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

Optimal. Leaf size=31 \[ \frac{8 \sin ^3(a+b x)}{3 b}-\frac{8 \sin ^5(a+b x)}{5 b} \]

[Out]

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

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Rubi [A]  time = 0.0513627, 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, 2564, 14} \[ \frac{8 \sin ^3(a+b x)}{3 b}-\frac{8 \sin ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0475814, size = 28, normalized size = 0.9 \[ \frac{8 \left (5 \sin ^3(a+b x)-3 \sin ^5(a+b x)\right )}{15 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*(5*Sin[a + b*x]^3 - 3*Sin[a + b*x]^5))/(15*b)

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Maple [A]  time = 0.047, size = 41, normalized size = 1.3 \begin{align*} 8\,{\frac{-1/5\,\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{4}+1/15\, \left ( 2+ \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) \sin \left ( bx+a \right ) }{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/b*(-1/5*sin(b*x+a)*cos(b*x+a)^4+1/15*(2+cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 1.20096, size = 49, normalized size = 1.58 \begin{align*} -\frac{3 \, \sin \left (5 \, b x + 5 \, a\right ) + 5 \, \sin \left (3 \, b x + 3 \, a\right ) - 30 \, \sin \left (b x + a\right )}{30 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.477091, size = 84, normalized size = 2.71 \begin{align*} -\frac{8 \,{\left (3 \, \cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2} - 2\right )} \sin \left (b x + a\right )}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/15*(3*cos(b*x + a)^4 - cos(b*x + a)^2 - 2)*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(csc(b*x+a)*sin(2*b*x+2*a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.34838, size = 35, normalized size = 1.13 \begin{align*} -\frac{8 \,{\left (3 \, \sin \left (b x + a\right )^{5} - 5 \, \sin \left (b x + a\right )^{3}\right )}}{15 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/15*(3*sin(b*x + a)^5 - 5*sin(b*x + a)^3)/b