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

Optimal. Leaf size=11 \[ \frac {\tanh (x)}{\sqrt {\text {sech}^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 11, 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, 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 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]

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 \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*}

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Mathematica [A]  time = 0.01, size = 11, normalized size = 1.00 \[ \frac {\tanh (x)}{\sqrt {\text {sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[1 - Tanh[x]^2],x]

[Out]

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

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fricas [A]  time = 0.55, size = 2, normalized size = 0.18 \[ \sinh \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sinh(x)

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giac [A]  time = 0.14, size = 11, normalized size = 1.00 \[ -\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

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*e^(-x) + 1/2*e^x

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maple [A]  time = 0.05, size = 14, normalized size = 1.27 \[ \frac {\tanh \relax (x )}{\sqrt {1-\left (\tanh ^{2}\relax (x )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 11, normalized size = 1.00 \[ -\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

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^(-x) + 1/2*e^x

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mupad [B]  time = 0.14, size = 2, normalized size = 0.18 \[ \mathrm {sinh}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - tanh(x)^2)^(1/2),x)

[Out]

sinh(x)

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sympy [A]  time = 0.41, size = 12, normalized size = 1.09 \[ \frac {\tanh {\relax (x )}}{\sqrt {1 - \tanh ^{2}{\relax (x )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tanh(x)/sqrt(1 - tanh(x)**2)

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