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

Optimal. Leaf size=46 \[ -\frac{4 \cos ^9(a+b x)}{9 b}+\frac{8 \cos ^7(a+b x)}{7 b}-\frac{4 \cos ^5(a+b x)}{5 b} \]

[Out]

(-4*Cos[a + b*x]^5)/(5*b) + (8*Cos[a + b*x]^7)/(7*b) - (4*Cos[a + b*x]^9)/(9*b)

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Rubi [A]  time = 0.0966612, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {4312, 2565, 270} \[ -\frac{4 \cos ^9(a+b x)}{9 b}+\frac{8 \cos ^7(a+b x)}{7 b}-\frac{4 \cos ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[a + b*x]^5)/(5*b) + (8*Cos[a + b*x]^7)/(7*b) - (4*Cos[a + b*x]^9)/(9*b)

Rule 4312

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.),
x_Symbol] :> Dist[2^p/(e^p*f^p), Int[(e*Cos[a + b*x])^(m + p)*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b
, c, d, e, f, m, 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 \cos ^2(a+b x) \sin ^3(a+b x) \sin ^2(2 a+2 b x) \, dx &=4 \int \cos ^4(a+b x) \sin ^5(a+b x) \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{4 \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{4 \cos ^5(a+b x)}{5 b}+\frac{8 \cos ^7(a+b x)}{7 b}-\frac{4 \cos ^9(a+b x)}{9 b}\\ \end{align*}

Mathematica [A]  time = 0.163096, size = 37, normalized size = 0.8 \[ \frac{\cos ^5(a+b x) (220 \cos (2 (a+b x))-35 \cos (4 (a+b x))-249)}{630 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[a + b*x]^5*(-249 + 220*Cos[2*(a + b*x)] - 35*Cos[4*(a + b*x)]))/(630*b)

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Maple [A]  time = 0.023, size = 69, normalized size = 1.5 \begin{align*} -{\frac{3\,\cos \left ( bx+a \right ) }{32\,b}}-{\frac{\cos \left ( 3\,bx+3\,a \right ) }{48\,b}}+{\frac{\cos \left ( 5\,bx+5\,a \right ) }{80\,b}}+{\frac{\cos \left ( 7\,bx+7\,a \right ) }{448\,b}}-{\frac{\cos \left ( 9\,bx+9\,a \right ) }{576\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/32*cos(b*x+a)/b-1/48*cos(3*b*x+3*a)/b+1/80*cos(5*b*x+5*a)/b+1/448*cos(7*b*x+7*a)/b-1/576*cos(9*b*x+9*a)/b

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Maxima [A]  time = 1.2228, size = 78, normalized size = 1.7 \begin{align*} -\frac{35 \, \cos \left (9 \, b x + 9 \, a\right ) - 45 \, \cos \left (7 \, b x + 7 \, a\right ) - 252 \, \cos \left (5 \, b x + 5 \, a\right ) + 420 \, \cos \left (3 \, b x + 3 \, a\right ) + 1890 \, \cos \left (b x + a\right )}{20160 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20160*(35*cos(9*b*x + 9*a) - 45*cos(7*b*x + 7*a) - 252*cos(5*b*x + 5*a) + 420*cos(3*b*x + 3*a) + 1890*cos(b
*x + a))/b

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Fricas [A]  time = 0.490691, size = 95, normalized size = 2.07 \begin{align*} -\frac{4 \,{\left (35 \, \cos \left (b x + a\right )^{9} - 90 \, \cos \left (b x + a\right )^{7} + 63 \, \cos \left (b x + a\right )^{5}\right )}}{315 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/315*(35*cos(b*x + a)^9 - 90*cos(b*x + a)^7 + 63*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(cos(b*x+a)**2*sin(b*x+a)**3*sin(2*b*x+2*a)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.85772, size = 92, normalized size = 2. \begin{align*} -\frac{\cos \left (9 \, b x + 9 \, a\right )}{576 \, b} + \frac{\cos \left (7 \, b x + 7 \, a\right )}{448 \, b} + \frac{\cos \left (5 \, b x + 5 \, a\right )}{80 \, b} - \frac{\cos \left (3 \, b x + 3 \, a\right )}{48 \, b} - \frac{3 \, \cos \left (b x + a\right )}{32 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/576*cos(9*b*x + 9*a)/b + 1/448*cos(7*b*x + 7*a)/b + 1/80*cos(5*b*x + 5*a)/b - 1/48*cos(3*b*x + 3*a)/b - 3/3
2*cos(b*x + a)/b