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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0496542, 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 = {4288, 2565, 14} \[ \frac{4 \cos ^5(a+b x)}{5 b}-\frac{4 \cos ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Cos[a + b*x]^3)/(3*b) + (4*Cos[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 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[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 \sin (a+b x) \sin ^2(2 a+2 b x) \, dx &=4 \int \cos ^2(a+b x) \sin ^3(a+b x) \, dx\\ &=-\frac{4 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{4 \operatorname{Subst}\left (\int \left (x^2-x^4\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{4 \cos ^3(a+b x)}{3 b}+\frac{4 \cos ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0662624, size = 27, normalized size = 0.87 \[ \frac{2 \cos ^3(a+b x) (3 \cos (2 (a+b x))-7)}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 41, normalized size = 1.3 \begin{align*} -{\frac{\cos \left ( bx+a \right ) }{2\,b}}-{\frac{\cos \left ( 3\,bx+3\,a \right ) }{12\,b}}+{\frac{\cos \left ( 5\,bx+5\,a \right ) }{20\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.23104, size = 49, normalized size = 1.58 \begin{align*} \frac{3 \, \cos \left (5 \, b x + 5 \, a\right ) - 5 \, \cos \left (3 \, b x + 3 \, a\right ) - 30 \, \cos \left (b x + a\right )}{60 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 0.478896, size = 62, normalized size = 2. \begin{align*} \frac{4 \,{\left (3 \, \cos \left (b x + a\right )^{5} - 5 \, \cos \left (b x + a\right )^{3}\right )}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/15*(3*cos(b*x + a)^5 - 5*cos(b*x + a)^3)/b

________________________________________________________________________________________

Sympy [A]  time = 4.35895, size = 92, normalized size = 2.97 \begin{align*} \begin{cases} - \frac{4 \sin{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos{\left (2 a + 2 b x \right )}}{15 b} - \frac{7 \sin ^{2}{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )}}{15 b} - \frac{8 \cos{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x \sin{\left (a \right )} \sin ^{2}{\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)*sin(2*b*x+2*a)**2,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.29399, size = 54, normalized size = 1.74 \begin{align*} \frac{\cos \left (5 \, b x + 5 \, a\right )}{20 \, b} - \frac{\cos \left (3 \, b x + 3 \, a\right )}{12 \, b} - \frac{\cos \left (b x + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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