3.3.6 \(\int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx\) [206]

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 = 0.01, 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 = {3738, 4207, 197} \begin {gather*} \frac {\tanh (x)}{\sqrt {\text {sech}^2(x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[1 - Tanh[x]^2],x]

[Out]

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

Rule 197

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 3738

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 4207

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\\ &=\text {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 \begin {gather*} \frac {\tanh (x)}{\sqrt {\text {sech}^2(x)}} \end {gather*}

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 = 0.30, size = 14, normalized size = 1.27

method	result	size
derivativedivides	\(\frac {\tanh \left (x \right )}{\sqrt {1-\left (\tanh ^{2}\left (x \right )\right )}}\)	\(14\)
default	\(\frac {\tanh \left (x \right )}{\sqrt {1-\left (\tanh ^{2}\left (x \right )\right )}}\)	\(14\)
risch	\(\frac {{\mathrm e}^{2 x}}{2 \sqrt {\frac {{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}\, \left (1+{\mathrm e}^{2 x}\right )}-\frac {1}{2 \left (1+{\mathrm e}^{2 x}\right ) \sqrt {\frac {{\mathrm e}^{2 x}}{\left (1+{\mathrm e}^{2 x}\right )^{2}}}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-tanh(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \end {gather*}

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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Fricas [A]
time = 0.38, size = 2, normalized size = 0.18 \begin {gather*} \sinh \left (x\right ) \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {1 - \tanh ^{2}{\left (x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(1 - tanh(x)**2), x)

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Giac [A]
time = 0.41, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \end {gather*}

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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Mupad [B]
time = 0.14, size = 2, normalized size = 0.18 \begin {gather*} \mathrm {sinh}\left (x\right ) \end {gather*}

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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