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]

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

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Rubi [A]  time = 0.02, antiderivative size = 14, normalized size of antiderivative = 1.00, 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 3770

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

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]

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*}

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Mathematica [B]  time = 0.01, size = 42, normalized size = 3.00 \[ \frac {1}{2} \left (\frac {\log \left (\sin \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b}-\frac {\log \left (\cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b}\right ) \]

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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fricas [B]  time = 0.60, size = 30, normalized size = 2.14 \[ -\frac {\log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) - \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right )}{4 \, b} \]

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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giac [A]  time = 0.22, size = 16, normalized size = 1.14 \[ \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) \right |}\right )}{2 \, b} \]

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

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maple [A]  time = 0.55, size = 22, normalized size = 1.57 \[ \frac {\ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{2 b} \]

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 = 0.34, size = 84, normalized size = 6.00 \[ -\frac {\log \left (\cos \left (b x\right )^{2} + 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} - 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right ) - \log \left (\cos \left (b x\right )^{2} - 2 \, \cos \left (b x\right ) \cos \relax (a) + \cos \relax (a)^{2} + \sin \left (b x\right )^{2} + 2 \, \sin \left (b x\right ) \sin \relax (a) + \sin \relax (a)^{2}\right )}{4 \, b} \]

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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mupad [B]  time = 0.02, size = 12, normalized size = 0.86 \[ -\frac {\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{2\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)/sin(2*a + 2*b*x),x)

[Out]

-atanh(cos(a + b*x))/(2*b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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