3.10.89 \(\int \frac {\text {sech}^2(x)}{11-5 \tanh (x)+5 \tanh ^2(x)} \, dx\) [989]

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

[Out]

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

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Rubi [A]
time = 0.04, 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 = {4427, 632, 210} \begin {gather*} -\frac {2 \text {ArcTan}\left (\sqrt {\frac {5}{39}} (1-2 \tanh (x))\right )}{\sqrt {195}} \end {gather*}

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 210

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

Rule 632

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 4427

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 &=\text {Subst}\left (\int \frac {1}{11-5 x+5 x^2} \, dx,x,\tanh (x)\right )\\ &=-\left (2 \text {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 \begin {gather*} \int \frac {\text {sech}^2(x)}{11-5 \tanh (x)+5 \tanh ^2(x)} \, dx \end {gather*}

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] Result contains complex when optimal does not.
time = 1.15, size = 62, normalized size = 2.82

method	result	size
risch	\(\frac {i \sqrt {195}\, \ln \left ({\mathrm e}^{2 x}+\frac {i \sqrt {195}}{11}+\frac {6}{11}\right )}{195}-\frac {i \sqrt {195}\, \ln \left ({\mathrm e}^{2 x}-\frac {i \sqrt {195}}{11}+\frac {6}{11}\right )}{195}\)	\(40\)
default	\(\frac {i \sqrt {195}\, \ln \left (11 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\left (-i \sqrt {195}-5\right ) \tanh \left (\frac {x}{2}\right )+11\right )}{195}-\frac {i \sqrt {195}\, \ln \left (11 \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )+\left (i \sqrt {195}-5\right ) \tanh \left (\frac {x}{2}\right )+11\right )}{195}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.33, size = 32, normalized size = 1.45 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {sech}^{2}{\left (x \right )}}{5 \tanh ^{2}{\left (x \right )} - 5 \tanh {\left (x \right )} + 11}\, dx \end {gather*}

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 = 0.40, size = 19, normalized size = 0.86 \begin {gather*} \frac {2}{195} \, \sqrt {195} \arctan \left (\frac {1}{195} \, \sqrt {195} {\left (11 \, e^{\left (2 \, x\right )} + 6\right )}\right ) \end {gather*}

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

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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