3.132 \(\int \cos (a+b x) \sin ^4(2 a+2 b x) \, dx\)

Optimal. Leaf size=46 \[ \frac{16 \sin ^9(a+b x)}{9 b}-\frac{32 \sin ^7(a+b x)}{7 b}+\frac{16 \sin ^5(a+b x)}{5 b} \]

[Out]

(16*Sin[a + b*x]^5)/(5*b) - (32*Sin[a + b*x]^7)/(7*b) + (16*Sin[a + b*x]^9)/(9*b)

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Rubi [A]  time = 0.0537951, antiderivative size = 46, 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, 270} \[ \frac{16 \sin ^9(a+b x)}{9 b}-\frac{32 \sin ^7(a+b x)}{7 b}+\frac{16 \sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(16*Sin[a + b*x]^5)/(5*b) - (32*Sin[a + b*x]^7)/(7*b) + (16*Sin[a + b*x]^9)/(9*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 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 (a+b x) \sin ^4(2 a+2 b x) \, dx &=16 \int \cos ^5(a+b x) \sin ^4(a+b x) \, dx\\ &=\frac{16 \operatorname{Subst}\left (\int x^4 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{16 \operatorname{Subst}\left (\int \left (x^4-2 x^6+x^8\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{16 \sin ^5(a+b x)}{5 b}-\frac{32 \sin ^7(a+b x)}{7 b}+\frac{16 \sin ^9(a+b x)}{9 b}\\ \end{align*}

Mathematica [A]  time = 0.131488, size = 37, normalized size = 0.8 \[ \frac{2 \sin ^5(a+b x) (220 \cos (2 (a+b x))+35 \cos (4 (a+b x))+249)}{315 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]*Sin[2*a + 2*b*x]^4,x]

[Out]

(2*(249 + 220*Cos[2*(a + b*x)] + 35*Cos[4*(a + b*x)])*Sin[a + b*x]^5)/(315*b)

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Maple [A]  time = 0.027, size = 69, normalized size = 1.5 \begin{align*}{\frac{3\,\sin \left ( bx+a \right ) }{8\,b}}-{\frac{\sin \left ( 3\,bx+3\,a \right ) }{12\,b}}-{\frac{\sin \left ( 5\,bx+5\,a \right ) }{20\,b}}+{\frac{\sin \left ( 7\,bx+7\,a \right ) }{112\,b}}+{\frac{\sin \left ( 9\,bx+9\,a \right ) }{144\,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)^4,x)

[Out]

3/8*sin(b*x+a)/b-1/12*sin(3*b*x+3*a)/b-1/20/b*sin(5*b*x+5*a)+1/112/b*sin(7*b*x+7*a)+1/144/b*sin(9*b*x+9*a)

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Maxima [A]  time = 1.05242, size = 78, normalized size = 1.7 \begin{align*} \frac{35 \, \sin \left (9 \, b x + 9 \, a\right ) + 45 \, \sin \left (7 \, b x + 7 \, a\right ) - 252 \, \sin \left (5 \, b x + 5 \, a\right ) - 420 \, \sin \left (3 \, b x + 3 \, a\right ) + 1890 \, \sin \left (b x + a\right )}{5040 \, 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)^4,x, algorithm="maxima")

[Out]

1/5040*(35*sin(9*b*x + 9*a) + 45*sin(7*b*x + 7*a) - 252*sin(5*b*x + 5*a) - 420*sin(3*b*x + 3*a) + 1890*sin(b*x
+ a))/b

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Fricas [A]  time = 0.504224, size = 142, normalized size = 3.09 \begin{align*} \frac{16 \,{\left (35 \, \cos \left (b x + a\right )^{8} - 50 \, \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 )}{315 \, 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)^4,x, algorithm="fricas")

[Out]

16/315*(35*cos(b*x + a)^8 - 50*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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Sympy [A]  time = 76.0474, size = 162, normalized size = 3.52 \begin{align*} \begin{cases} \frac{107 \sin{\left (a + b x \right )} \sin ^{4}{\left (2 a + 2 b x \right )}}{315 b} + \frac{16 \sin{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{21 b} + \frac{128 \sin{\left (a + b x \right )} \cos ^{4}{\left (2 a + 2 b x \right )}}{315 b} - \frac{104 \sin ^{3}{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{315 b} - \frac{64 \sin{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{315 b} & \text{for}\: b \neq 0 \\x \sin ^{4}{\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)**4,x)

[Out]

Piecewise((107*sin(a + b*x)*sin(2*a + 2*b*x)**4/(315*b) + 16*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(2*a + 2*b*x)
**2/(21*b) + 128*sin(a + b*x)*cos(2*a + 2*b*x)**4/(315*b) - 104*sin(2*a + 2*b*x)**3*cos(a + b*x)*cos(2*a + 2*b
*x)/(315*b) - 64*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**3/(315*b), Ne(b, 0)), (x*sin(2*a)**4*cos(a),
True))

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Giac [A]  time = 1.2651, size = 92, normalized size = 2. \begin{align*} \frac{\sin \left (9 \, b x + 9 \, a\right )}{144 \, b} + \frac{\sin \left (7 \, b x + 7 \, a\right )}{112 \, b} - \frac{\sin \left (5 \, b x + 5 \, a\right )}{20 \, b} - \frac{\sin \left (3 \, b x + 3 \, a\right )}{12 \, b} + \frac{3 \, \sin \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)^4,x, algorithm="giac")

[Out]

1/144*sin(9*b*x + 9*a)/b + 1/112*sin(7*b*x + 7*a)/b - 1/20*sin(5*b*x + 5*a)/b - 1/12*sin(3*b*x + 3*a)/b + 3/8*
sin(b*x + a)/b