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

Optimal. Leaf size=31 \[ \frac{8 \cos ^7(a+b x)}{7 b}-\frac{8 \cos ^5(a+b x)}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.0500029, antiderivative size = 31, 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 = {4287, 2565, 14} \[ \frac{8 \cos ^7(a+b x)}{7 b}-\frac{8 \cos ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \cos (a+b x) \sin ^3(2 a+2 b x) \, dx &=8 \int \cos ^4(a+b x) \sin ^3(a+b x) \, dx\\ &=-\frac{8 \operatorname{Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{8 \operatorname{Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{8 \cos ^5(a+b x)}{5 b}+\frac{8 \cos ^7(a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.0899755, size = 27, normalized size = 0.87 \[ \frac{4 \cos ^5(a+b x) (5 \cos (2 (a+b x))-9)}{35 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[a + b*x]^5*(-9 + 5*Cos[2*(a + b*x)]))/(35*b)

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Maple [A]  time = 0.019, size = 55, normalized size = 1.8 \begin{align*} -{\frac{3\,\cos \left ( bx+a \right ) }{8\,b}}-{\frac{\cos \left ( 3\,bx+3\,a \right ) }{8\,b}}+{\frac{\cos \left ( 5\,bx+5\,a \right ) }{40\,b}}+{\frac{\cos \left ( 7\,bx+7\,a \right ) }{56\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*cos(b*x+a)/b-1/8*cos(3*b*x+3*a)/b+1/40*cos(5*b*x+5*a)/b+1/56*cos(7*b*x+7*a)/b

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Maxima [A]  time = 1.01116, size = 63, normalized size = 2.03 \begin{align*} \frac{5 \, \cos \left (7 \, b x + 7 \, a\right ) + 7 \, \cos \left (5 \, b x + 5 \, a\right ) - 35 \, \cos \left (3 \, b x + 3 \, a\right ) - 105 \, \cos \left (b x + a\right )}{280 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/280*(5*cos(7*b*x + 7*a) + 7*cos(5*b*x + 5*a) - 35*cos(3*b*x + 3*a) - 105*cos(b*x + a))/b

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Fricas [A]  time = 0.478136, size = 62, normalized size = 2. \begin{align*} \frac{8 \,{\left (5 \, \cos \left (b x + a\right )^{7} - 7 \, \cos \left (b x + a\right )^{5}\right )}}{35 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/35*(5*cos(b*x + a)^7 - 7*cos(b*x + a)^5)/b

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Sympy [A]  time = 22.7035, size = 128, normalized size = 4.13 \begin{align*} \begin{cases} - \frac{9 \sin{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{35 b} - \frac{8 \sin{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac{22 \sin ^{2}{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{35 b} - \frac{16 \cos{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{35 b} & \text{for}\: b \neq 0 \\x \sin ^{3}{\left (2 a \right )} \cos{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-9*sin(a + b*x)*sin(2*a + 2*b*x)**3/(35*b) - 8*sin(a + b*x)*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)**2/(3
5*b) - 22*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)/(35*b) - 16*cos(a + b*x)*cos(2*a + 2*b*x)**3/(35*b
), Ne(b, 0)), (x*sin(2*a)**3*cos(a), True))

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Giac [B]  time = 1.2452, size = 73, normalized size = 2.35 \begin{align*} \frac{\cos \left (7 \, b x + 7 \, a\right )}{56 \, b} + \frac{\cos \left (5 \, b x + 5 \, a\right )}{40 \, b} - \frac{\cos \left (3 \, b x + 3 \, a\right )}{8 \, b} - \frac{3 \, \cos \left (b x + a\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56*cos(7*b*x + 7*a)/b + 1/40*cos(5*b*x + 5*a)/b - 1/8*cos(3*b*x + 3*a)/b - 3/8*cos(b*x + a)/b