3.136 \(\int \cos (a+b x) \csc (2 a+2 b x) \, dx\)

Optimal. Leaf size=14 \[ -\frac{\tanh ^{-1}(\cos (a+b x))}{2 b} \]

[Out]

-ArcTanh[Cos[a + b*x]]/(2*b)

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Rubi [A]  time = 0.0163863, antiderivative size = 14, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {4287, 3770} \[ -\frac{\tanh ^{-1}(\cos (a+b x))}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]*Csc[2*a + 2*b*x],x]

[Out]

-ArcTanh[Cos[a + b*x]]/(2*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 3770

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

Rubi steps

\begin{align*} \int \cos (a+b x) \csc (2 a+2 b x) \, dx &=\frac{1}{2} \int \csc (a+b x) \, dx\\ &=-\frac{\tanh ^{-1}(\cos (a+b x))}{2 b}\\ \end{align*}

Mathematica [B]  time = 0.0124396, size = 42, normalized size = 3. \[ \frac{1}{2} \left (\frac{\log \left (\sin \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}-\frac{\log \left (\cos \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.019, size = 22, normalized size = 1.6 \begin{align*}{\frac{\ln \left ( \csc \left ( bx+a \right ) -\cot \left ( bx+a \right ) \right ) }{2\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/b*ln(csc(b*x+a)-cot(b*x+a))

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Maxima [B]  time = 1.15764, size = 113, normalized size = 8.07 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/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

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Fricas [B]  time = 0.491409, size = 93, normalized size = 6.64 \begin{align*} -\frac{\log \left (\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right ) - \log \left (-\frac{1}{2} \, \cos \left (b x + a\right ) + \frac{1}{2}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.19284, size = 22, normalized size = 1.57 \begin{align*} \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) \right |}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(abs(tan(1/2*b*x + 1/2*a)))/b