3.10.81 \(\int \frac {\text {sech}^2(x)}{9+\tanh ^2(x)} \, dx\) [981]

Optimal. Leaf size=11 \[ \frac {1}{3} \text {ArcTan}\left (\frac {\tanh (x)}{3}\right ) \]

[Out]

1/3*arctan(1/3*tanh(x))

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Rubi [A]
time = 0.02, antiderivative size = 11, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3756, 209} \begin {gather*} \frac {1}{3} \text {ArcTan}\left (\frac {\tanh (x)}{3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(9 + Tanh[x]^2),x]

[Out]

ArcTan[Tanh[x]/3]/3

Rule 209

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

Rule 3756

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)}{9+\tanh ^2(x)} \, dx &=\text {Subst}\left (\int \frac {1}{9+x^2} \, dx,x,\tanh (x)\right )\\ &=\frac {1}{3} \tan ^{-1}\left (\frac {\tanh (x)}{3}\right )\\ \end {align*}

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Mathematica [F]
time = 0.02, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\text {sech}^2(x)}{9+\tanh ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Sech[x]^2/(9 + Tanh[x]^2),x]

[Out]

Integrate[Sech[x]^2/(9 + Tanh[x]^2), x]

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(71\) vs. \(2(7)=14\).
time = 1.39, size = 72, normalized size = 6.55

method	result	size
risch	\(\frac {i \ln \left ({\mathrm e}^{2 x}+\frac {4}{5}+\frac {3 i}{5}\right )}{6}-\frac {i \ln \left ({\mathrm e}^{2 x}+\frac {4}{5}-\frac {3 i}{5}\right )}{6}\)	\(26\)
default	\(-\frac {\left (-10+\sqrt {10}\right ) \sqrt {10}\, \arctan \left (\frac {18 \tanh \left (\frac {x}{2}\right )}{6 \sqrt {10}-6}\right )}{5 \left (6 \sqrt {10}-6\right )}-\frac {\left (10+\sqrt {10}\right ) \sqrt {10}\, \arctan \left (\frac {18 \tanh \left (\frac {x}{2}\right )}{6 \sqrt {10}+6}\right )}{5 \left (6 \sqrt {10}+6\right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(9+tanh(x)^2),x,method=_RETURNVERBOSE)

[Out]

-1/5*(-10+10^(1/2))*10^(1/2)/(6*10^(1/2)-6)*arctan(18*tanh(1/2*x)/(6*10^(1/2)-6))-1/5*(10+10^(1/2))*10^(1/2)/(
6*10^(1/2)+6)*arctan(18*tanh(1/2*x)/(6*10^(1/2)+6))

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Maxima [A]
time = 0.47, size = 11, normalized size = 1.00 \begin {gather*} -\frac {1}{3} \, \arctan \left (\frac {5}{3} \, e^{\left (-2 \, x\right )} + \frac {4}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(9+tanh(x)^2),x, algorithm="maxima")

[Out]

-1/3*arctan(5/3*e^(-2*x) + 4/3)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 21 vs. \(2 (7) = 14\).
time = 0.35, size = 21, normalized size = 1.91 \begin {gather*} -\frac {1}{3} \, \arctan \left (-\frac {9 \, \cosh \left (x\right ) + \sinh \left (x\right )}{3 \, {\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/(9+tanh(x)^2),x, algorithm="fricas")

[Out]

-1/3*arctan(-1/3*(9*cosh(x) + 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 )}}{\tanh ^{2}{\left (x \right )} + 9}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(9+tanh(x)**2),x)

[Out]

Integral(sech(x)**2/(tanh(x)**2 + 9), x)

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Giac [A]
time = 0.41, size = 11, normalized size = 1.00 \begin {gather*} \frac {1}{3} \, \arctan \left (\frac {5}{3} \, e^{\left (2 \, x\right )} + \frac {4}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(9+tanh(x)^2),x, algorithm="giac")

[Out]

1/3*arctan(5/3*e^(2*x) + 4/3)

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Mupad [B]
time = 1.64, size = 11, normalized size = 1.00 \begin {gather*} \frac {\mathrm {atan}\left (\frac {5\,{\mathrm {e}}^{2\,x}}{3}+\frac {4}{3}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(tanh(x)^2 + 9)),x)

[Out]

atan((5*exp(2*x))/3 + 4/3)/3

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