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

Optimal. Leaf size=15 \[ \frac{2 \sin ^5(a+b x)}{5 b} \]

[Out]

(2*Sin[a + b*x]^5)/(5*b)

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Rubi [A]  time = 0.0330478, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4288, 2564, 30} \[ \frac{2 \sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sin[a + b*x]^5)/(5*b)

Rule 4288

Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/f^p, Int[Cos[a
+ b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b, c, d, f, n}, 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0073833, size = 15, normalized size = 1. \[ \frac{2 \sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sin[a + b*x]^5)/(5*b)

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Maple [B]  time = 0.009, size = 41, normalized size = 2.7 \begin{align*}{\frac{\sin \left ( bx+a \right ) }{4\,b}}-{\frac{\sin \left ( 3\,bx+3\,a \right ) }{8\,b}}+{\frac{\sin \left ( 5\,bx+5\,a \right ) }{40\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sin(b*x+a)/b-1/8*sin(3*b*x+3*a)/b+1/40/b*sin(5*b*x+5*a)

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Maxima [B]  time = 1.07632, size = 46, normalized size = 3.07 \begin{align*} \frac{\sin \left (5 \, b x + 5 \, a\right ) - 5 \, \sin \left (3 \, b x + 3 \, a\right ) + 10 \, \sin \left (b x + a\right )}{40 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(sin(5*b*x + 5*a) - 5*sin(3*b*x + 3*a) + 10*sin(b*x + a))/b

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Fricas [B]  time = 0.470941, size = 81, normalized size = 5.4 \begin{align*} \frac{2 \,{\left (\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(cos(b*x + a)^4 - 2*cos(b*x + a)^2 + 1)*sin(b*x + a)/b

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Sympy [A]  time = 19.1654, size = 117, normalized size = 7.8 \begin{align*} \begin{cases} - \frac{2 \sin ^{3}{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{5 b} - \frac{\sin ^{2}{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )}}{5 b} - \frac{4 \sin{\left (a + b x \right )} \cos ^{2}{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{5 b} + \frac{2 \sin{\left (2 a + 2 b x \right )} \cos ^{3}{\left (a + b x \right )}}{5 b} & \text{for}\: b \neq 0 \\x \sin ^{3}{\left (a \right )} \sin{\left (2 a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*sin(a + b*x)**3*cos(2*a + 2*b*x)/(5*b) - sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(a + b*x)/(5*b) - 4
*sin(a + b*x)*cos(a + b*x)**2*cos(2*a + 2*b*x)/(5*b) + 2*sin(2*a + 2*b*x)*cos(a + b*x)**3/(5*b), Ne(b, 0)), (x
*sin(a)**3*sin(2*a), True))

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Giac [B]  time = 1.2782, size = 54, normalized size = 3.6 \begin{align*} \frac{\sin \left (5 \, b x + 5 \, a\right )}{40 \, b} - \frac{\sin \left (3 \, b x + 3 \, a\right )}{8 \, b} + \frac{\sin \left (b x + a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*sin(5*b*x + 5*a)/b - 1/8*sin(3*b*x + 3*a)/b + 1/4*sin(b*x + a)/b