3.999 \(\int \frac {\text {sech}^2(x)}{\sqrt {1-4 \tanh ^2(x)}} \, dx\)

Optimal. Leaf size=9 \[ \frac {1}{2} \sin ^{-1}(2 \tanh (x)) \]

[Out]

1/2*arcsin(2*tanh(x))

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Rubi [A]  time = 0.05, antiderivative size = 9, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3675, 216} \[ \frac {1}{2} \sin ^{-1}(2 \tanh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/Sqrt[1 - 4*Tanh[x]^2],x]

[Out]

ArcSin[2*Tanh[x]]/2

Rule 216

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

Rule 3675

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/(c^(m - 1)*f), Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)
^n)^p, x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2] && (IntegersQ[n, p
] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {\text {sech}^2(x)}{\sqrt {1-4 \tanh ^2(x)}} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\sqrt {1-4 x^2}} \, dx,x,\tanh (x)\right )\\ &=\frac {1}{2} \sin ^{-1}(2 \tanh (x))\\ \end {align*}

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Mathematica [B]  time = 0.06, size = 47, normalized size = 5.22 \[ \frac {\sqrt {3 \cosh (2 x)-5} \text {sech}(x) \tanh ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {3 \sinh ^2(x)-1}}\right )}{2 \sqrt {2-8 \tanh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/Sqrt[1 - 4*Tanh[x]^2],x]

[Out]

(ArcTanh[(2*Sinh[x])/Sqrt[-1 + 3*Sinh[x]^2]]*Sqrt[-5 + 3*Cosh[2*x]]*Sech[x])/(2*Sqrt[2 - 8*Tanh[x]^2])

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fricas [B]  time = 0.46, size = 118, normalized size = 13.11 \[ -\frac {1}{2} \, \arctan \left (\frac {2 \, \sqrt {2} {\left (\cosh \relax (x)^{2} + 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2} - 1\right )} \sqrt {-\frac {3 \, \cosh \relax (x)^{2} + 3 \, \sinh \relax (x)^{2} - 5}{\cosh \relax (x)^{2} - 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2}}}}{3 \, \cosh \relax (x)^{4} + 12 \, \cosh \relax (x) \sinh \relax (x)^{3} + 3 \, \sinh \relax (x)^{4} + 2 \, {\left (9 \, \cosh \relax (x)^{2} - 5\right )} \sinh \relax (x)^{2} - 10 \, \cosh \relax (x)^{2} + 4 \, {\left (3 \, \cosh \relax (x)^{3} - 5 \, \cosh \relax (x)\right )} \sinh \relax (x) + 3}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(1-4*tanh(x)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*arctan(2*sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-(3*cosh(x)^2 + 3*sinh(x)^2 - 5)/(c
osh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(3*cosh(x)^4 + 12*cosh(x)*sinh(x)^3 + 3*sinh(x)^4 + 2*(9*cosh(x)^2
- 5)*sinh(x)^2 - 10*cosh(x)^2 + 4*(3*cosh(x)^3 - 5*cosh(x))*sinh(x) + 3))

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giac [B]  time = 0.16, size = 44, normalized size = 4.89 \[ -\arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (\sqrt {3} \sqrt {-3 \, e^{\left (4 \, x\right )} + 10 \, e^{\left (2 \, x\right )} - 3} - 4\right )}}{3 \, e^{\left (2 \, x\right )} - 5} - 1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(1-4*tanh(x)^2)^(1/2),x, algorithm="giac")

[Out]

-arctan(1/3*sqrt(3)*(2*(sqrt(3)*sqrt(-3*e^(4*x) + 10*e^(2*x) - 3) - 4)/(3*e^(2*x) - 5) - 1))

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maple [F]  time = 0.52, size = 0, normalized size = 0.00 \[ \int \frac {\mathrm {sech}\relax (x )^{2}}{\sqrt {1-4 \left (\tanh ^{2}\relax (x )\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(1-4*tanh(x)^2)^(1/2),x)

[Out]

int(sech(x)^2/(1-4*tanh(x)^2)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}\relax (x)^{2}}{\sqrt {-4 \, \tanh \relax (x)^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(1-4*tanh(x)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sech(x)^2/sqrt(-4*tanh(x)^2 + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.11 \[ \int \frac {1}{{\mathrm {cosh}\relax (x)}^2\,\sqrt {1-4\,{\mathrm {tanh}\relax (x)}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(1 - 4*tanh(x)^2)^(1/2)),x)

[Out]

int(1/(cosh(x)^2*(1 - 4*tanh(x)^2)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(1-4*tanh(x)**2)**(1/2),x)

[Out]

Integral(sech(x)**2/sqrt(-(2*tanh(x) - 1)*(2*tanh(x) + 1)), x)

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