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

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

[Out]

-32/9*cos(b*x+a)^9/b+64/11*cos(b*x+a)^11/b-32/13*cos(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 = {4287, 2565, 270} \[ -\frac {32 \cos ^{13}(a+b x)}{13 b}+\frac {64 \cos ^{11}(a+b x)}{11 b}-\frac {32 \cos ^9(a+b x)}{9 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 0.39, size = 37, normalized size = 0.80 \[ \frac {4 \cos ^9(a+b x) (540 \cos (2 (a+b x))-99 \cos (4 (a+b x))-505)}{1287 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 36, normalized size = 0.78 \[ -\frac {32 \, {\left (99 \, \cos \left (b x + a\right )^{13} - 234 \, \cos \left (b x + a\right )^{11} + 143 \, \cos \left (b x + a\right )^{9}\right )}}{1287 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.60, size = 96, normalized size = 2.09 \[ -\frac {\cos \left (13 \, b x + 13 \, a\right )}{1664 \, b} - \frac {3 \, \cos \left (11 \, b x + 11 \, a\right )}{1408 \, b} + \frac {\cos \left (9 \, b x + 9 \, a\right )}{576 \, b} + \frac {\cos \left (7 \, b x + 7 \, a\right )}{64 \, b} + \frac {\cos \left (5 \, b x + 5 \, a\right )}{128 \, b} - \frac {25 \, \cos \left (3 \, b x + 3 \, a\right )}{384 \, b} - \frac {5 \, \cos \left (b x + a\right )}{32 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.40, size = 97, normalized size = 2.11 \[ -\frac {5 \cos \left (b x +a \right )}{32 b}-\frac {25 \cos \left (3 b x +3 a \right )}{384 b}+\frac {\cos \left (5 b x +5 a \right )}{128 b}+\frac {\cos \left (7 b x +7 a \right )}{64 b}+\frac {\cos \left (9 b x +9 a \right )}{576 b}-\frac {3 \cos \left (11 b x +11 a \right )}{1408 b}-\frac {\cos \left (13 b x +13 a \right )}{1664 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 80, normalized size = 1.74 \[ -\frac {99 \, \cos \left (13 \, b x + 13 \, a\right ) + 351 \, \cos \left (11 \, b x + 11 \, a\right ) - 286 \, \cos \left (9 \, b x + 9 \, a\right ) - 2574 \, \cos \left (7 \, b x + 7 \, a\right ) - 1287 \, \cos \left (5 \, b x + 5 \, a\right ) + 10725 \, \cos \left (3 \, b x + 3 \, a\right ) + 25740 \, \cos \left (b x + a\right )}{164736 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/164736*(99*cos(13*b*x + 13*a) + 351*cos(11*b*x + 11*a) - 286*cos(9*b*x + 9*a) - 2574*cos(7*b*x + 7*a) - 128
7*cos(5*b*x + 5*a) + 10725*cos(3*b*x + 3*a) + 25740*cos(b*x + a))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(32*(143*cos(a + b*x)^9 - 234*cos(a + b*x)^11 + 99*cos(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(cos(b*x+a)**3*sin(2*b*x+2*a)**5,x)

[Out]

Timed out

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