3.989 \(\int \frac{\text{sech}^2(x)}{11-5 \tanh (x)+5 \tanh ^2(x)} \, dx\)

Optimal. Leaf size=22 \[ -\frac{2 \tan ^{-1}\left (\sqrt{\frac{5}{39}} (1-2 \tanh (x))\right )}{\sqrt{195}} \]

[Out]

(-2*ArcTan[Sqrt[5/39]*(1 - 2*Tanh[x])])/Sqrt[195]

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Rubi [A]  time = 0.0680059, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {4342, 618, 204} \[ -\frac{2 \tan ^{-1}\left (\sqrt{\frac{5}{39}} (1-2 \tanh (x))\right )}{\sqrt{195}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(11 - 5*Tanh[x] + 5*Tanh[x]^2),x]

[Out]

(-2*ArcTan[Sqrt[5/39]*(1 - 2*Tanh[x])])/Sqrt[195]

Rule 4342

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\text{sech}^2(x)}{11-5 \tanh (x)+5 \tanh ^2(x)} \, dx &=\operatorname{Subst}\left (\int \frac{1}{11-5 x+5 x^2} \, dx,x,\tanh (x)\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{1}{-195-x^2} \, dx,x,-5+10 \tanh (x)\right )\right )\\ &=-\frac{2 \tan ^{-1}\left (\sqrt{\frac{5}{39}} (1-2 \tanh (x))\right )}{\sqrt{195}}\\ \end{align*}

Mathematica [F]  time = 0.0307312, size = 0, normalized size = 0. \[ \int \frac{\text{sech}^2(x)}{11-5 \tanh (x)+5 \tanh ^2(x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sech[x]^2/(11 - 5*Tanh[x] + 5*Tanh[x]^2),x]

[Out]

Integrate[Sech[x]^2/(11 - 5*Tanh[x] + 5*Tanh[x]^2), x]

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Maple [C]  time = 0.069, size = 62, normalized size = 2.8 \begin{align*}{\frac{i}{195}}\sqrt{195}\ln \left ( 11\, \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+ \left ( -i\sqrt{195}-5 \right ) \tanh \left ({\frac{x}{2}} \right ) +11 \right ) -{\frac{i}{195}}\sqrt{195}\ln \left ( 11\, \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+ \left ( i\sqrt{195}-5 \right ) \tanh \left ({\frac{x}{2}} \right ) +11 \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(11-5*tanh(x)+5*tanh(x)^2),x)

[Out]

1/195*I*195^(1/2)*ln(11*tanh(1/2*x)^2+(-I*195^(1/2)-5)*tanh(1/2*x)+11)-1/195*I*195^(1/2)*ln(11*tanh(1/2*x)^2+(
I*195^(1/2)-5)*tanh(1/2*x)+11)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}\left (x\right )^{2}}{5 \, \tanh \left (x\right )^{2} - 5 \, \tanh \left (x\right ) + 11}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(11-5*tanh(x)+5*tanh(x)^2),x, algorithm="maxima")

[Out]

integrate(sech(x)^2/(5*tanh(x)^2 - 5*tanh(x) + 11), x)

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Fricas [A]  time = 2.09234, size = 132, normalized size = 6. \begin{align*} -\frac{2}{195} \, \sqrt{195} \arctan \left (-\frac{17 \, \sqrt{195} \cosh \left (x\right ) + 5 \, \sqrt{195} \sinh \left (x\right )}{195 \,{\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(11-5*tanh(x)+5*tanh(x)^2),x, algorithm="fricas")

[Out]

-2/195*sqrt(195)*arctan(-1/195*(17*sqrt(195)*cosh(x) + 5*sqrt(195)*sinh(x))/(cosh(x) - sinh(x)))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{2}{\left (x \right )}}{5 \tanh ^{2}{\left (x \right )} - 5 \tanh{\left (x \right )} + 11}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(11-5*tanh(x)+5*tanh(x)**2),x)

[Out]

Integral(sech(x)**2/(5*tanh(x)**2 - 5*tanh(x) + 11), x)

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Giac [A]  time = 1.15518, size = 26, normalized size = 1.18 \begin{align*} \frac{2}{195} \, \sqrt{195} \arctan \left (\frac{1}{195} \, \sqrt{195}{\left (11 \, e^{\left (2 \, x\right )} + 6\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(11-5*tanh(x)+5*tanh(x)^2),x, algorithm="giac")

[Out]

2/195*sqrt(195)*arctan(1/195*sqrt(195)*(11*e^(2*x) + 6))