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

Optimal. Leaf size=13 \[ -\frac {2 \cos ^4(a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4288, 2565, 30} \[ -\frac {2 \cos ^4(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x]^4)/b

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[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 \csc ^2(a+b x) \sin ^3(2 a+2 b x) \, dx &=8 \int \cos ^3(a+b x) \sin (a+b x) \, dx\\ &=-\frac {8 \operatorname {Subst}\left (\int x^3 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac {2 \cos ^4(a+b x)}{b}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 1.00 \[ -\frac {2 \cos ^4(a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[a + b*x]^4)/b

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fricas [A]  time = 0.45, size = 13, normalized size = 1.00 \[ -\frac {2 \, \cos \left (b x + a\right )^{4}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*cos(b*x + a)^4/b

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giac [B]  time = 0.49, size = 69, normalized size = 5.31 \[ -\frac {16 \, {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + \frac {{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}}\right )}}{b {\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) + (cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3)/(b*((cos(b*x + a) - 1)
/(cos(b*x + a) + 1) - 1)^4)

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maple [A]  time = 0.57, size = 14, normalized size = 1.08 \[ -\frac {2 \left (\cos ^{4}\left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 26, normalized size = 2.00 \[ -\frac {\cos \left (4 \, b x + 4 \, a\right ) + 4 \, \cos \left (2 \, b x + 2 \, a\right )}{4 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(cos(4*b*x + 4*a) + 4*cos(2*b*x + 2*a))/b

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mupad [B]  time = 0.11, size = 13, normalized size = 1.00 \[ -\frac {2\,{\cos \left (a+b\,x\right )}^4}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*cos(a + b*x)^4)/b

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

[Out]

Timed out

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