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

Optimal. Leaf size=60 \[ -\frac{8 \sin (a+b x) \cos ^5(a+b x)}{3 b}+\frac{2 \sin (a+b x) \cos ^3(a+b x)}{3 b}+\frac{\sin (a+b x) \cos (a+b x)}{b}+x \]

[Out]

x + (Cos[a + b*x]*Sin[a + b*x])/b + (2*Cos[a + b*x]^3*Sin[a + b*x])/(3*b) - (8*Cos[a + b*x]^5*Sin[a + b*x])/(3
*b)

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Rubi [A]  time = 0.0743127, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4288, 2568, 2635, 8} \[ -\frac{8 \sin (a+b x) \cos ^5(a+b x)}{3 b}+\frac{2 \sin (a+b x) \cos ^3(a+b x)}{3 b}+\frac{\sin (a+b x) \cos (a+b x)}{b}+x \]

Antiderivative was successfully verified.

[In]

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

[Out]

x + (Cos[a + b*x]*Sin[a + b*x])/b + (2*Cos[a + b*x]^3*Sin[a + b*x])/(3*b) - (8*Cos[a + b*x]^5*Sin[a + b*x])/(3
*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 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[2*
m, 2*n]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \csc ^2(a+b x) \sin ^4(2 a+2 b x) \, dx &=16 \int \cos ^4(a+b x) \sin ^2(a+b x) \, dx\\ &=-\frac{8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+\frac{8}{3} \int \cos ^4(a+b x) \, dx\\ &=\frac{2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac{8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+2 \int \cos ^2(a+b x) \, dx\\ &=\frac{\cos (a+b x) \sin (a+b x)}{b}+\frac{2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac{8 \cos ^5(a+b x) \sin (a+b x)}{3 b}+\int 1 \, dx\\ &=x+\frac{\cos (a+b x) \sin (a+b x)}{b}+\frac{2 \cos ^3(a+b x) \sin (a+b x)}{3 b}-\frac{8 \cos ^5(a+b x) \sin (a+b x)}{3 b}\\ \end{align*}

Mathematica [A]  time = 0.099401, size = 40, normalized size = 0.67 \[ -\frac{-3 \sin (2 (a+b x))+3 \sin (4 (a+b x))+\sin (6 (a+b x))-12 b x}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-12*b*x - 3*Sin[2*(a + b*x)] + 3*Sin[4*(a + b*x)] + Sin[6*(a + b*x)])/(12*b)

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Maple [A]  time = 0.05, size = 55, normalized size = 0.9 \begin{align*} 16\,{\frac{-1/6\,\sin \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{5}+1/24\, \left ( \left ( \cos \left ( bx+a \right ) \right ) ^{3}+3/2\,\cos \left ( bx+a \right ) \right ) \sin \left ( bx+a \right ) +1/16\,bx+a/16}{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)^4,x)

[Out]

16/b*(-1/6*sin(b*x+a)*cos(b*x+a)^5+1/24*(cos(b*x+a)^3+3/2*cos(b*x+a))*sin(b*x+a)+1/16*b*x+1/16*a)

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Maxima [A]  time = 1.19902, size = 58, normalized size = 0.97 \begin{align*} \frac{12 \, b x - \sin \left (6 \, b x + 6 \, a\right ) - 3 \, \sin \left (4 \, b x + 4 \, a\right ) + 3 \, \sin \left (2 \, b x + 2 \, a\right )}{12 \, 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)^4,x, algorithm="maxima")

[Out]

1/12*(12*b*x - sin(6*b*x + 6*a) - 3*sin(4*b*x + 4*a) + 3*sin(2*b*x + 2*a))/b

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Fricas [A]  time = 0.485046, size = 115, normalized size = 1.92 \begin{align*} \frac{3 \, b x -{\left (8 \, \cos \left (b x + a\right )^{5} - 2 \, \cos \left (b x + a\right )^{3} - 3 \, \cos \left (b x + a\right )\right )} \sin \left (b x + a\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)^4,x, algorithm="fricas")

[Out]

1/3*(3*b*x - (8*cos(b*x + a)^5 - 2*cos(b*x + a)^3 - 3*cos(b*x + a))*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)**2*sin(2*b*x+2*a)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.38985, size = 74, normalized size = 1.23 \begin{align*} \frac{3 \, b x + 3 \, a + \frac{3 \, \tan \left (b x + a\right )^{5} + 8 \, \tan \left (b x + a\right )^{3} - 3 \, \tan \left (b x + a\right )}{{\left (\tan \left (b x + a\right )^{2} + 1\right )}^{3}}}{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)^4,x, algorithm="giac")

[Out]

1/3*(3*b*x + 3*a + (3*tan(b*x + a)^5 + 8*tan(b*x + a)^3 - 3*tan(b*x + a))/(tan(b*x + a)^2 + 1)^3)/b