\(\int \frac {1}{\sqrt {-1-\tanh ^2(x)}} \, dx\) [256]

Optimal result

Integrand size = 12, antiderivative size = 27 \[ \int \frac {1}{\sqrt {-1-\tanh ^2(x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} \tanh (x)}{\sqrt {-1-\tanh ^2(x)}}\right )}{\sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, 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 = {3742, 385, 209} \[ \int \frac {1}{\sqrt {-1-\tanh ^2(x)}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} \tanh (x)}{\sqrt {-\tanh ^2(x)-1}}\right )}{\sqrt {2}} \]

[In]

Int[1/Sqrt[-1 - Tanh[x]^2],x]

[Out]

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

Rule 209

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

Rule 385

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 3742

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])

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\sqrt {-1-x^2} \left (1-x^2\right )} \, dx,x,\tanh (x)\right ) \\ & = \text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {\tanh (x)}{\sqrt {-1-\tanh ^2(x)}}\right ) \\ & = \frac {\arctan \left (\frac {\sqrt {2} \tanh (x)}{\sqrt {-1-\tanh ^2(x)}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {1}{\sqrt {-1-\tanh ^2(x)}} \, dx=\frac {\text {arcsinh}\left (\sqrt {2} \sinh (x)\right ) \sqrt {\cosh (2 x)} \text {sech}(x)}{\sqrt {2} \sqrt {-1-\tanh ^2(x)}} \]

[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])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(65\) vs. \(2(22)=44\).

Time = 0.17 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.44

method	result	size
derivativedivides	\(-\frac {\sqrt {2}\, \arctan \left (\frac {\left (-2-2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {-\left (\tanh \left (x \right )-1\right )^{2}-2 \tanh \left (x \right )}}\right )}{4}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (-2+2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {-\left (1+\tanh \left (x \right )\right )^{2}+2 \tanh \left (x \right )}}\right )}{4}\)	\(66\)
default	\(-\frac {\sqrt {2}\, \arctan \left (\frac {\left (-2-2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {-\left (\tanh \left (x \right )-1\right )^{2}-2 \tanh \left (x \right )}}\right )}{4}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (-2+2 \tanh \left (x \right )\right ) \sqrt {2}}{4 \sqrt {-\left (1+\tanh \left (x \right )\right )^{2}+2 \tanh \left (x \right )}}\right )}{4}\)	\(66\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.25 (sec) , antiderivative size = 175, normalized size of antiderivative = 6.48 \[ \int \frac {1}{\sqrt {-1-\tanh ^2(x)}} \, dx=\frac {1}{8} i \, \sqrt {2} \log \left (\frac {1}{2} \, {\left (i \, \sqrt {2} \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} + 2 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )}\right ) - \frac {1}{8} i \, \sqrt {2} \log \left (\frac {1}{2} \, {\left (-i \, \sqrt {2} \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} + 2 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )}\right ) - \frac {1}{8} i \, \sqrt {2} \log \left ({\left (\sqrt {-2 \, e^{\left (4 \, x\right )} - 2} {\left (e^{\left (2 \, x\right )} - 2\right )} + i \, \sqrt {2} e^{\left (4 \, x\right )} - i \, \sqrt {2} e^{\left (2 \, x\right )} + 2 i \, \sqrt {2}\right )} e^{\left (-4 \, x\right )}\right ) + \frac {1}{8} i \, \sqrt {2} \log \left ({\left (\sqrt {-2 \, e^{\left (4 \, x\right )} - 2} {\left (e^{\left (2 \, x\right )} - 2\right )} - i \, \sqrt {2} e^{\left (4 \, x\right )} + i \, \sqrt {2} e^{\left (2 \, x\right )} - 2 i \, \sqrt {2}\right )} e^{\left (-4 \, x\right )}\right ) \]

[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))

Sympy [F]

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

[In]

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

[Out]

Integral(1/sqrt(-tanh(x)**2 - 1), x)

Maxima [F]

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

[In]

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

[Out]

integrate(1/sqrt(-tanh(x)^2 - 1), x)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {1}{\sqrt {-1-\tanh ^2(x)}} \, dx=\frac {1}{4} i \, \sqrt {2} {\left (\log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) + \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) - \log \left (-\sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right )\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.87 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\sqrt {-1-\tanh ^2(x)}} \, dx=\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tanh}\left (x\right )}{\sqrt {-{\mathrm {tanh}\left (x\right )}^2-1}}\right )}{2} \]

[In]

int(1/(- tanh(x)^2 - 1)^(1/2),x)

[Out]

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