3.23 \(\int \sin ^3(a+b x) \sin ^5(2 a+2 b x) \, dx\)

Optimal. Leaf size=46 \[ \frac {32 \sin ^{13}(a+b x)}{13 b}-\frac {64 \sin ^{11}(a+b x)}{11 b}+\frac {32 \sin ^9(a+b x)}{9 b} \]

[Out]

32/9*sin(b*x+a)^9/b-64/11*sin(b*x+a)^11/b+32/13*sin(b*x+a)^13/b

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Rubi [A]  time = 0.06, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4288, 2564, 270} \[ \frac {32 \sin ^{13}(a+b x)}{13 b}-\frac {64 \sin ^{11}(a+b x)}{11 b}+\frac {32 \sin ^9(a+b x)}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(32*Sin[a + b*x]^9)/(9*b) - (64*Sin[a + b*x]^11)/(11*b) + (32*Sin[a + b*x]^13)/(13*b)

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]

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 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]

Rubi steps

\begin {align*} \int \sin ^3(a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^8(a+b x) \, dx\\ &=\frac {32 \operatorname {Subst}\left (\int x^8 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {32 \operatorname {Subst}\left (\int \left (x^8-2 x^{10}+x^{12}\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac {32 \sin ^9(a+b x)}{9 b}-\frac {64 \sin ^{11}(a+b x)}{11 b}+\frac {32 \sin ^{13}(a+b x)}{13 b}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 37, normalized size = 0.80 \[ \frac {4 \sin ^9(a+b x) (540 \cos (2 (a+b x))+99 \cos (4 (a+b x))+505)}{1287 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*(505 + 540*Cos[2*(a + b*x)] + 99*Cos[4*(a + b*x)])*Sin[a + b*x]^9)/(1287*b)

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fricas [A]  time = 0.42, size = 73, normalized size = 1.59 \[ \frac {32 \, {\left (99 \, \cos \left (b x + a\right )^{12} - 360 \, \cos \left (b x + a\right )^{10} + 458 \, \cos \left (b x + a\right )^{8} - 212 \, \cos \left (b x + a\right )^{6} + 3 \, \cos \left (b x + a\right )^{4} + 4 \, \cos \left (b x + a\right )^{2} + 8\right )} \sin \left (b x + a\right )}{1287 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="fricas")

[Out]

32/1287*(99*cos(b*x + a)^12 - 360*cos(b*x + a)^10 + 458*cos(b*x + a)^8 - 212*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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giac [B]  time = 0.62, size = 96, normalized size = 2.09 \[ \frac {\sin \left (13 \, b x + 13 \, a\right )}{1664 \, b} - \frac {3 \, \sin \left (11 \, b x + 11 \, a\right )}{1408 \, b} - \frac {\sin \left (9 \, b x + 9 \, a\right )}{576 \, b} + \frac {\sin \left (7 \, b x + 7 \, a\right )}{64 \, b} - \frac {\sin \left (5 \, b x + 5 \, a\right )}{128 \, b} - \frac {25 \, \sin \left (3 \, b x + 3 \, a\right )}{384 \, b} + \frac {5 \, \sin \left (b x + a\right )}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="giac")

[Out]

1/1664*sin(13*b*x + 13*a)/b - 3/1408*sin(11*b*x + 11*a)/b - 1/576*sin(9*b*x + 9*a)/b + 1/64*sin(7*b*x + 7*a)/b
- 1/128*sin(5*b*x + 5*a)/b - 25/384*sin(3*b*x + 3*a)/b + 5/32*sin(b*x + a)/b

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maple [B]  time = 1.48, size = 97, normalized size = 2.11 \[ \frac {5 \sin \left (b x +a \right )}{32 b}-\frac {25 \sin \left (3 b x +3 a \right )}{384 b}-\frac {\sin \left (5 b x +5 a \right )}{128 b}+\frac {\sin \left (7 b x +7 a \right )}{64 b}-\frac {\sin \left (9 b x +9 a \right )}{576 b}-\frac {3 \sin \left (11 b x +11 a \right )}{1408 b}+\frac {\sin \left (13 b x +13 a \right )}{1664 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x)

[Out]

5/32*sin(b*x+a)/b-25/384*sin(3*b*x+3*a)/b-1/128/b*sin(5*b*x+5*a)+1/64/b*sin(7*b*x+7*a)-1/576/b*sin(9*b*x+9*a)-
3/1408/b*sin(11*b*x+11*a)+1/1664/b*sin(13*b*x+13*a)

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maxima [A]  time = 0.35, size = 80, normalized size = 1.74 \[ \frac {99 \, \sin \left (13 \, b x + 13 \, a\right ) - 351 \, \sin \left (11 \, b x + 11 \, a\right ) - 286 \, \sin \left (9 \, b x + 9 \, a\right ) + 2574 \, \sin \left (7 \, b x + 7 \, a\right ) - 1287 \, \sin \left (5 \, b x + 5 \, a\right ) - 10725 \, \sin \left (3 \, b x + 3 \, a\right ) + 25740 \, \sin \left (b x + a\right )}{164736 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^3*sin(2*b*x+2*a)^5,x, algorithm="maxima")

[Out]

1/164736*(99*sin(13*b*x + 13*a) - 351*sin(11*b*x + 11*a) - 286*sin(9*b*x + 9*a) + 2574*sin(7*b*x + 7*a) - 1287
*sin(5*b*x + 5*a) - 10725*sin(3*b*x + 3*a) + 25740*sin(b*x + a))/b

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mupad [B]  time = 0.07, size = 36, normalized size = 0.78 \[ \frac {32\,\left (99\,{\sin \left (a+b\,x\right )}^{13}-234\,{\sin \left (a+b\,x\right )}^{11}+143\,{\sin \left (a+b\,x\right )}^9\right )}{1287\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^3*sin(2*a + 2*b*x)^5,x)

[Out]

(32*(143*sin(a + b*x)^9 - 234*sin(a + b*x)^11 + 99*sin(a + b*x)^13))/(1287*b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**3*sin(2*b*x+2*a)**5,x)

[Out]

Timed out

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