3.203 \(\int \sqrt {-1+\tanh ^2(x)} \, dx\)

Optimal. Leaf size=16 \[ -\tanh ^{-1}\left (\frac {\tanh (x)}{\sqrt {-\text {sech}^2(x)}}\right ) \]

[Out]

-arctanh(tanh(x)/(-sech(x)^2)^(1/2))

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Rubi [A]  time = 0.02, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3657, 4122, 217, 206} \[ -\tanh ^{-1}\left (\frac {\tanh (x)}{\sqrt {-\text {sech}^2(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + Tanh[x]^2],x]

[Out]

-ArcTanh[Tanh[x]/Sqrt[-Sech[x]^2]]

Rule 206

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

Rule 217

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

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+\tanh ^2(x)} \, dx &=\int \sqrt {-\text {sech}^2(x)} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {1}{\sqrt {-1+x^2}} \, dx,x,\tanh (x)\right )\\ &=-\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {-\text {sech}^2(x)}}\right )\\ &=-\tanh ^{-1}\left (\frac {\tanh (x)}{\sqrt {-\text {sech}^2(x)}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.31 \[ 2 \cosh (x) \sqrt {-\text {sech}^2(x)} \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + Tanh[x]^2],x]

[Out]

2*ArcTan[Tanh[x/2]]*Cosh[x]*Sqrt[-Sech[x]^2]

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fricas [A]  time = 2.08, size = 1, normalized size = 0.06 \[ 0 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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giac [A]  time = 0.13, size = 1, normalized size = 0.06 \[ 0 \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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maple [A]  time = 0.11, size = 15, normalized size = 0.94 \[ -\ln \left (\tanh \relax (x )+\sqrt {-1+\tanh ^{2}\relax (x )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1+tanh(x)^2)^(1/2),x)

[Out]

-ln(tanh(x)+(-1+tanh(x)^2)^(1/2))

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maxima [C]  time = 0.49, size = 5, normalized size = 0.31 \[ 2 i \, \arctan \left (e^{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I*arctan(e^x)

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mupad [B]  time = 0.25, size = 14, normalized size = 0.88 \[ -\ln \left (\mathrm {tanh}\relax (x)+\sqrt {{\mathrm {tanh}\relax (x)}^2-1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((tanh(x)^2 - 1)^(1/2),x)

[Out]

-log(tanh(x) + (tanh(x)^2 - 1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1+tanh(x)**2)**(1/2),x)

[Out]

Integral(sqrt(tanh(x)**2 - 1), x)

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