3.64 \(\int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx\)

Optimal. Leaf size=46 \[ \frac{32 \sin ^7(a+b x)}{7 b}-\frac{64 \sin ^5(a+b x)}{5 b}+\frac{32 \sin ^3(a+b x)}{3 b} \]

[Out]

(32*Sin[a + b*x]^3)/(3*b) - (64*Sin[a + b*x]^5)/(5*b) + (32*Sin[a + b*x]^7)/(7*b)

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Rubi [A]  time = 0.0656032, antiderivative size = 46, 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, 2564, 270} \[ \frac{32 \sin ^7(a+b x)}{7 b}-\frac{64 \sin ^5(a+b x)}{5 b}+\frac{32 \sin ^3(a+b x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(32*Sin[a + b*x]^3)/(3*b) - (64*Sin[a + b*x]^5)/(5*b) + (32*Sin[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 2564

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \csc ^3(a+b x) \sin ^5(2 a+2 b x) \, dx &=32 \int \cos ^5(a+b x) \sin ^2(a+b x) \, dx\\ &=\frac{32 \operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{32 \operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (a+b x)\right )}{b}\\ &=\frac{32 \sin ^3(a+b x)}{3 b}-\frac{64 \sin ^5(a+b x)}{5 b}+\frac{32 \sin ^7(a+b x)}{7 b}\\ \end{align*}

Mathematica [A]  time = 0.108067, size = 37, normalized size = 0.8 \[ \frac{4 \sin ^3(a+b x) (108 \cos (2 (a+b x))+15 \cos (4 (a+b x))+157)}{105 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^5,x]

[Out]

(4*(157 + 108*Cos[2*(a + b*x)] + 15*Cos[4*(a + b*x)])*Sin[a + b*x]^3)/(105*b)

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Maple [A]  time = 0.029, size = 51, normalized size = 1.1 \begin{align*} 32\,{\frac{-1/7\,\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{6}+1/35\, \left ( 8/3+ \left ( \cos \left ( bx+a \right ) \right ) ^{4}+4/3\, \left ( \cos \left ( bx+a \right ) \right ) ^{2} \right ) \sin \left ( bx+a \right ) }{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)^5,x)

[Out]

32/b*(-1/7*sin(b*x+a)*cos(b*x+a)^6+1/35*(8/3+cos(b*x+a)^4+4/3*cos(b*x+a)^2)*sin(b*x+a))

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Maxima [A]  time = 1.03676, size = 63, normalized size = 1.37 \begin{align*} -\frac{15 \, \sin \left (7 \, b x + 7 \, a\right ) + 63 \, \sin \left (5 \, b x + 5 \, a\right ) + 35 \, \sin \left (3 \, b x + 3 \, a\right ) - 525 \, \sin \left (b x + a\right )}{210 \, 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)^5,x, algorithm="maxima")

[Out]

-1/210*(15*sin(7*b*x + 7*a) + 63*sin(5*b*x + 5*a) + 35*sin(3*b*x + 3*a) - 525*sin(b*x + a))/b

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Fricas [A]  time = 0.484142, size = 116, normalized size = 2.52 \begin{align*} -\frac{32 \,{\left (15 \, \cos \left (b x + a\right )^{6} - 3 \, \cos \left (b x + a\right )^{4} - 4 \, \cos \left (b x + a\right )^{2} - 8\right )} \sin \left (b x + a\right )}{105 \, 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)^5,x, algorithm="fricas")

[Out]

-32/105*(15*cos(b*x + a)^6 - 3*cos(b*x + a)^4 - 4*cos(b*x + a)^2 - 8)*sin(b*x + a)/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)**5,x)

[Out]

Timed out

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Giac [A]  time = 1.36548, size = 49, normalized size = 1.07 \begin{align*} \frac{32 \,{\left (15 \, \sin \left (b x + a\right )^{7} - 42 \, \sin \left (b x + a\right )^{5} + 35 \, \sin \left (b x + a\right )^{3}\right )}}{105 \, 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)^5,x, algorithm="giac")

[Out]

32/105*(15*sin(b*x + a)^7 - 42*sin(b*x + a)^5 + 35*sin(b*x + a)^3)/b