{288\,b}}-{\frac{\sin \left ( 11\,bx+11\,a \right ) }{352\,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)^5,x)

[Out]

5/16*sin(b*x+a)/b-5/48*sin(3*b*x+3*a)/b-1/32/b*sin(5*b*x+5*a)+5/224/b*sin(7*b*x+7*a)+1/288/b*sin(9*b*x+9*a)-1/
352/b*sin(11*b*x+11*a)

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Maxima [A]  time = 1.21303, size = 93, normalized size = 2.02 \begin{align*} -\frac{63 \, \sin \left (11 \, b x + 11 \, a\right ) - 77 \, \sin \left (9 \, b x + 9 \, a\right ) - 495 \, \sin \left (7 \, b x + 7 \, a\right ) + 693 \, \sin \left (5 \, b x + 5 \, a\right ) + 2310 \, \sin \left (3 \, b x + 3 \, a\right ) - 6930 \, \sin \left (b x + a\right )}{22176 \, 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)^5,x, algorithm="maxima")

[Out]

-1/22176*(63*sin(11*b*x + 11*a) - 77*sin(9*b*x + 9*a) - 495*sin(7*b*x + 7*a) + 693*sin(5*b*x + 5*a) + 2310*sin
(3*b*x + 3*a) - 6930*sin(b*x + a))/b

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Fricas [A]  time = 0.504112, size = 174, normalized size = 3.78 \begin{align*} -\frac{32 \,{\left (63 \, \cos \left (b x + a\right )^{10} - 161 \, \cos \left (b x + a\right )^{8} + 113 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{693 \, 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)^5,x, algorithm="fricas")

[Out]

-32/693*(63*cos(b*x + a)^10 - 161*cos(b*x + a)^8 + 113*cos(b*x + a)^6 - 3*cos(b*x + a)^4 - 4*cos(b*x + a)^2 -
8)*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(sin(b*x+a)*sin(2*b*x+2*a)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.38042, size = 111, normalized size = 2.41 \begin{align*} -\frac{\sin \left (11 \, b x + 11 \, a\right )}{352 \, b} + \frac{\sin \left (9 \, b x + 9 \, a\right )}{288 \, b} + \frac{5 \, \sin \left (7 \, b x + 7 \, a\right )}{224 \, b} - \frac{\sin \left (5 \, b x + 5 \, a\right )}{32 \, b} - \frac{5 \, \sin \left (3 \, b x + 3 \, a\right )}{48 \, 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(sin(b*x+a)*sin(2*b*x+2*a)^5,x, algorithm="giac")

[Out]

-1/352*sin(11*b*x + 11*a)/b + 1/288*sin(9*b*x + 9*a)/b + 5/224*sin(7*b*x + 7*a)/b - 1/32*sin(5*b*x + 5*a)/b -
5/48*sin(3*b*x + 3*a)/b + 5/16*sin(b*x + a)/b