### 3.984 $$\int \frac{\text{sech}^2(x) (2-\tanh ^2(x))}{1-\tanh ^2(x)} \, dx$$

Optimal. Leaf size=4 $x+\tanh (x)$

[Out]

x + Tanh[x]

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Rubi [A]  time = 0.0742743, antiderivative size = 4, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.13, Rules used = {3657, 3473, 8} $x+\tanh (x)$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sech[x]^2*(2 - Tanh[x]^2))/(1 - Tanh[x]^2),x]

[Out]

x + Tanh[x]

Rule 3657

Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.0031483, size = 4, normalized size = 1. $x+\tanh (x)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sech[x]^2*(2 - Tanh[x]^2))/(1 - Tanh[x]^2),x]

[Out]

x + Tanh[x]

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Maple [B]  time = 0.033, size = 34, normalized size = 8.5 \begin{align*} \ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) -\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +2\,{\frac{\tanh \left ( x/2 \right ) }{ \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2*(2-tanh(x)^2)/(1-tanh(x)^2),x)

[Out]

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

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Maxima [B]  time = 1.06472, size = 16, normalized size = 4. \begin{align*} x + \frac{2}{e^{\left (-2 \, x\right )} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*(2-tanh(x)^2)/(1-tanh(x)^2),x, algorithm="maxima")

[Out]

x + 2/(e^(-2*x) + 1)

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Fricas [B]  time = 2.10975, size = 50, normalized size = 12.5 \begin{align*} \frac{{\left (x - 1\right )} \cosh \left (x\right ) + \sinh \left (x\right )}{\cosh \left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*(2-tanh(x)^2)/(1-tanh(x)^2),x, algorithm="fricas")

[Out]

((x - 1)*cosh(x) + sinh(x))/cosh(x)

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Sympy [B]  time = 1.42465, size = 29, normalized size = 7.25 \begin{align*} - \frac{x \operatorname{sech}^{2}{\left (x \right )}}{\tanh ^{2}{\left (x \right )} - 1} - \frac{\tanh{\left (x \right )} \operatorname{sech}^{2}{\left (x \right )}}{\tanh ^{2}{\left (x \right )} - 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2*(2-tanh(x)**2)/(1-tanh(x)**2),x)

[Out]

-x*sech(x)**2/(tanh(x)**2 - 1) - tanh(x)*sech(x)**2/(tanh(x)**2 - 1)

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Giac [B]  time = 1.14009, size = 16, normalized size = 4. \begin{align*} x - \frac{2}{e^{\left (2 \, x\right )} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*(2-tanh(x)^2)/(1-tanh(x)^2),x, algorithm="giac")

[Out]

x - 2/(e^(2*x) + 1)