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

Optimal result

Integrand size = 12, antiderivative size = 11 \[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=\frac {\tanh (x)}{\sqrt {\text {sech}^2(x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3738, 4207, 197} \[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=\frac {\tanh (x)}{\sqrt {\text {sech}^2(x)}} \]

[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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=\frac {\tanh (x)}{\sqrt {\text {sech}^2(x)}} \]

[In]

Integrate[1/Sqrt[1 - Tanh[x]^2],x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.27

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.18 \[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=\sinh \left (x\right ) \]

[In]

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

[Out]

sinh(x)

Sympy [F]

\[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=\int \frac {1}{\sqrt {1 - \tanh ^{2}{\left (x \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(1 - tanh(x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=-\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

[In]

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

[Out]

-1/2*e^(-x) + 1/2*e^x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=-\frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

[In]

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

[Out]

-1/2*e^(-x) + 1/2*e^x

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 2, normalized size of antiderivative = 0.18 \[ \int \frac {1}{\sqrt {1-\tanh ^2(x)}} \, dx=\mathrm {sinh}\left (x\right ) \]

[In]

int(1/(1 - tanh(x)^2)^(1/2),x)

[Out]

sinh(x)