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

Optimal. Leaf size=46 \[ \frac{4 \sin ^7(a+b x)}{7 b}-\frac{8 \sin ^5(a+b x)}{5 b}+\frac{4 \sin ^3(a+b x)}{3 b} \]

[Out]

(4*Sin[a + b*x]^3)/(3*b) - (8*Sin[a + b*x]^5)/(5*b) + (4*Sin[a + b*x]^7)/(7*b)

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Rubi [A]  time = 0.0603713, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4287, 2564, 270} \[ \frac{4 \sin ^7(a+b x)}{7 b}-\frac{8 \sin ^5(a+b x)}{5 b}+\frac{4 \sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0918017, size = 37, normalized size = 0.8 \[ \frac{\sin ^3(a+b x) (108 \cos (2 (a+b x))+15 \cos (4 (a+b x))+157)}{210 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((157 + 108*Cos[2*(a + b*x)] + 15*Cos[4*(a + b*x)])*Sin[a + b*x]^3)/(210*b)

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Maple [A]  time = 0.026, size = 55, normalized size = 1.2 \begin{align*}{\frac{5\,\sin \left ( bx+a \right ) }{16\,b}}-{\frac{\sin \left ( 3\,bx+3\,a \right ) }{48\,b}}-{\frac{3\,\sin \left ( 5\,bx+5\,a \right ) }{80\,b}}-{\frac{\sin \left ( 7\,bx+7\,a \right ) }{112\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/16*sin(b*x+a)/b-1/48*sin(3*b*x+3*a)/b-3/80/b*sin(5*b*x+5*a)-1/112/b*sin(7*b*x+7*a)

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Maxima [A]  time = 1.09732, size = 63, normalized size = 1.37 \begin{align*} -\frac{15 \, \sin \left (7 \, b x + 7 \, a\right ) + 63 \, \sin \left (5 \, b x + 5 \, a\right ) + 35 \, \sin \left (3 \, b x + 3 \, a\right ) - 525 \, \sin \left (b x + a\right )}{1680 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1680*(15*sin(7*b*x + 7*a) + 63*sin(5*b*x + 5*a) + 35*sin(3*b*x + 3*a) - 525*sin(b*x + a))/b

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Fricas [A]  time = 0.480916, size = 115, normalized size = 2.5 \begin{align*} -\frac{4 \,{\left (15 \, \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 )}{105 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/105*(15*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 = 77.6936, size = 202, normalized size = 4.39 \begin{align*} \begin{cases} \frac{38 \sin ^{3}{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )}}{105 b} + \frac{32 \sin ^{3}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{105 b} + \frac{8 \sin ^{2}{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{35 b} + \frac{11 \sin{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{35 b} + \frac{24 \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} - \frac{12 \sin{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{35 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\left (2 a \right )} \cos ^{3}{\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)**3*sin(2*b*x+2*a)**2,x)

[Out]

Piecewise((38*sin(a + b*x)**3*sin(2*a + 2*b*x)**2/(105*b) + 32*sin(a + b*x)**3*cos(2*a + 2*b*x)**2/(105*b) + 8
*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)/(35*b) + 11*sin(a + b*x)*sin(2*a + 2*b*x)**2*c
os(a + b*x)**2/(35*b) + 24*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)**2/(35*b) - 12*sin(2*a + 2*b*x)*cos(a
+ b*x)**3*cos(2*a + 2*b*x)/(35*b), Ne(b, 0)), (x*sin(2*a)**2*cos(a)**3, True))

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Giac [A]  time = 1.22179, size = 73, normalized size = 1.59 \begin{align*} -\frac{\sin \left (7 \, b x + 7 \, a\right )}{112 \, b} - \frac{3 \, \sin \left (5 \, b x + 5 \, a\right )}{80 \, b} - \frac{\sin \left (3 \, b x + 3 \, a\right )}{48 \, b} + \frac{5 \, \sin \left (b x + a\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/112*sin(7*b*x + 7*a)/b - 3/80*sin(5*b*x + 5*a)/b - 1/48*sin(3*b*x + 3*a)/b + 5/16*sin(b*x + a)/b