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

Optimal. Leaf size=14 \[ \frac {x}{a}-\frac {\tan ^{-1}(\sinh (x))}{a} \]

[Out]

x/a-arctan(sinh(x))/a

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Rubi [A]  time = 0.05, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3888, 3770} \[ \frac {x}{a}-\frac {\tan ^{-1}(\sinh (x))}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + a*Sech[x]),x]

[Out]

x/a - ArcTan[Sinh[x]]/a

Rule 3770

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

Rule 3888

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 15, normalized size = 1.07 \[ \frac {x-2 \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x - 2*ArcTan[Tanh[x/2]])/a

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fricas [A]  time = 0.39, size = 14, normalized size = 1.00 \[ \frac {x - 2 \, \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x - 2*arctan(cosh(x) + sinh(x)))/a

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giac [A]  time = 0.13, size = 14, normalized size = 1.00 \[ \frac {x}{a} - \frac {2 \, \arctan \left (e^{x}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a - 2*arctan(e^x)/a

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maple [B]  time = 0.10, size = 35, normalized size = 2.50 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}-\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 16, normalized size = 1.14 \[ \frac {x}{a} + \frac {2 \, \arctan \left (e^{\left (-x\right )}\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/a + 2*arctan(e^(-x))/a

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mupad [B]  time = 1.32, size = 25, normalized size = 1.79 \[ \frac {x}{a}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {a^2}}{a}\right )}{\sqrt {a^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(x)^2/(a + a/cosh(x)),x)

[Out]

x/a - (2*atan((exp(x)*(a^2)^(1/2))/a))/(a^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\tanh ^{2}{\relax (x )}}{\operatorname {sech}{\relax (x )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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