3.97 \(\int \frac{\text{sech}^3(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=6 \[ \text{sech}(x)+\tan ^{-1}(\sinh (x)) \]

[Out]

ArcTan[Sinh[x]] + Sech[x]

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Rubi [A]  time = 0.035025, antiderivative size = 6, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {3501, 3770} \[ \text{sech}(x)+\tan ^{-1}(\sinh (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^3/(1 + Tanh[x]),x]

[Out]

ArcTan[Sinh[x]] + Sech[x]

Rule 3501

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d^2*
(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(d^2*(m - 2))/(a*(m + n -
1)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2
+ b^2, 0] && LtQ[n, 0] && GtQ[m, 1] &&  !ILtQ[m + n, 0] && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\text{sech}^3(x)}{1+\tanh (x)} \, dx &=\text{sech}(x)+\int \text{sech}(x) \, dx\\ &=\tan ^{-1}(\sinh (x))+\text{sech}(x)\\ \end{align*}

Mathematica [A]  time = 0.0245494, size = 12, normalized size = 2. \[ \text{sech}(x)+2 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^3/(1 + Tanh[x]),x]

[Out]

2*ArcTan[Tanh[x/2]] + Sech[x]

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Maple [B]  time = 0.022, size = 21, normalized size = 3.5 \begin{align*} 2\, \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{-1}+2\,\arctan \left ( \tanh \left ( x/2 \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^3/(1+tanh(x)),x)

[Out]

2/(tanh(1/2*x)^2+1)+2*arctan(tanh(1/2*x))

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Maxima [B]  time = 1.58321, size = 30, normalized size = 5. \begin{align*} \frac{2 \, e^{\left (-x\right )}}{e^{\left (-2 \, x\right )} + 1} - 2 \, \arctan \left (e^{\left (-x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(1+tanh(x)),x, algorithm="maxima")

[Out]

2*e^(-x)/(e^(-2*x) + 1) - 2*arctan(e^(-x))

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Fricas [B]  time = 2.08175, size = 198, normalized size = 33. \begin{align*} \frac{2 \,{\left ({\left (\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + \cosh \left (x\right ) + \sinh \left (x\right )\right )}}{\cosh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(1+tanh(x)),x, algorithm="fricas")

[Out]

2*((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + 1)*arctan(cosh(x) + sinh(x)) + cosh(x) + sinh(x))/(cosh(x)^2 +
2*cosh(x)*sinh(x) + sinh(x)^2 + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{3}{\left (x \right )}}{\tanh{\left (x \right )} + 1}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**3/(1+tanh(x)),x)

[Out]

Integral(sech(x)**3/(tanh(x) + 1), x)

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Giac [B]  time = 1.22233, size = 24, normalized size = 4. \begin{align*} \frac{2 \, e^{x}}{e^{\left (2 \, x\right )} + 1} + 2 \, \arctan \left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^3/(1+tanh(x)),x, algorithm="giac")

[Out]

2*e^x/(e^(2*x) + 1) + 2*arctan(e^x)