3.207 $$\int \frac{1}{\sqrt{-1+\tanh ^2(x)}} \, dx$$

Optimal. Leaf size=13 $\frac{\tanh (x)}{\sqrt{-\text{sech}^2(x)}}$

[Out]

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

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Rubi [A]  time = 0.0210973, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.3, Rules used = {3657, 4122, 191} $\frac{\tanh (x)}{\sqrt{-\text{sech}^2(x)}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/Sqrt[-1 + Tanh[x]^2],x]

[Out]

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

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
]

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-1+\tanh ^2(x)}} \, dx &=\int \frac{1}{\sqrt{-\text{sech}^2(x)}} \, dx\\ &=-\operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right )^{3/2}} \, dx,x,\tanh (x)\right )\\ &=\frac{\tanh (x)}{\sqrt{-\text{sech}^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.0067899, size = 13, normalized size = 1. $\frac{\tanh (x)}{\sqrt{-\text{sech}^2(x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/Sqrt[-1 + Tanh[x]^2],x]

[Out]

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

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Maple [A]  time = 0.014, size = 12, normalized size = 0.9 \begin{align*}{\tanh \left ( x \right ){\frac{1}{\sqrt{-1+ \left ( \tanh \left ( x \right ) \right ) ^{2}}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.5597, size = 34, normalized size = 2.62 \begin{align*} -\frac{e^{\left (-2 \, x\right )}}{2 \, \sqrt{-e^{\left (-2 \, x\right )}}} + \frac{1}{2 \, \sqrt{-e^{\left (-2 \, x\right )}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e^(-2*x)/sqrt(-e^(-2*x)) + 1/2/sqrt(-e^(-2*x))

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Fricas [A]  time = 2.28971, size = 4, normalized size = 0.31 \begin{align*} 0 \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\tanh ^{2}{\left (x \right )} - 1}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(tanh(x)**2 - 1), x)

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Giac [C]  time = 1.18516, size = 15, normalized size = 1.15 \begin{align*} \frac{1}{2} i \, e^{\left (-x\right )} - \frac{1}{2} i \, e^{x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*e^(-x) - 1/2*I*e^x