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

Optimal. Leaf size=13 \[ \frac{\tan (a+b x)}{4 b} \]

[Out]

Tan[a + b*x]/(4*b)

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Rubi [A]  time = 0.034776, antiderivative size = 13, 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, 3767, 8} \[ \frac{\tan (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[a + b*x]/(4*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 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

\begin{align*} \int \csc ^2(2 a+2 b x) \sin ^2(a+b x) \, dx &=\frac{1}{4} \int \sec ^2(a+b x) \, dx\\ &=-\frac{\operatorname{Subst}(\int 1 \, dx,x,-\tan (a+b x))}{4 b}\\ &=\frac{\tan (a+b x)}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.0075078, size = 13, normalized size = 1. \[ \frac{\tan (a+b x)}{4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Tan[a + b*x]/(4*b)

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Maple [A]  time = 0.091, size = 12, normalized size = 0.9 \begin{align*}{\frac{\tan \left ( bx+a \right ) }{4\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*tan(b*x+a)/b

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Maxima [B]  time = 1.11319, size = 72, normalized size = 5.54 \begin{align*} \frac{\sin \left (2 \, b x + 2 \, a\right )}{2 \,{\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.453275, size = 47, normalized size = 3.62 \begin{align*} \frac{\sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sin(b*x + a)/(b*cos(b*x + a))

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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(2*b*x+2*a)**2*sin(b*x+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.47716, size = 205, normalized size = 15.77 \begin{align*} -\frac{\tan \left (\frac{1}{2} \, a\right )^{12} + 6 \, \tan \left (\frac{1}{2} \, a\right )^{10} + 15 \, \tan \left (\frac{1}{2} \, a\right )^{8} + 20 \, \tan \left (\frac{1}{2} \, a\right )^{6} + 15 \, \tan \left (\frac{1}{2} \, a\right )^{4} + 6 \, \tan \left (\frac{1}{2} \, a\right )^{2} + 1}{8 \,{\left (6 \, \tan \left (b x + 4 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{5} - \tan \left (\frac{1}{2} \, a\right )^{6} - 20 \, \tan \left (b x + 4 \, a\right ) \tan \left (\frac{1}{2} \, a\right )^{3} + 15 \, \tan \left (\frac{1}{2} \, a\right )^{4} + 6 \, \tan \left (b x + 4 \, a\right ) \tan \left (\frac{1}{2} \, a\right ) - 15 \, \tan \left (\frac{1}{2} \, a\right )^{2} + 1\right )}{\left (3 \, \tan \left (\frac{1}{2} \, a\right )^{5} - 10 \, \tan \left (\frac{1}{2} \, a\right )^{3} + 3 \, \tan \left (\frac{1}{2} \, a\right )\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(tan(1/2*a)^12 + 6*tan(1/2*a)^10 + 15*tan(1/2*a)^8 + 20*tan(1/2*a)^6 + 15*tan(1/2*a)^4 + 6*tan(1/2*a)^2 +
1)/((6*tan(b*x + 4*a)*tan(1/2*a)^5 - tan(1/2*a)^6 - 20*tan(b*x + 4*a)*tan(1/2*a)^3 + 15*tan(1/2*a)^4 + 6*tan(
b*x + 4*a)*tan(1/2*a) - 15*tan(1/2*a)^2 + 1)*(3*tan(1/2*a)^5 - 10*tan(1/2*a)^3 + 3*tan(1/2*a))*b)