\(\int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx\) [157]

Optimal result

Integrand size = 20, antiderivative size = 13 \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\csc (a+b x)}{4 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (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, 2686, 8} \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\csc (a+b x)}{4 b} \]

[In]

Int[Cos[a + b*x]^3*Csc[2*a + 2*b*x]^2,x]

[Out]

-1/4*Csc[a + b*x]/b

Rule 8

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

Rule 2686

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\csc (a+b x)}{4 b} \]

[In]

Integrate[Cos[a + b*x]^3*Csc[2*a + 2*b*x]^2,x]

[Out]

-1/4*Csc[a + b*x]/b

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
default	\(-\frac {1}{4 \sin \left (x b +a \right ) b}\)	\(14\)
risch	\(-\frac {i {\mathrm e}^{i \left (x b +a \right )}}{2 b \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}\)	\(29\)

[In]

int(cos(b*x+a)^3/sin(2*b*x+2*a)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {1}{4 \, b \sin \left (b x + a\right )} \]

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^2,x, algorithm="fricas")

[Out]

-1/4/(b*sin(b*x + a))

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

integrate(cos(b*x+a)**3/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. 84 vs. \(2 (11) = 22\).

Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 6.46 \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {\cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right ) - \cos \left (2 \, b x + 2 \, a\right ) \sin \left (b x + a\right ) + \sin \left (b x + a\right )}{2 \, {\left (b \cos \left (2 \, b x + 2 \, a\right )^{2} + b \sin \left (2 \, b x + 2 \, a\right )^{2} - 2 \, b \cos \left (2 \, b x + 2 \, a\right ) + b\right )}} \]

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^2,x, algorithm="maxima")

[Out]

-1/2*(cos(b*x + a)*sin(2*b*x + 2*a) - cos(2*b*x + 2*a)*sin(b*x + a) + sin(b*x + 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 \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {1}{4 \, b \sin \left (b x + a\right )} \]

[In]

integrate(cos(b*x+a)^3/sin(2*b*x+2*a)^2,x, algorithm="giac")

[Out]

-1/4/(b*sin(b*x + a))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \cos ^3(a+b x) \csc ^2(2 a+2 b x) \, dx=-\frac {1}{4\,b\,\sin \left (a+b\,x\right )} \]

[In]

int(cos(a + b*x)^3/sin(2*a + 2*b*x)^2,x)

[Out]

-1/(4*b*sin(a + b*x))