\(\int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx\) [147]

Optimal result

Integrand size = 20, antiderivative size = 13 \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\cot (a+b x)}{4 b} \]

[Out]

-1/4*cot(b*x+a)/b

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4372, 3852, 8} \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\cot (a+b x)}{4 b} \]

[In]

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

[Out]

-1/4*Cot[a + b*x]/b

Rule 8

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

Rule 3852

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 4372

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \int \csc ^2(a+b x) \, dx \\ & = -\frac {\text {Subst}(\int 1 \, dx,x,\cot (a+b x))}{4 b} \\ & = -\frac {\cot (a+b x)}{4 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\cot (a+b x)}{4 b} \]

[In]

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

[Out]

-1/4*Cot[a + b*x]/b

Maple [A] (verified)

Time = 1.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {\cot \left (x b +a \right )}{4 b}\)	\(12\)
risch	\(-\frac {i}{2 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}\)	\(20\)

[In]

int(cos(b*x+a)^2/sin(2*b*x+2*a)^2,x,method=_RETURNVERBOSE)

[Out]

-1/4*cot(b*x+a)/b

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\cos \left (b x + a\right )}{4 \, b \sin \left (b x + a\right )} \]

[In]

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

[Out]

-1/4*cos(b*x + a)/(b*sin(b*x + a))

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)**2/sin(2*b*x+2*a)**2,x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 53 vs. \(2 (11) = 22\).

Time = 0.24 (sec) , antiderivative size = 53, normalized size of antiderivative = 4.08 \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\sin \left (2 \, b x + 2 \, a\right )}{2 \, {\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {1}{4 \, b \tan \left (b x + a\right )} \]

[In]

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

[Out]

-1/4/(b*tan(b*x + a))

Mupad [B] (verification not implemented)

Time = 19.81 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \cos ^2(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\mathrm {cot}\left (a+b\,x\right )}{4\,b} \]

[In]

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

[Out]

-cot(a + b*x)/(4*b)