3.65 \(\int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx\)

Optimal. Leaf size=15 \[ -\frac{16 \cos ^5(a+b x)}{5 b} \]

[Out]

(-16*Cos[a + b*x]^5)/(5*b)

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Rubi [A]  time = 0.0444241, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4288, 2565, 30} \[ -\frac{16 \cos ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

(-16*Cos[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 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 30

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

Rubi steps

\begin{align*} \int \csc ^3(a+b x) \sin ^4(2 a+2 b x) \, dx &=16 \int \cos ^4(a+b x) \sin (a+b x) \, dx\\ &=-\frac{16 \operatorname{Subst}\left (\int x^4 \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{16 \cos ^5(a+b x)}{5 b}\\ \end{align*}

Mathematica [A]  time = 0.0096055, size = 15, normalized size = 1. \[ -\frac{16 \cos ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^4,x]

[Out]

(-16*Cos[a + b*x]^5)/(5*b)

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Maple [A]  time = 0.023, size = 14, normalized size = 0.9 \begin{align*} -{\frac{16\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}{5\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x)

[Out]

-16/5*cos(b*x+a)^5/b

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Maxima [B]  time = 1.06807, size = 46, normalized size = 3.07 \begin{align*} -\frac{\cos \left (5 \, b x + 5 \, a\right ) + 5 \, \cos \left (3 \, b x + 3 \, a\right ) + 10 \, \cos \left (b x + a\right )}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="maxima")

[Out]

-1/5*(cos(5*b*x + 5*a) + 5*cos(3*b*x + 3*a) + 10*cos(b*x + a))/b

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Fricas [A]  time = 0.479734, size = 32, normalized size = 2.13 \begin{align*} -\frac{16 \, \cos \left (b x + a\right )^{5}}{5 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="fricas")

[Out]

-16/5*cos(b*x + a)^5/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)**3*sin(2*b*x+2*a)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.42222, size = 100, normalized size = 6.67 \begin{align*} \frac{32 \,{\left (\frac{10 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{5 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + 1\right )}}{5 \, b{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^3*sin(2*b*x+2*a)^4,x, algorithm="giac")

[Out]

32/5*(10*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 5*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 1)/(b*((cos
(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^5)