3.98 \(\int \frac{\text{sech}^4(x)}{1+\tanh (x)} \, dx\)

Optimal. Leaf size=11 \[ \tanh (x)-\frac{\tanh ^2(x)}{2} \]

[Out]

Tanh[x] - Tanh[x]^2/2

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Rubi [A]  time = 0.032956, antiderivative size = 11, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {3487} \[ \tanh (x)-\frac{\tanh ^2(x)}{2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(1 + Tanh[x]),x]

[Out]

Tanh[x] - Tanh[x]^2/2

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps

\begin{align*} \int \frac{\text{sech}^4(x)}{1+\tanh (x)} \, dx &=\operatorname{Subst}(\int (1-x) \, dx,x,\tanh (x))\\ &=\tanh (x)-\frac{\tanh ^2(x)}{2}\\ \end{align*}

Mathematica [A]  time = 0.0236339, size = 11, normalized size = 1. \[ \tanh (x)+\frac{\text{sech}^2(x)}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^4/(1 + Tanh[x]),x]

[Out]

Sech[x]^2/2 + Tanh[x]

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Maple [B]  time = 0.025, size = 34, normalized size = 3.1 \begin{align*} -2\,{\frac{- \left ( \tanh \left ( x/2 \right ) \right ) ^{3}+ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}-\tanh \left ( x/2 \right ) }{ \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^4/(1+tanh(x)),x)

[Out]

-2*(-tanh(1/2*x)^3+tanh(1/2*x)^2-tanh(1/2*x))/(tanh(1/2*x)^2+1)^2

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Maxima [B]  time = 1.1833, size = 50, normalized size = 4.55 \begin{align*} \frac{4 \, e^{\left (-2 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} + \frac{2}{2 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(1+tanh(x)),x, algorithm="maxima")

[Out]

4*e^(-2*x)/(2*e^(-2*x) + e^(-4*x) + 1) + 2/(2*e^(-2*x) + e^(-4*x) + 1)

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Fricas [B]  time = 2.02179, size = 181, normalized size = 16.45 \begin{align*} -\frac{2}{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(1+tanh(x)),x, algorithm="fricas")

[Out]

-2/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 1)*sinh(x)^2 + 2*cosh(x)^2 + 4*(cosh(x)^3 +
cosh(x))*sinh(x) + 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\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)**4/(1+tanh(x)),x)

[Out]

Integral(sech(x)**4/(tanh(x) + 1), x)

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Giac [A]  time = 1.2545, size = 14, normalized size = 1.27 \begin{align*} -\frac{2}{{\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(1+tanh(x)),x, algorithm="giac")

[Out]

-2/(e^(2*x) + 1)^2