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

Optimal. Leaf size=45 \[ \tan ^{-1}\left (\frac {\tanh (x)}{\sqrt {-\tanh ^2(x)-1}}\right )-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \tanh (x)}{\sqrt {-\tanh ^2(x)-1}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3661, 402, 217, 203, 377} \[ \tan ^{-1}\left (\frac {\tanh (x)}{\sqrt {-\tanh ^2(x)-1}}\right )-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} \tanh (x)}{\sqrt {-\tanh ^2(x)-1}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 - Tanh[x]^2],x]

[Out]

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

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])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

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])

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 53, normalized size = 1.18 \[ \frac {\cosh (x) \sqrt {-\tanh ^2(x)-1} \left (\sqrt {2} \sinh ^{-1}\left (\sqrt {2} \sinh (x)\right )-\tanh ^{-1}\left (\frac {\sinh (x)}{\sqrt {\cosh (2 x)}}\right )\right )}{\sqrt {\cosh (2 x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 - Tanh[x]^2],x]

[Out]

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

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fricas [C]  time = 0.43, size = 228, normalized size = 5.07 \[ -\frac {1}{4} \, \sqrt {-2} \log \left (-{\left (\sqrt {-2} \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} + 2 \, e^{\left (2 \, x\right )} + 2\right )} e^{\left (-2 \, x\right )}\right ) + \frac {1}{4} \, \sqrt {-2} \log \left ({\left (\sqrt {-2} \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} - 2 \, e^{\left (2 \, x\right )} - 2\right )} e^{\left (-2 \, x\right )}\right ) + \frac {1}{4} \, \sqrt {-2} \log \left (-2 \, {\left (\sqrt {-2 \, e^{\left (4 \, x\right )} - 2} {\left (e^{\left (2 \, x\right )} - 2\right )} + \sqrt {-2} e^{\left (4 \, x\right )} - \sqrt {-2} e^{\left (2 \, x\right )} + 2 \, \sqrt {-2}\right )} e^{\left (-4 \, x\right )}\right ) - \frac {1}{4} \, \sqrt {-2} \log \left (-2 \, {\left (\sqrt {-2 \, e^{\left (4 \, x\right )} - 2} {\left (e^{\left (2 \, x\right )} - 2\right )} - \sqrt {-2} e^{\left (4 \, x\right )} + \sqrt {-2} e^{\left (2 \, x\right )} - 2 \, \sqrt {-2}\right )} e^{\left (-4 \, x\right )}\right ) - \frac {1}{2} i \, \log \left ({\left (4 i \, \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} - 4 \, e^{\left (2 \, x\right )} + 4\right )} e^{\left (-2 \, x\right )}\right ) + \frac {1}{2} i \, \log \left ({\left (-4 i \, \sqrt {-2 \, e^{\left (4 \, x\right )} - 2} - 4 \, e^{\left (2 \, x\right )} + 4\right )} e^{\left (-2 \, x\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [C]  time = 0.17, size = 104, normalized size = 2.31 \[ -\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} \log \left (\frac {\sqrt {2} - \sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1}{\sqrt {2} + \sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} - 1}\right ) + \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )} + 1\right ) + \log \left (\sqrt {e^{\left (4 \, x\right )} + 1} - e^{\left (2 \, x\right )}\right ) - \log \left (-\sqrt {e^{\left (4 \, x\right )} + 1} + e^{\left (2 \, x\right )} + 1\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.12, size = 142, normalized size = 3.16 \[ -\frac {\sqrt {-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )}}{2}+\frac {\arctan \left (\frac {\tanh \relax (x )}{\sqrt {-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )}}\right )}{2}+\frac {\sqrt {2}\, \arctan \left (\frac {\left (-2-2 \tanh \relax (x )\right ) \sqrt {2}}{4 \sqrt {-\left (\tanh \relax (x )-1\right )^{2}-2 \tanh \relax (x )}}\right )}{2}+\frac {\sqrt {-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )}}{2}+\frac {\arctan \left (\frac {\tanh \relax (x )}{\sqrt {-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )}}\right )}{2}-\frac {\sqrt {2}\, \arctan \left (\frac {\left (-2+2 \tanh \relax (x )\right ) \sqrt {2}}{4 \sqrt {-\left (1+\tanh \relax (x )\right )^{2}+2 \tanh \relax (x )}}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {-\tanh \relax (x)^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-tanh(x)^2 - 1), x)

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mupad [B]  time = 1.34, size = 43, normalized size = 0.96 \[ -\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\mathrm {tanh}\relax (x)}{\sqrt {-{\mathrm {tanh}\relax (x)}^2-1}}\right )-\ln \left (\mathrm {tanh}\relax (x)-\sqrt {-{\mathrm {tanh}\relax (x)}^2-1}\,1{}\mathrm {i}\right )\,1{}\mathrm {i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((- tanh(x)^2 - 1)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- \tanh ^{2}{\relax (x )} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-tanh(x)**2 - 1), x)

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