3.1.66 \(\int \csc ^3(a+b x) \sin ^3(2 a+2 b x) \, dx\) [66]

Optimal. Leaf size=27 \[ \frac {8 \sin (a+b x)}{b}-\frac {8 \sin ^3(a+b x)}{3 b} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4373, 2713} \begin {gather*} \frac {8 \sin (a+b x)}{b}-\frac {8 \sin ^3(a+b x)}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 4373

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 \csc ^3(a+b x) \sin ^3(2 a+2 b x) \, dx &=8 \int \cos ^3(a+b x) \, dx\\ &=-\frac {8 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (a+b x)\right )}{b}\\ &=\frac {8 \sin (a+b x)}{b}-\frac {8 \sin ^3(a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 28, normalized size = 1.04 \begin {gather*} 8 \left (\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 22, normalized size = 0.81

method	result	size
default	\(\frac {8 \left (2+\cos ^{2}\left (x b +a \right )\right ) \sin \left (x b +a \right )}{3 b}\)	\(22\)
risch	\(\frac {6 \sin \left (x b +a \right )}{b}+\frac {2 \sin \left (3 x b +3 a \right )}{3 b}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/3/b*(2+cos(b*x+a)^2)*sin(b*x+a)

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Maxima [A]
time = 0.27, size = 23, normalized size = 0.85 \begin {gather*} \frac {2 \, {\left (\sin \left (3 \, b x + 3 \, a\right ) + 9 \, \sin \left (b x + a\right )\right )}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(sin(3*b*x + 3*a) + 9*sin(b*x + a))/b

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Fricas [A]
time = 2.80, size = 21, normalized size = 0.78 \begin {gather*} \frac {8 \, {\left (\cos \left (b x + a\right )^{2} + 2\right )} \sin \left (b x + a\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8/3*(cos(b*x + a)^2 + 2)*sin(b*x + a)/b

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 0.62, size = 22, normalized size = 0.81 \begin {gather*} -\frac {8 \, {\left (\sin \left (b x + a\right )^{3} - 3 \, \sin \left (b x + a\right )\right )}}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.10, size = 24, normalized size = 0.89 \begin {gather*} \frac {8\,\left (3\,\sin \left (a+b\,x\right )-{\sin \left (a+b\,x\right )}^3\right )}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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