### 3.256 $$\int \frac{1}{\sqrt{-1-\tanh ^2(x)}} \, dx$$

Optimal. Leaf size=27 $\frac{\tan ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{-\tanh ^2(x)-1}}\right )}{\sqrt{2}}$

[Out]

ArcTan[(Sqrt[2]*Tanh[x])/Sqrt[-1 - Tanh[x]^2]]/Sqrt[2]

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Rubi [A]  time = 0.0201954, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.25, Rules used = {3661, 377, 203} $\frac{\tan ^{-1}\left (\frac{\sqrt{2} \tanh (x)}{\sqrt{-\tanh ^2(x)-1}}\right )}{\sqrt{2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/Sqrt[-1 - Tanh[x]^2],x]

[Out]

ArcTan[(Sqrt[2]*Tanh[x])/Sqrt[-1 - Tanh[x]^2]]/Sqrt[2]

Rule 3661

Int[((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x]
, x]}, Dist[(c*ff)/f, Subst[Int[(a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ
[{a, b, c, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

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

Rubi steps

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

Mathematica [A]  time = 0.0208512, size = 37, normalized size = 1.37 $\frac{\sinh ^{-1}\left (\sqrt{2} \sinh (x)\right ) \sqrt{\cosh (2 x)} \text{sech}(x)}{\sqrt{2} \sqrt{-\tanh ^2(x)-1}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/Sqrt[-1 - Tanh[x]^2],x]

[Out]

(ArcSinh[Sqrt[2]*Sinh[x]]*Sqrt[Cosh[2*x]]*Sech[x])/(Sqrt[2]*Sqrt[-1 - Tanh[x]^2])

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Maple [B]  time = 0.049, size = 66, normalized size = 2.4 \begin{align*}{\frac{\sqrt{2}}{4}\arctan \left ({\frac{ \left ( 2\,\tanh \left ( x \right ) -2 \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{- \left ( 1+\tanh \left ( x \right ) \right ) ^{2}+2\,\tanh \left ( x \right ) }}}} \right ) }-{\frac{\sqrt{2}}{4}\arctan \left ({\frac{ \left ( -2-2\,\tanh \left ( x \right ) \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{- \left ( \tanh \left ( x \right ) -1 \right ) ^{2}-2\,\tanh \left ( x \right ) }}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*2^(1/2)*arctan(1/4*(2*tanh(x)-2)*2^(1/2)/(-(1+tanh(x))^2+2*tanh(x))^(1/2))-1/4*2^(1/2)*arctan(1/4*(-2-2*ta
nh(x))*2^(1/2)/(-(tanh(x)-1)^2-2*tanh(x))^(1/2))

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

integrate(1/sqrt(-tanh(x)^2 - 1), x)

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Fricas [C]  time = 2.32529, size = 551, normalized size = 20.41 \begin{align*} \frac{1}{8} i \, \sqrt{2} \log \left (\frac{1}{2} \,{\left (i \, \sqrt{2} \sqrt{-2 \, e^{\left (4 \, x\right )} - 2} + 2 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )}\right ) - \frac{1}{8} i \, \sqrt{2} \log \left (\frac{1}{2} \,{\left (-i \, \sqrt{2} \sqrt{-2 \, e^{\left (4 \, x\right )} - 2} + 2 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )}\right ) - \frac{1}{8} i \, \sqrt{2} \log \left ({\left (\sqrt{-2 \, e^{\left (4 \, x\right )} - 2}{\left (e^{\left (2 \, x\right )} - 2\right )} + i \, \sqrt{2} e^{\left (4 \, x\right )} - i \, \sqrt{2} e^{\left (2 \, x\right )} + 2 i \, \sqrt{2}\right )} e^{\left (-4 \, x\right )}\right ) + \frac{1}{8} i \, \sqrt{2} \log \left ({\left (\sqrt{-2 \, e^{\left (4 \, x\right )} - 2}{\left (e^{\left (2 \, x\right )} - 2\right )} - i \, \sqrt{2} e^{\left (4 \, x\right )} + i \, \sqrt{2} e^{\left (2 \, x\right )} - 2 i \, \sqrt{2}\right )} e^{\left (-4 \, x\right )}\right ) \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]

1/8*I*sqrt(2)*log(1/2*(I*sqrt(2)*sqrt(-2*e^(4*x) - 2) + 2*e^(2*x) + 2)*e^(-2*x)) - 1/8*I*sqrt(2)*log(1/2*(-I*s
qrt(2)*sqrt(-2*e^(4*x) - 2) + 2*e^(2*x) + 2)*e^(-2*x)) - 1/8*I*sqrt(2)*log((sqrt(-2*e^(4*x) - 2)*(e^(2*x) - 2)
+ I*sqrt(2)*e^(4*x) - I*sqrt(2)*e^(2*x) + 2*I*sqrt(2))*e^(-4*x)) + 1/8*I*sqrt(2)*log((sqrt(-2*e^(4*x) - 2)*(e
^(2*x) - 2) - I*sqrt(2)*e^(4*x) + I*sqrt(2)*e^(2*x) - 2*I*sqrt(2))*e^(-4*x))

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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 = 1.33031, size = 96, normalized size = 3.56 \begin{align*} \frac{1}{4} \, \sqrt{2}{\left (i \, \log \left (-{\left (\sqrt{e^{\left (4 \, x\right )} + 1} + 1\right )} e^{\left (-2 \, x\right )}\right ) - i \, \log \left (-{\left (-i \, \sqrt{e^{\left (4 \, x\right )} + 1} - i\right )} e^{\left (-2 \, x\right )} + i\right ) + i \, \log \left (-{\left (-i \, \sqrt{e^{\left (4 \, x\right )} + 1} - i\right )} e^{\left (-2 \, x\right )} - i\right )\right )} \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]

1/4*sqrt(2)*(I*log(-(sqrt(e^(4*x) + 1) + 1)*e^(-2*x)) - I*log(-(-I*sqrt(e^(4*x) + 1) - I)*e^(-2*x) + I) + I*lo
g(-(-I*sqrt(e^(4*x) + 1) - I)*e^(-2*x) - I))