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

Optimal. Leaf size=29 \[ \frac{4 \sin ^6(a+b x)}{3 b}-\frac{\sin ^8(a+b x)}{b} \]

[Out]

(4*Sin[a + b*x]^6)/(3*b) - Sin[a + b*x]^8/b

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Rubi [A]  time = 0.0591716, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4288, 2564, 14} \[ \frac{4 \sin ^6(a+b x)}{3 b}-\frac{\sin ^8(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sin[a + b*x]^6)/(3*b) - Sin[a + b*x]^8/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 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 ^2(a+b x) \sin ^3(2 a+2 b x) \, dx &=8 \int \cos ^3(a+b x) \sin ^5(a+b x) \, dx\\ &=\frac{8 \operatorname{Subst}\left (\int x^5 \left (1-x^2\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{8 \operatorname{Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{4 \sin ^6(a+b x)}{3 b}-\frac{\sin ^8(a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.113724, size = 48, normalized size = 1.66 \[ \frac{-72 \cos (2 (a+b x))+12 \cos (4 (a+b x))+8 \cos (6 (a+b x))-3 \cos (8 (a+b x))}{384 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-72*Cos[2*(a + b*x)] + 12*Cos[4*(a + b*x)] + 8*Cos[6*(a + b*x)] - 3*Cos[8*(a + b*x)])/(384*b)

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Maple [B]  time = 0.01, size = 58, normalized size = 2. \begin{align*} -{\frac{3\,\cos \left ( 2\,bx+2\,a \right ) }{16\,b}}+{\frac{\cos \left ( 4\,bx+4\,a \right ) }{32\,b}}+{\frac{\cos \left ( 6\,bx+6\,a \right ) }{48\,b}}-{\frac{\cos \left ( 8\,bx+8\,a \right ) }{128\,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)^3,x)

[Out]

-3/16*cos(2*b*x+2*a)/b+1/32*cos(4*b*x+4*a)/b+1/48*cos(6*b*x+6*a)/b-1/128*cos(8*b*x+8*a)/b

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Maxima [A]  time = 1.0841, size = 68, normalized size = 2.34 \begin{align*} -\frac{3 \, \cos \left (8 \, b x + 8 \, a\right ) - 8 \, \cos \left (6 \, b x + 6 \, a\right ) - 12 \, \cos \left (4 \, b x + 4 \, a\right ) + 72 \, \cos \left (2 \, b x + 2 \, a\right )}{384 \, 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)^3,x, algorithm="maxima")

[Out]

-1/384*(3*cos(8*b*x + 8*a) - 8*cos(6*b*x + 6*a) - 12*cos(4*b*x + 4*a) + 72*cos(2*b*x + 2*a))/b

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Fricas [A]  time = 0.483444, size = 88, normalized size = 3.03 \begin{align*} -\frac{3 \, \cos \left (b x + a\right )^{8} - 8 \, \cos \left (b x + a\right )^{6} + 6 \, \cos \left (b x + a\right )^{4}}{3 \, 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)^3,x, algorithm="fricas")

[Out]

-1/3*(3*cos(b*x + a)^8 - 8*cos(b*x + a)^6 + 6*cos(b*x + a)^4)/b

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Sympy [A]  time = 63.2081, size = 362, normalized size = 12.48 \begin{align*} \begin{cases} \frac{3 x \sin ^{2}{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )}}{16} + \frac{3 x \sin ^{2}{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac{3 x \sin{\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{8} + \frac{3 x \sin{\left (a + b x \right )} \cos{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{8} - \frac{3 x \sin ^{3}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )}}{16} - \frac{3 x \sin{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{16} + \frac{17 \sin ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} - \frac{13 \sin{\left (a + b x \right )} \sin ^{3}{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )}}{16 b} - \frac{7 \sin{\left (a + b x \right )} \sin{\left (2 a + 2 b x \right )} \cos{\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{8 b} - \frac{\sin ^{2}{\left (2 a + 2 b x \right )} \cos ^{2}{\left (a + b x \right )} \cos{\left (2 a + 2 b x \right )}}{2 b} - \frac{49 \cos ^{2}{\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{96 b} & \text{for}\: b \neq 0 \\x \sin ^{2}{\left (a \right )} \sin ^{3}{\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)**3,x)

[Out]

Piecewise((3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)**3/16 + 3*x*sin(a + b*x)**2*sin(2*a + 2*b*x)*cos(2*a + 2*b*x)*
*2/16 + 3*x*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(a + b*x)*cos(2*a + 2*b*x)/8 + 3*x*sin(a + b*x)*cos(a + b*x)*c
os(2*a + 2*b*x)**3/8 - 3*x*sin(2*a + 2*b*x)**3*cos(a + b*x)**2/16 - 3*x*sin(2*a + 2*b*x)*cos(a + b*x)**2*cos(2
*a + 2*b*x)**2/16 + 17*sin(a + b*x)**2*cos(2*a + 2*b*x)**3/(96*b) - 13*sin(a + b*x)*sin(2*a + 2*b*x)**3*cos(a
+ b*x)/(16*b) - 7*sin(a + b*x)*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**2/(8*b) - sin(2*a + 2*b*x)**2*c
os(a + b*x)**2*cos(2*a + 2*b*x)/(2*b) - 49*cos(a + b*x)**2*cos(2*a + 2*b*x)**3/(96*b), Ne(b, 0)), (x*sin(a)**2
*sin(2*a)**3, True))

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Giac [B]  time = 1.37836, size = 77, normalized size = 2.66 \begin{align*} -\frac{\cos \left (8 \, b x + 8 \, a\right )}{128 \, b} + \frac{\cos \left (6 \, b x + 6 \, a\right )}{48 \, b} + \frac{\cos \left (4 \, b x + 4 \, a\right )}{32 \, b} - \frac{3 \, \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)^3,x, algorithm="giac")

[Out]

-1/128*cos(8*b*x + 8*a)/b + 1/48*cos(6*b*x + 6*a)/b + 1/32*cos(4*b*x + 4*a)/b - 3/16*cos(2*b*x + 2*a)/b