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

Optimal. Leaf size=31 \[ \frac{4 \sin ^3(a+b x)}{3 b}-\frac{4 \sin ^5(a+b x)}{5 b} \]

[Out]

(4*Sin[a + b*x]^3)/(3*b) - (4*Sin[a + b*x]^5)/(5*b)

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Rubi [A]  time = 0.0483909, 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, 2564, 14} \[ \frac{4 \sin ^3(a+b x)}{3 b}-\frac{4 \sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sin[a + b*x]^3)/(3*b) - (4*Sin[a + b*x]^5)/(5*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 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 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 ^2(2 a+2 b x) \, dx &=4 \int \cos ^3(a+b x) \sin ^2(a+b x) \, dx\\ &=\frac{4 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{4 \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{4 \sin ^3(a+b x)}{3 b}-\frac{4 \sin ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0630134, size = 27, normalized size = 0.87 \[ \frac{2 \sin ^3(a+b x) (3 \cos (2 (a+b x))+7)}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(7 + 3*Cos[2*(a + b*x)])*Sin[a + b*x]^3)/(15*b)

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Maple [A]  time = 0.022, size = 41, normalized size = 1.3 \begin{align*}{\frac{\sin \left ( bx+a \right ) }{2\,b}}-{\frac{\sin \left ( 3\,bx+3\,a \right ) }{12\,b}}-{\frac{\sin \left ( 5\,bx+5\,a \right ) }{20\,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)^2,x)

[Out]

1/2*sin(b*x+a)/b-1/12*sin(3*b*x+3*a)/b-1/20/b*sin(5*b*x+5*a)

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Maxima [A]  time = 1.03342, size = 49, normalized size = 1.58 \begin{align*} -\frac{3 \, \sin \left (5 \, b x + 5 \, a\right ) + 5 \, \sin \left (3 \, b x + 3 \, a\right ) - 30 \, \sin \left (b x + a\right )}{60 \, 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)^2,x, algorithm="maxima")

[Out]

-1/60*(3*sin(5*b*x + 5*a) + 5*sin(3*b*x + 3*a) - 30*sin(b*x + a))/b

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Fricas [A]  time = 0.478375, size = 84, normalized size = 2.71 \begin{align*} -\frac{4 \,{\left (3 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{15 \, 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)^2,x, algorithm="fricas")

[Out]

-4/15*(3*cos(b*x + a)^4 - cos(b*x + a)^2 - 2)*sin(b*x + a)/b

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Sympy [A]  time = 6.3123, size = 90, normalized size = 2.9 \begin{align*} \begin{cases} \frac{7 \sin{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{15 b} + \frac{8 \sin{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{15 b} - \frac{4 \sin{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\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)**2,x)

[Out]

Piecewise((7*sin(a + b*x)*sin(2*a + 2*b*x)**2/(15*b) + 8*sin(a + b*x)*cos(2*a + 2*b*x)**2/(15*b) - 4*sin(2*a +
2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)/(15*b), Ne(b, 0)), (x*sin(2*a)**2*cos(a), True))

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Giac [A]  time = 1.25192, size = 54, normalized size = 1.74 \begin{align*} -\frac{\sin \left (5 \, b x + 5 \, a\right )}{20 \, b} - \frac{\sin \left (3 \, b x + 3 \, a\right )}{12 \, b} + \frac{\sin \left (b x + a\right )}{2 \, 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)^2,x, algorithm="giac")

[Out]

-1/20*sin(5*b*x + 5*a)/b - 1/12*sin(3*b*x + 3*a)/b + 1/2*sin(b*x + a)/b