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

Optimal. Leaf size=15 \[ \frac{\sin ^4(a+b x)}{2 b} \]

[Out]

Sin[a + b*x]^4/(2*b)

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Rubi [A]  time = 0.0332715, 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} \[ \frac{\sin ^4(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0045543, size = 15, normalized size = 1. \[ \frac{\sin ^4(a+b x)}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[a + b*x]^4/(2*b)

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Maple [B]  time = 0.009, size = 30, normalized size = 2. \begin{align*} -{\frac{\cos \left ( 2\,bx+2\,a \right ) }{4\,b}}+{\frac{\cos \left ( 4\,bx+4\,a \right ) }{16\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*cos(2*b*x+2*a)/b+1/16*cos(4*b*x+4*a)/b

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Maxima [A]  time = 1.19698, size = 35, normalized size = 2.33 \begin{align*} \frac{\cos \left (4 \, b x + 4 \, a\right ) - 4 \, \cos \left (2 \, b x + 2 \, a\right )}{16 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(cos(4*b*x + 4*a) - 4*cos(2*b*x + 2*a))/b

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Fricas [A]  time = 0.474385, size = 58, normalized size = 3.87 \begin{align*} \frac{\cos \left (b x + a\right )^{4} - 2 \, \cos \left (b x + a\right )^{2}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.49128, size = 133, normalized size = 8.87 \begin{align*} \begin{cases} \frac{x \sin ^{2}{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )}}{4} + \frac{x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{2} - \frac{x \sin{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{4} - \frac{3 \sin{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )}}{4 b} - \frac{\cos ^{2}{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{2 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\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)**2*sin(2*b*x+2*a),x)

[Out]

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

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Giac [B]  time = 1.40023, size = 39, normalized size = 2.6 \begin{align*} \frac{\cos \left (4 \, b x + 4 \, a\right )}{16 \, b} - \frac{\cos \left (2 \, b x + 2 \, a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*cos(4*b*x + 4*a)/b - 1/4*cos(2*b*x + 2*a)/b