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

Optimal. Leaf size=31 \[ \frac{16 \cos ^7(a+b x)}{7 b}-\frac{16 \cos ^5(a+b x)}{5 b} \]

[Out]

(-16*Cos[a + b*x]^5)/(5*b) + (16*Cos[a + b*x]^7)/(7*b)

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Rubi [A]  time = 0.0519787, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4288, 2565, 14} \[ \frac{16 \cos ^7(a+b x)}{7 b}-\frac{16 \cos ^5(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0380123, size = 59, normalized size = 1.9 \[ -\frac{3 \cos (a+b x)}{4 b}-\frac{\cos (3 (a+b x))}{4 b}+\frac{\cos (5 (a+b x))}{20 b}+\frac{\cos (7 (a+b x))}{28 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[a + b*x])/(4*b) - Cos[3*(a + b*x)]/(4*b) + Cos[5*(a + b*x)]/(20*b) + Cos[7*(a + b*x)]/(28*b)

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Maple [A]  time = 0.029, size = 35, normalized size = 1.1 \begin{align*} 16\,{\frac{1}{b} \left ( -1/7\, \left ( \sin \left ( bx+a \right ) \right ) ^{2} \left ( \cos \left ( bx+a \right ) \right ) ^{5}-{\frac{2\, \left ( \cos \left ( bx+a \right ) \right ) ^{5}}{35}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/b*(-1/7*sin(b*x+a)^2*cos(b*x+a)^5-2/35*cos(b*x+a)^5)

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Maxima [A]  time = 1.14041, size = 63, normalized size = 2.03 \begin{align*} \frac{5 \, \cos \left (7 \, b x + 7 \, a\right ) + 7 \, \cos \left (5 \, b x + 5 \, a\right ) - 35 \, \cos \left (3 \, b x + 3 \, a\right ) - 105 \, \cos \left (b x + a\right )}{140 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/140*(5*cos(7*b*x + 7*a) + 7*cos(5*b*x + 5*a) - 35*cos(3*b*x + 3*a) - 105*cos(b*x + a))/b

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Fricas [A]  time = 0.482037, size = 63, normalized size = 2.03 \begin{align*} \frac{16 \,{\left (5 \, \cos \left (b x + a\right )^{7} - 7 \, \cos \left (b x + a\right )^{5}\right )}}{35 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/35*(5*cos(b*x + a)^7 - 7*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)*sin(2*b*x+2*a)**4,x)

[Out]

Timed out

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Giac [B]  time = 1.40043, size = 186, normalized size = 6. \begin{align*} -\frac{64 \,{\left (\frac{7 \,{\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac{14 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac{70 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{3}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{3}} + \frac{35 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{4}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{4}} + \frac{35 \,{\left (\cos \left (b x + a\right ) - 1\right )}^{5}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{5}} - 1\right )}}{35 \, b{\left (\frac{\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} - 1\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-64/35*(7*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 14*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 70*(cos(b*x +
a) - 1)^3/(cos(b*x + a) + 1)^3 + 35*(cos(b*x + a) - 1)^4/(cos(b*x + a) + 1)^4 + 35*(cos(b*x + a) - 1)^5/(cos(
b*x + a) + 1)^5 - 1)/(b*((cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 1)^7)