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

Optimal. Leaf size=31 \[ \frac {8 \sin ^5(a+b x)}{5 b}-\frac {8 \sin ^7(a+b x)}{7 b} \]

[Out]

8/5*sin(b*x+a)^5/b-8/7*sin(b*x+a)^7/b

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Rubi [A]  time = 0.05, antiderivative size = 31, normalized size of antiderivative = 1.00, 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, 2564, 14} \[ \frac {8 \sin ^5(a+b x)}{5 b}-\frac {8 \sin ^7(a+b x)}{7 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sin[a + b*x]^5)/(5*b) - (8*Sin[a + b*x]^7)/(7*b)

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]]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 27, normalized size = 0.87 \[ \frac {4 \sin ^5(a+b x) (5 \cos (2 (a+b x))+9)}{35 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*(9 + 5*Cos[2*(a + b*x)])*Sin[a + b*x]^5)/(35*b)

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fricas [A]  time = 0.47, size = 41, normalized size = 1.32 \[ \frac {8 \, {\left (5 \, \cos \left (b x + a\right )^{6} - 8 \, \cos \left (b x + a\right )^{4} + \cos \left (b x + a\right )^{2} + 2\right )} \sin \left (b x + a\right )}{35 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^3,x, algorithm="fricas")

[Out]

8/35*(5*cos(b*x + a)^6 - 8*cos(b*x + a)^4 + cos(b*x + a)^2 + 2)*sin(b*x + a)/b

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giac [A]  time = 0.37, size = 54, normalized size = 1.74 \[ \frac {\sin \left (7 \, b x + 7 \, a\right )}{56 \, b} - \frac {\sin \left (5 \, b x + 5 \, a\right )}{40 \, b} - \frac {\sin \left (3 \, b x + 3 \, a\right )}{8 \, b} + \frac {3 \, \sin \left (b x + a\right )}{8 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^3,x, algorithm="giac")

[Out]

1/56*sin(7*b*x + 7*a)/b - 1/40*sin(5*b*x + 5*a)/b - 1/8*sin(3*b*x + 3*a)/b + 3/8*sin(b*x + a)/b

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maple [A]  time = 0.93, size = 55, normalized size = 1.77 \[ \frac {3 \sin \left (b x +a \right )}{8 b}-\frac {\sin \left (3 b x +3 a \right )}{8 b}-\frac {\sin \left (5 b x +5 a \right )}{40 b}+\frac {\sin \left (7 b x +7 a \right )}{56 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)*sin(2*b*x+2*a)^3,x)

[Out]

3/8*sin(b*x+a)/b-1/8*sin(3*b*x+3*a)/b-1/40/b*sin(5*b*x+5*a)+1/56/b*sin(7*b*x+7*a)

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maxima [A]  time = 0.34, size = 47, normalized size = 1.52 \[ \frac {5 \, \sin \left (7 \, b x + 7 \, a\right ) - 7 \, \sin \left (5 \, b x + 5 \, a\right ) - 35 \, \sin \left (3 \, b x + 3 \, a\right ) + 105 \, \sin \left (b x + a\right )}{280 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^3,x, algorithm="maxima")

[Out]

1/280*(5*sin(7*b*x + 7*a) - 7*sin(5*b*x + 5*a) - 35*sin(3*b*x + 3*a) + 105*sin(b*x + a))/b

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mupad [B]  time = 0.05, size = 26, normalized size = 0.84 \[ \frac {8\,\left (7\,{\sin \left (a+b\,x\right )}^5-5\,{\sin \left (a+b\,x\right )}^7\right )}{35\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*sin(2*a + 2*b*x)^3,x)

[Out]

(8*(7*sin(a + b*x)^5 - 5*sin(a + b*x)^7))/(35*b)

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sympy [A]  time = 11.91, size = 126, normalized size = 4.06 \[ \begin {cases} - \frac {22 \sin {\left (a + b x \right )} \sin ^{2}{\left (2 a + 2 b x \right )} \cos {\left (2 a + 2 b x \right )}}{35 b} - \frac {16 \sin {\left (a + b x \right )} \cos ^{3}{\left (2 a + 2 b x \right )}}{35 b} + \frac {9 \sin ^{3}{\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )}}{35 b} + \frac {8 \sin {\left (2 a + 2 b x \right )} \cos {\left (a + b x \right )} \cos ^{2}{\left (2 a + 2 b x \right )}}{35 b} & \text {for}\: b \neq 0 \\x \sin {\relax (a )} \sin ^{3}{\left (2 a \right )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)**3,x)

[Out]

Piecewise((-22*sin(a + b*x)*sin(2*a + 2*b*x)**2*cos(2*a + 2*b*x)/(35*b) - 16*sin(a + b*x)*cos(2*a + 2*b*x)**3/
(35*b) + 9*sin(2*a + 2*b*x)**3*cos(a + b*x)/(35*b) + 8*sin(2*a + 2*b*x)*cos(a + b*x)*cos(2*a + 2*b*x)**2/(35*b
), Ne(b, 0)), (x*sin(a)*sin(2*a)**3, True))

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