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

Optimal. Leaf size=11 \[ -\frac {2 \csc (a+b x)}{b} \]

[Out]

-2*csc(b*x+a)/b

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Rubi [A]  time = 0.03, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4288, 2606, 8} \[ -\frac {2 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Csc[a + b*x])/b

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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 \csc ^3(a+b x) \sin (2 a+2 b x) \, dx &=2 \int \cot (a+b x) \csc (a+b x) \, dx\\ &=-\frac {2 \operatorname {Subst}(\int 1 \, dx,x,\csc (a+b x))}{b}\\ &=-\frac {2 \csc (a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 11, normalized size = 1.00 \[ -\frac {2 \csc (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Csc[a + b*x])/b

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fricas [A]  time = 0.52, size = 13, normalized size = 1.18 \[ -\frac {2}{b \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(b*sin(b*x + a))

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giac [A]  time = 0.35, size = 13, normalized size = 1.18 \[ -\frac {2}{b \sin \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(b*sin(b*x + a))

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maple [A]  time = 0.29, size = 14, normalized size = 1.27 \[ -\frac {2}{b \sin \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/b/sin(b*x+a)

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maxima [B]  time = 0.35, size = 84, normalized size = 7.64 \[ -\frac {4 \, {\left (\cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + \sin \left (b x + a\right )\right )}}{b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(cos(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a)*sin(b*x + a) + sin(b*x + a))/(b*cos(2*b*x + 2*a)^2 + b*si
n(2*b*x + 2*a)^2 - 2*b*cos(2*b*x + 2*a) + b)

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mupad [B]  time = 0.04, size = 13, normalized size = 1.18 \[ -\frac {2}{b\,\sin \left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/(b*sin(a + b*x))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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