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

Optimal. Leaf size=76 \[ -\frac{\sin ^5(2 a+2 b x)}{20 b}-\frac{\sin ^3(2 a+2 b x) \cos (2 a+2 b x)}{16 b}-\frac{3 \sin (2 a+2 b x) \cos (2 a+2 b x)}{32 b}+\frac{3 x}{16} \]

[Out]

(3*x)/16 - (3*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(32*b) - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^3)/(16*b) - Sin[2
*a + 2*b*x]^5/(20*b)

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Rubi [A]  time = 0.0676803, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4286, 2635, 8, 2564, 30} \[ -\frac{\sin ^5(2 a+2 b x)}{20 b}-\frac{\sin ^3(2 a+2 b x) \cos (2 a+2 b x)}{16 b}-\frac{3 \sin (2 a+2 b x) \cos (2 a+2 b x)}{32 b}+\frac{3 x}{16} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*x)/16 - (3*Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x])/(32*b) - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^3)/(16*b) - Sin[2
*a + 2*b*x]^5/(20*b)

Rule 4286

Int[sin[(a_.) + (b_.)*(x_)]^2*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[1/2, Int[(g*Sin[c + d*x]
)^p, x], x] - Dist[1/2, Int[Cos[c + d*x]*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, g}, x] && EqQ[b*c -
a*d, 0] && EqQ[d/b, 2] && IGtQ[p/2, 0]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 ^4(2 a+2 b x) \, dx &=\frac{1}{2} \int \sin ^4(2 a+2 b x) \, dx-\frac{1}{2} \int \cos (2 a+2 b x) \sin ^4(2 a+2 b x) \, dx\\ &=-\frac{\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}+\frac{3}{8} \int \sin ^2(2 a+2 b x) \, dx-\frac{\operatorname{Subst}\left (\int x^4 \, dx,x,\sin (2 a+2 b x)\right )}{4 b}\\ &=-\frac{3 \cos (2 a+2 b x) \sin (2 a+2 b x)}{32 b}-\frac{\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}-\frac{\sin ^5(2 a+2 b x)}{20 b}+\frac{3 \int 1 \, dx}{16}\\ &=\frac{3 x}{16}-\frac{3 \cos (2 a+2 b x) \sin (2 a+2 b x)}{32 b}-\frac{\cos (2 a+2 b x) \sin ^3(2 a+2 b x)}{16 b}-\frac{\sin ^5(2 a+2 b x)}{20 b}\\ \end{align*}

Mathematica [A]  time = 0.193295, size = 62, normalized size = 0.82 \[ \frac{-20 \sin (2 (a+b x))-40 \sin (4 (a+b x))+10 \sin (6 (a+b x))+5 \sin (8 (a+b x))-2 \sin (10 (a+b x))+120 b x}{640 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(120*b*x - 20*Sin[2*(a + b*x)] - 40*Sin[4*(a + b*x)] + 10*Sin[6*(a + b*x)] + 5*Sin[8*(a + b*x)] - 2*Sin[10*(a
+ b*x)])/(640*b)

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Maple [A]  time = 0.027, size = 75, normalized size = 1. \begin{align*}{\frac{3\,x}{16}}-{\frac{\sin \left ( 2\,bx+2\,a \right ) }{32\,b}}-{\frac{\sin \left ( 4\,bx+4\,a \right ) }{16\,b}}+{\frac{\sin \left ( 6\,bx+6\,a \right ) }{64\,b}}+{\frac{\sin \left ( 8\,bx+8\,a \right ) }{128\,b}}-{\frac{\sin \left ( 10\,bx+10\,a \right ) }{320\,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)^4,x)

[Out]

3/16*x-1/32*sin(2*b*x+2*a)/b-1/16/b*sin(4*b*x+4*a)+1/64/b*sin(6*b*x+6*a)+1/128/b*sin(8*b*x+8*a)-1/320/b*sin(10
*b*x+10*a)

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Maxima [A]  time = 1.12836, size = 88, normalized size = 1.16 \begin{align*} \frac{120 \, b x - 2 \, \sin \left (10 \, b x + 10 \, a\right ) + 5 \, \sin \left (8 \, b x + 8 \, a\right ) + 10 \, \sin \left (6 \, b x + 6 \, a\right ) - 40 \, \sin \left (4 \, b x + 4 \, a\right ) - 20 \, \sin \left (2 \, b x + 2 \, a\right )}{640 \, 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)^4,x, algorithm="maxima")

[Out]

1/640*(120*b*x - 2*sin(10*b*x + 10*a) + 5*sin(8*b*x + 8*a) + 10*sin(6*b*x + 6*a) - 40*sin(4*b*x + 4*a) - 20*si
n(2*b*x + 2*a))/b

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Fricas [A]  time = 0.511443, size = 180, normalized size = 2.37 \begin{align*} \frac{15 \, b x -{\left (128 \, \cos \left (b x + a\right )^{9} - 336 \, \cos \left (b x + a\right )^{7} + 248 \, \cos \left (b x + a\right )^{5} - 10 \, \cos \left (b x + a\right )^{3} - 15 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )}{80 \, 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)^4,x, algorithm="fricas")

[Out]

1/80*(15*b*x - (128*cos(b*x + a)^9 - 336*cos(b*x + a)^7 + 248*cos(b*x + a)^5 - 10*cos(b*x + a)^3 - 15*cos(b*x
+ a))*sin(b*x + a))/b

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.35376, size = 100, normalized size = 1.32 \begin{align*} \frac{3}{16} \, x - \frac{\sin \left (10 \, b x + 10 \, a\right )}{320 \, b} + \frac{\sin \left (8 \, b x + 8 \, a\right )}{128 \, b} + \frac{\sin \left (6 \, b x + 6 \, a\right )}{64 \, b} - \frac{\sin \left (4 \, b x + 4 \, a\right )}{16 \, b} - \frac{\sin \left (2 \, b x + 2 \, a\right )}{32 \, 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)^4,x, algorithm="giac")

[Out]

3/16*x - 1/320*sin(10*b*x + 10*a)/b + 1/128*sin(8*b*x + 8*a)/b + 1/64*sin(6*b*x + 6*a)/b - 1/16*sin(4*b*x + 4*
a)/b - 1/32*sin(2*b*x + 2*a)/b