\(\int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx\) [146]

Optimal result

Integrand size = 18, antiderivative size = 14 \[ \int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx=\frac {\log (\sin (a+b x))}{2 b} \]

[Out]

1/2*ln(sin(b*x+a))/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4372, 3556} \[ \int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx=\frac {\log (\sin (a+b x))}{2 b} \]

[In]

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

[Out]

Log[Sin[a + b*x]]/(2*b)

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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}{2} \int \cot (a+b x) \, dx \\ & = \frac {\log (\sin (a+b x))}{2 b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.93 \[ \int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx=\frac {1}{2} \left (\frac {\log (\cos (a+b x))}{b}+\frac {\log (\tan (a+b x))}{b}\right ) \]

[In]

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

[Out]

(Log[Cos[a + b*x]]/b + Log[Tan[a + b*x]]/b)/2

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {\ln \left (\sin \left (x b +a \right )\right )}{2 b}\)	\(13\)
risch	\(-\frac {i x}{2}-\frac {i a}{b}+\frac {\ln \left ({\mathrm e}^{2 i \left (x b +a \right )}-1\right )}{2 b}\)	\(30\)

[In]

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

[Out]

1/2*ln(sin(b*x+a))/b

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx=\frac {\log \left (\frac {1}{2} \, \sin \left (b x + a\right )\right )}{2 \, b} \]

[In]

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

[Out]

1/2*log(1/2*sin(b*x + a))/b

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (12) = 24\).

Time = 0.23 (sec) , antiderivative size = 82, normalized size of antiderivative = 5.86 \[ \int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx=\frac {\log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\right ) + \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \left (a\right ) + \cos \left (a\right )^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \left (a\right ) + \sin \left (a\right )^{2}\right )}{4 \, b} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93 \[ \int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx=\frac {\log \left ({\left | \sin \left (b x + a\right ) \right |}\right )}{2 \, b} \]

[In]

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

[Out]

1/2*log(abs(sin(b*x + a)))/b

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \cos ^2(a+b x) \csc (2 a+2 b x) \, dx=\frac {\ln \left ({\sin \left (a+b\,x\right )}^2\right )}{4\,b} \]

[In]

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

[Out]

log(sin(a + b*x)^2)/(4*b)