3.48 \(\int \csc ^2(a+b x) \sin ^5(2 a+2 b x) \, dx\)

Optimal. Leaf size=29 \[ \frac{4 \cos ^8(a+b x)}{b}-\frac{16 \cos ^6(a+b x)}{3 b} \]

[Out]

(-16*Cos[a + b*x]^6)/(3*b) + (4*Cos[a + b*x]^8)/b

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Rubi [A]  time = 0.0567422, antiderivative size = 29, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4288, 2565, 14} \[ \frac{4 \cos ^8(a+b x)}{b}-\frac{16 \cos ^6(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]

[Out]

(-16*Cos[a + b*x]^6)/(3*b) + (4*Cos[a + b*x]^8)/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 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 ^2(a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^3(a+b x) \, dx\\ &=-\frac{32 \operatorname{Subst}\left (\int x^5 \left (1-x^2\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{32 \operatorname{Subst}\left (\int \left (x^5-x^7\right ) \, dx,x,\cos (a+b x)\right )}{b}\\ &=-\frac{16 \cos ^6(a+b x)}{3 b}+\frac{4 \cos ^8(a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.124245, size = 48, normalized size = 1.66 \[ \frac{-72 \cos (2 (a+b x))-12 \cos (4 (a+b x))+8 \cos (6 (a+b x))+3 \cos (8 (a+b x))}{96 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2*Sin[2*a + 2*b*x]^5,x]

[Out]

(-72*Cos[2*(a + b*x)] - 12*Cos[4*(a + b*x)] + 8*Cos[6*(a + b*x)] + 3*Cos[8*(a + b*x)])/(96*b)

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Maple [A]  time = 0.025, size = 35, normalized size = 1.2 \begin{align*} 32\,{\frac{-1/8\, \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{6}-1/24\, \left ( \cos \left ( bx+a \right ) \right ) ^{6}}{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2*sin(2*b*x+2*a)^5,x)

[Out]

32/b*(-1/8*sin(b*x+a)^2*cos(b*x+a)^6-1/24*cos(b*x+a)^6)

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Maxima [A]  time = 1.12087, size = 68, normalized size = 2.34 \begin{align*} \frac{3 \, \cos \left (8 \, b x + 8 \, a\right ) + 8 \, \cos \left (6 \, b x + 6 \, a\right ) - 12 \, \cos \left (4 \, b x + 4 \, a\right ) - 72 \, \cos \left (2 \, b x + 2 \, a\right )}{96 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^5,x, algorithm="maxima")

[Out]

1/96*(3*cos(8*b*x + 8*a) + 8*cos(6*b*x + 6*a) - 12*cos(4*b*x + 4*a) - 72*cos(2*b*x + 2*a))/b

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Fricas [A]  time = 0.489855, size = 61, normalized size = 2.1 \begin{align*} \frac{4 \,{\left (3 \, \cos \left (b x + a\right )^{8} - 4 \, \cos \left (b x + a\right )^{6}\right )}}{3 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^5,x, algorithm="fricas")

[Out]

4/3*(3*cos(b*x + a)^8 - 4*cos(b*x + a)^6)/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)**2*sin(2*b*x+2*a)**5,x)

[Out]

Timed out

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Giac [B]  time = 1.45094, size = 188, normalized size = 6.48 \begin{align*} \frac{128 \,{\left (\frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{4 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{10 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac{4 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} + \frac{3 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{6}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{6}}\right )}}{3 \, b{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2*sin(2*b*x+2*a)^5,x, algorithm="giac")

[Out]

128/3*(3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 4*(cos(b*x + a) - 1)^3/(cos(b*x + a) + 1)^3 + 10*(cos(b*x
+ a) - 1)^4/(cos(b*x + a) + 1)^4 + 4*(cos(b*x + a) - 1)^5/(cos(b*x + a) + 1)^5 + 3*(cos(b*x + a) - 1)^6/(cos(
b*x + a) + 1)^6)/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^8)