3.10 \(\int \sqrt {1-\coth ^2(x)} \, dx\)

Optimal. Leaf size=3 \[ \sin ^{-1}(\coth (x)) \]

[Out]

arcsin(coth(x))

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Rubi [A]  time = 0.02, antiderivative size = 3, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3657, 4122, 216} \[ \sin ^{-1}(\coth (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - Coth[x]^2],x]

[Out]

ArcSin[Coth[x]]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 4122

Int[((b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(b*ff)
/f, Subst[Int[(b + b*ff^2*x^2)^(p - 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{b, e, f, p}, x] &&  !IntegerQ[p
]

Rubi steps

\begin {align*} \int \sqrt {1-\coth ^2(x)} \, dx &=\int \sqrt {-\text {csch}^2(x)} \, dx\\ &=\operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2}} \, dx,x,\coth (x)\right )\\ &=\sin ^{-1}(\coth (x))\\ \end {align*}

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Mathematica [B]  time = 0.01, size = 20, normalized size = 6.67 \[ \sinh (x) \sqrt {-\text {csch}^2(x)} \log \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[1 - Coth[x]^2],x]

[Out]

Sqrt[-Csch[x]^2]*Log[Tanh[x/2]]*Sinh[x]

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fricas [A]  time = 0.48, size = 1, normalized size = 0.33 \[ 0 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-coth(x)^2)^(1/2),x, algorithm="fricas")

[Out]

0

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giac [C]  time = 0.14, size = 26, normalized size = 8.67 \[ {\left (i \, \log \left (e^{x} + 1\right ) - i \, \log \left ({\left | e^{x} - 1 \right |}\right )\right )} \mathrm {sgn}\left (-e^{\left (2 \, x\right )} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-coth(x)^2)^(1/2),x, algorithm="giac")

[Out]

(I*log(e^x + 1) - I*log(abs(e^x - 1)))*sgn(-e^(2*x) + 1)

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maple [A]  time = 0.08, size = 4, normalized size = 1.33 \[ \arcsin \left (\coth \relax (x )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-coth(x)^2)^(1/2),x)

[Out]

arcsin(coth(x))

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maxima [C]  time = 0.42, size = 19, normalized size = 6.33 \[ i \, \log \left (e^{\left (-x\right )} + 1\right ) - i \, \log \left (e^{\left (-x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-coth(x)^2)^(1/2),x, algorithm="maxima")

[Out]

I*log(e^(-x) + 1) - I*log(e^(-x) - 1)

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mupad [B]  time = 1.18, size = 3, normalized size = 1.00 \[ \mathrm {asin}\left (\mathrm {coth}\relax (x)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - coth(x)^2)^(1/2),x)

[Out]

asin(coth(x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {1 - \coth ^{2}{\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-coth(x)**2)**(1/2),x)

[Out]

Integral(sqrt(1 - coth(x)**2), x)

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