3.2.7 \(\int \frac {\tanh ^2(x)}{a+a \text {sech}(x)} \, dx\) [107]

Optimal. Leaf size=14 \[ \frac {x}{a}-\frac {\text {ArcTan}(\sinh (x))}{a} \]

[Out]

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

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Rubi [A]
time = 0.03, 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 = {3973, 3855} \begin {gather*} \frac {x}{a}-\frac {\text {ArcTan}(\sinh (x))}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tanh[x]^2/(a + a*Sech[x]),x]

[Out]

x/a - ArcTan[Sinh[x]]/a

Rule 3855

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

Rule 3973

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 = 0.02, size = 15, normalized size = 1.07 \begin {gather*} \frac {x-2 \text {ArcTan}\left (\tanh \left (\frac {x}{2}\right )\right )}{a} \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(31\) vs. \(2(14)=28\).
time = 0.60, size = 32, normalized size = 2.29

method	result	size
risch	\(\frac {x}{a}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{a}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{a}\)	\(31\)
default	\(\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )-2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )-\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/a*(1/4*ln(tanh(1/2*x)+1)-1/2*arctan(tanh(1/2*x))-1/4*ln(tanh(1/2*x)-1))

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Maxima [A]
time = 0.47, size = 16, normalized size = 1.14 \begin {gather*} \frac {x}{a} + \frac {2 \, \arctan \left (e^{\left (-x\right )}\right )}{a} \end {gather*}

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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Fricas [A]
time = 0.37, size = 14, normalized size = 1.00 \begin {gather*} \frac {x - 2 \, \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}{a} \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\tanh ^{2}{\left (x \right )}}{\operatorname {sech}{\left (x \right )} + 1}\, dx}{a} \end {gather*}

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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Giac [A]
time = 0.39, size = 14, normalized size = 1.00 \begin {gather*} \frac {x}{a} - \frac {2 \, \arctan \left (e^{x}\right )}{a} \end {gather*}

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

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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