3.13 \(\int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx\)

Optimal. Leaf size=44 \[ \frac{8 \sin ^{12}(a+b x)}{3 b}-\frac{32 \sin ^{10}(a+b x)}{5 b}+\frac{4 \sin ^8(a+b x)}{b} \]

[Out]

(4*Sin[a + b*x]^8)/b - (32*Sin[a + b*x]^10)/(5*b) + (8*Sin[a + b*x]^12)/(3*b)

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Rubi [A]  time = 0.0687566, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4288, 2564, 266, 43} \[ \frac{8 \sin ^{12}(a+b x)}{3 b}-\frac{32 \sin ^{10}(a+b x)}{5 b}+\frac{4 \sin ^8(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*Sin[a + b*x]^8)/b - (32*Sin[a + b*x]^10)/(5*b) + (8*Sin[a + b*x]^12)/(3*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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \sin ^2(a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^7(a+b x) \, dx\\ &=\frac{32 \operatorname{Subst}\left (\int x^7 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{16 \operatorname{Subst}\left (\int (1-x)^2 x^3 \, dx,x,\sin ^2(a+b x)\right )}{b}\\ &=\frac{16 \operatorname{Subst}\left (\int \left (x^3-2 x^4+x^5\right ) \, dx,x,\sin ^2(a+b x)\right )}{b}\\ &=\frac{4 \sin ^8(a+b x)}{b}-\frac{32 \sin ^{10}(a+b x)}{5 b}+\frac{8 \sin ^{12}(a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.34191, size = 68, normalized size = 1.55 \[ \frac{-600 \cos (2 (a+b x))+75 \cos (4 (a+b x))+100 \cos (6 (a+b x))-30 \cos (8 (a+b x))-12 \cos (10 (a+b x))+5 \cos (12 (a+b x))}{3840 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]

[Out]

(-600*Cos[2*(a + b*x)] + 75*Cos[4*(a + b*x)] + 100*Cos[6*(a + b*x)] - 30*Cos[8*(a + b*x)] - 12*Cos[10*(a + b*x
)] + 5*Cos[12*(a + b*x)])/(3840*b)

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Maple [B]  time = 0.022, size = 86, normalized size = 2. \begin{align*} -{\frac{5\,\cos \left ( 2\,bx+2\,a \right ) }{32\,b}}+{\frac{5\,\cos \left ( 4\,bx+4\,a \right ) }{256\,b}}+{\frac{5\,\cos \left ( 6\,bx+6\,a \right ) }{192\,b}}-{\frac{\cos \left ( 8\,bx+8\,a \right ) }{128\,b}}-{\frac{\cos \left ( 10\,bx+10\,a \right ) }{320\,b}}+{\frac{\cos \left ( 12\,bx+12\,a \right ) }{768\,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)^5,x)

[Out]

-5/32*cos(2*b*x+2*a)/b+5/256*cos(4*b*x+4*a)/b+5/192*cos(6*b*x+6*a)/b-1/128*cos(8*b*x+8*a)/b-1/320*cos(10*b*x+1
0*a)/b+1/768*cos(12*b*x+12*a)/b

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Maxima [A]  time = 1.19228, size = 97, normalized size = 2.2 \begin{align*} \frac{5 \, \cos \left (12 \, b x + 12 \, a\right ) - 12 \, \cos \left (10 \, b x + 10 \, a\right ) - 30 \, \cos \left (8 \, b x + 8 \, a\right ) + 100 \, \cos \left (6 \, b x + 6 \, a\right ) + 75 \, \cos \left (4 \, b x + 4 \, a\right ) - 600 \, \cos \left (2 \, b x + 2 \, a\right )}{3840 \, 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)^5,x, algorithm="maxima")

[Out]

1/3840*(5*cos(12*b*x + 12*a) - 12*cos(10*b*x + 10*a) - 30*cos(8*b*x + 8*a) + 100*cos(6*b*x + 6*a) + 75*cos(4*b
*x + 4*a) - 600*cos(2*b*x + 2*a))/b

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Fricas [A]  time = 0.504133, size = 122, normalized size = 2.77 \begin{align*} \frac{4 \,{\left (10 \, \cos \left (b x + a\right )^{12} - 36 \, \cos \left (b x + a\right )^{10} + 45 \, \cos \left (b x + a\right )^{8} - 20 \, \cos \left (b x + a\right )^{6}\right )}}{15 \, 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)^5,x, algorithm="fricas")

[Out]

4/15*(10*cos(b*x + a)^12 - 36*cos(b*x + a)^10 + 45*cos(b*x + a)^8 - 20*cos(b*x + a)^6)/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)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.44942, size = 115, normalized size = 2.61 \begin{align*} \frac{\cos \left (12 \, b x + 12 \, a\right )}{768 \, b} - \frac{\cos \left (10 \, b x + 10 \, a\right )}{320 \, b} - \frac{\cos \left (8 \, b x + 8 \, a\right )}{128 \, b} + \frac{5 \, \cos \left (6 \, b x + 6 \, a\right )}{192 \, b} + \frac{5 \, \cos \left (4 \, b x + 4 \, a\right )}{256 \, b} - \frac{5 \, \cos \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)^5,x, algorithm="giac")

[Out]

1/768*cos(12*b*x + 12*a)/b - 1/320*cos(10*b*x + 10*a)/b - 1/128*cos(8*b*x + 8*a)/b + 5/192*cos(6*b*x + 6*a)/b
+ 5/256*cos(4*b*x + 4*a)/b - 5/32*cos(2*b*x + 2*a)/b