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]

-2/195*arctan(1/39*195^(1/2)*(1-2*tanh(x)))*195^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 22, normalized size of antiderivative = 1.00, 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 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])

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 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])

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*}

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Mathematica [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {\text {sech}^2(x)}{11-5 \tanh (x)+5 \tanh ^2(x)} \, dx \]

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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fricas [A]  time = 0.48, size = 32, normalized size = 1.45 \[ -\frac {2}{195} \, \sqrt {195} \arctan \left (-\frac {17 \, \sqrt {195} \cosh \relax (x) + 5 \, \sqrt {195} \sinh \relax (x)}{195 \, {\left (\cosh \relax (x) - \sinh \relax (x)\right )}}\right ) \]

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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giac [A]  time = 0.13, size = 19, normalized size = 0.86 \[ \frac {2}{195} \, \sqrt {195} \arctan \left (\frac {1}{195} \, \sqrt {195} {\left (11 \, e^{\left (2 \, x\right )} + 6\right )}\right ) \]

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))

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maple [C]  time = 0.25, size = 62, normalized size = 2.82 \[ \frac {i \sqrt {195}\, \ln \left (11 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\left (-i \sqrt {195}-5\right ) \tanh \left (\frac {x}{2}\right )+11\right )}{195}-\frac {i \sqrt {195}\, \ln \left (11 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\left (i \sqrt {195}-5\right ) \tanh \left (\frac {x}{2}\right )+11\right )}{195} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}\relax (x)^{2}}{5 \, \tanh \relax (x)^{2} - 5 \, \tanh \relax (x) + 11}\,{d x} \]

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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mupad [B]  time = 0.11, size = 19, normalized size = 0.86 \[ \frac {2\,\sqrt {195}\,\mathrm {atan}\left (\frac {\sqrt {195}\,\left (11\,{\mathrm {e}}^{2\,x}+6\right )}{195}\right )}{195} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(5*tanh(x)^2 - 5*tanh(x) + 11)),x)

[Out]

(2*195^(1/2)*atan((195^(1/2)*(11*exp(2*x) + 6))/195))/195

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{2}{\relax (x )}}{5 \tanh ^{2}{\relax (x )} - 5 \tanh {\relax (x )} + 11}\, dx \]

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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