3.149 \(\int \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx\)

Optimal. Leaf size=42 \[ \frac{\tan (a+b x)}{16 b}-\frac{\cot ^3(a+b x)}{48 b}-\frac{\cot (a+b x)}{8 b} \]

[Out]

-Cot[a + b*x]/(8*b) - Cot[a + b*x]^3/(48*b) + Tan[a + b*x]/(16*b)

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Rubi [A]  time = 0.0595376, antiderivative size = 42, 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 = {4287, 2620, 270} \[ \frac{\tan (a+b x)}{16 b}-\frac{\cot ^3(a+b x)}{48 b}-\frac{\cot (a+b x)}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Cot[a + b*x]/(8*b) - Cot[a + b*x]^3/(48*b) + Tan[a + b*x]/(16*b)

Rule 4287

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_.)*sin[(c_.) + (d_.)*(x_)]^(p_.), x_Symbol] :> Dist[2^p/e^p, Int[(e*Cos
[a + b*x])^(m + p)*Sin[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2]
&& IntegerQ[p]

Rule 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

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 \cos ^2(a+b x) \csc ^4(2 a+2 b x) \, dx &=\frac{1}{16} \int \csc ^4(a+b x) \sec ^2(a+b x) \, dx\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^2}{x^4} \, dx,x,\tan (a+b x)\right )}{16 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}+\frac{2}{x^2}\right ) \, dx,x,\tan (a+b x)\right )}{16 b}\\ &=-\frac{\cot (a+b x)}{8 b}-\frac{\cot ^3(a+b x)}{48 b}+\frac{\tan (a+b x)}{16 b}\\ \end{align*}

Mathematica [A]  time = 0.0520714, size = 48, normalized size = 1.14 \[ \frac{\tan (a+b x)}{16 b}-\frac{5 \cot (a+b x)}{48 b}-\frac{\cot (a+b x) \csc ^2(a+b x)}{48 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*Cot[a + b*x])/(48*b) - (Cot[a + b*x]*Csc[a + b*x]^2)/(48*b) + Tan[a + b*x]/(16*b)

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Maple [A]  time = 0.03, size = 51, normalized size = 1.2 \begin{align*}{\frac{1}{16\,b} \left ( -{\frac{1}{3\, \left ( \sin \left ( bx+a \right ) \right ) ^{3}\cos \left ( bx+a \right ) }}+{\frac{4}{3\,\cos \left ( bx+a \right ) \sin \left ( bx+a \right ) }}-{\frac{8\,\cot \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/b*(-1/3/sin(b*x+a)^3/cos(b*x+a)+4/3/sin(b*x+a)/cos(b*x+a)-8/3*cot(b*x+a))

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Maxima [B]  time = 1.13744, size = 416, normalized size = 9.9 \begin{align*} \frac{{\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \sin \left (8 \, b x + 8 \, a\right ) - 2 \,{\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \sin \left (6 \, b x + 6 \, a\right ) - 2 \, \cos \left (8 \, b x + 8 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, \cos \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right )}{3 \,{\left (b \cos \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \cos \left (6 \, b x + 6 \, a\right )^{2} + 4 \, b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (8 \, b x + 8 \, a\right )^{2} + 4 \, b \sin \left (6 \, b x + 6 \, a\right )^{2} - 8 \, b \sin \left (6 \, b x + 6 \, a\right ) \sin \left (2 \, b x + 2 \, a\right ) + 4 \, b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \,{\left (2 \, b \cos \left (6 \, b x + 6 \, a\right ) - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )} \cos \left (8 \, b x + 8 \, a\right ) - 4 \,{\left (2 \, b \cos \left (2 \, b x + 2 \, a\right ) - b\right )} \cos \left (6 \, b x + 6 \, a\right ) - 4 \, b \cos \left (2 \, b x + 2 \, a\right ) - 4 \,{\left (b \sin \left (6 \, b x + 6 \, a\right ) - b \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (8 \, b x + 8 \, a\right ) + b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.46893, size = 136, normalized size = 3.24 \begin{align*} -\frac{8 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} + 3}{48 \,{\left (b \cos \left (b x + a\right )^{3} - b \cos \left (b x + a\right )\right )} \sin \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(8*cos(b*x + a)^4 - 12*cos(b*x + a)^2 + 3)/((b*cos(b*x + a)^3 - b*cos(b*x + a))*sin(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(cos(b*x+a)**2/sin(2*b*x+2*a)**4,x)

[Out]

Timed out

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Giac [A]  time = 1.26408, size = 47, normalized size = 1.12 \begin{align*} -\frac{\frac{6 \, \tan \left (b x + a\right )^{2} + 1}{\tan \left (b x + a\right )^{3}} - 3 \, \tan \left (b x + a\right )}{48 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*((6*tan(b*x + a)^2 + 1)/tan(b*x + a)^3 - 3*tan(b*x + a))/b