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

Optimal. Leaf size=21 \[ \frac{2 \sin (a+b x) \cos (a+b x)}{b}+2 x \]

[Out]

2*x + (2*Cos[a + b*x]*Sin[a + b*x])/b

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Rubi [A]  time = 0.0343239, antiderivative size = 21, 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, 2635, 8} \[ \frac{2 \sin (a+b x) \cos (a+b x)}{b}+2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

2*x + (2*Cos[a + b*x]*Sin[a + b*x])/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 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 ^2(2 a+2 b x) \, dx &=4 \int \cos ^2(a+b x) \, dx\\ &=\frac{2 \cos (a+b x) \sin (a+b x)}{b}+2 \int 1 \, dx\\ &=2 x+\frac{2 \cos (a+b x) \sin (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.0243541, size = 20, normalized size = 0.95 \[ \frac{2 (a+b x)+\sin (2 (a+b x))}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*x) + Sin[2*(a + b*x)])/b

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Maple [A]  time = 0.025, size = 28, normalized size = 1.3 \begin{align*} 4\,{\frac{1/2\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) +1/2\,bx+a/2}{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)^2,x)

[Out]

4/b*(1/2*cos(b*x+a)*sin(b*x+a)+1/2*b*x+1/2*a)

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Maxima [A]  time = 1.12614, size = 24, normalized size = 1.14 \begin{align*} \frac{2 \, b x + \sin \left (2 \, b x + 2 \, a\right )}{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)^2,x, algorithm="maxima")

[Out]

(2*b*x + sin(2*b*x + 2*a))/b

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Fricas [A]  time = 0.476171, size = 53, normalized size = 2.52 \begin{align*} \frac{2 \,{\left (b x + \cos \left (b x + a\right ) \sin \left (b x + a\right )\right )}}{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)^2,x, algorithm="fricas")

[Out]

2*(b*x + 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)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.29708, size = 39, normalized size = 1.86 \begin{align*} \frac{2 \,{\left (b x + a + \frac{\tan \left (b x + a\right )}{\tan \left (b x + a\right )^{2} + 1}\right )}}{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)^2,x, algorithm="giac")

[Out]

2*(b*x + a + tan(b*x + a)/(tan(b*x + a)^2 + 1))/b