### 3.575 $$\int \frac{1-\sinh ^2(x)}{1+\sinh ^2(x)} \, dx$$

Optimal. Leaf size=8 $2 \tanh (x)-x$

[Out]

-x + 2*Tanh[x]

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Rubi [A]  time = 0.0418972, antiderivative size = 8, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {3171, 3175, 3767, 8} $2 \tanh (x)-x$

Antiderivative was successfully veriﬁed.

[In]

Int[(1 - Sinh[x]^2)/(1 + Sinh[x]^2),x]

[Out]

-x + 2*Tanh[x]

Rule 3171

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(B*x
)/b, x] + Dist[(A*b - a*B)/b, Int[1/(a + b*Sin[e + f*x]^2), x], x] /; FreeQ[{a, b, e, f, A, B}, x]

Rule 3175

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0149371, size = 8, normalized size = 1. $2 \tanh (x)-x$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(1 - Sinh[x]^2)/(1 + Sinh[x]^2),x]

[Out]

-x + 2*Tanh[x]

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Maple [B]  time = 0.024, size = 34, normalized size = 4.3 \begin{align*} -\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) +\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) +4\,{\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((1-sinh(x)^2)/(1+sinh(x)^2),x)

[Out]

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

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Maxima [A]  time = 1.06353, size = 19, normalized size = 2.38 \begin{align*} -x + \frac{4}{e^{\left (-2 \, x\right )} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-sinh(x)^2)/(1+sinh(x)^2),x, algorithm="maxima")

[Out]

-x + 4/(e^(-2*x) + 1)

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Fricas [B]  time = 2.17497, size = 54, normalized size = 6.75 \begin{align*} -\frac{{\left (x + 2\right )} \cosh \left (x\right ) - 2 \, \sinh \left (x\right )}{\cosh \left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-sinh(x)^2)/(1+sinh(x)^2),x, algorithm="fricas")

[Out]

-((x + 2)*cosh(x) - 2*sinh(x))/cosh(x)

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Sympy [B]  time = 1.71308, size = 41, normalized size = 5.12 \begin{align*} - \frac{x \tanh ^{2}{\left (\frac{x}{2} \right )}}{\tanh ^{2}{\left (\frac{x}{2} \right )} + 1} - \frac{x}{\tanh ^{2}{\left (\frac{x}{2} \right )} + 1} + \frac{4 \tanh{\left (\frac{x}{2} \right )}}{\tanh ^{2}{\left (\frac{x}{2} \right )} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-sinh(x)**2)/(1+sinh(x)**2),x)

[Out]

-x*tanh(x/2)**2/(tanh(x/2)**2 + 1) - x/(tanh(x/2)**2 + 1) + 4*tanh(x/2)/(tanh(x/2)**2 + 1)

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Giac [A]  time = 1.16762, size = 19, normalized size = 2.38 \begin{align*} -x - \frac{4}{e^{\left (2 \, x\right )} + 1} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-sinh(x)^2)/(1+sinh(x)^2),x, algorithm="giac")

[Out]

-x - 4/(e^(2*x) + 1)