3.73 \(\int \frac{\text{sech}^2(x)}{a+a \text{sech}(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac{\tan ^{-1}(\sinh (x))}{a}-\frac{\tanh (x)}{a \text{sech}(x)+a} \]

[Out]

ArcTan[Sinh[x]]/a - Tanh[x]/(a + a*Sech[x])

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Rubi [A]  time = 0.0658841, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3789, 3770, 3794} \[ \frac{\tan ^{-1}(\sinh (x))}{a}-\frac{\tanh (x)}{a \text{sech}(x)+a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(a + a*Sech[x]),x]

[Out]

ArcTan[Sinh[x]]/a - Tanh[x]/(a + a*Sech[x])

Rule 3789

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3770

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

Rule 3794

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> -Simp[Cot[e + f*x]/(f*(b + a*
Csc[e + f*x])), x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

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

Mathematica [A]  time = 0.028501, size = 22, normalized size = 1.1 \[ \frac{2 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )-\tanh \left (\frac{x}{2}\right )}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^2/(a + a*Sech[x]),x]

[Out]

(2*ArcTan[Tanh[x/2]] - Tanh[x/2])/a

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Maple [A]  time = 0.01, size = 21, normalized size = 1.1 \begin{align*} -{\frac{1}{a}\tanh \left ({\frac{x}{2}} \right ) }+2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(a+a*sech(x)),x)

[Out]

-1/a*tanh(1/2*x)+2/a*arctan(tanh(1/2*x))

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Maxima [A]  time = 1.75343, size = 31, normalized size = 1.55 \begin{align*} -\frac{2 \, \arctan \left (e^{\left (-x\right )}\right )}{a} - \frac{2}{a e^{\left (-x\right )} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+a*sech(x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.39204, size = 117, normalized size = 5.85 \begin{align*} \frac{2 \,{\left ({\left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + 1\right )}}{a \cosh \left (x\right ) + a \sinh \left (x\right ) + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+a*sech(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\operatorname{sech}^{2}{\left (x \right )}}{\operatorname{sech}{\left (x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2/(a+a*sech(x)),x)

[Out]

Integral(sech(x)**2/(sech(x) + 1), x)/a

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Giac [A]  time = 1.11875, size = 27, normalized size = 1.35 \begin{align*} \frac{2 \, \arctan \left (e^{x}\right )}{a} + \frac{2}{a{\left (e^{x} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+a*sech(x)),x, algorithm="giac")

[Out]

2*arctan(e^x)/a + 2/(a*(e^x + 1))