### 3.577 $$\int \frac{1-\cosh ^2(x)}{1+\cosh ^2(x)} \, dx$$

Optimal. Leaf size=19 $\sqrt{2} \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )-x$

[Out]

-x + Sqrt[2]*ArcTanh[Tanh[x]/Sqrt[2]]

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Rubi [A]  time = 0.0394022, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.176, Rules used = {3171, 3181, 206} $\sqrt{2} \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )-x$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-x + Sqrt[2]*ArcTanh[Tanh[x]/Sqrt[2]]

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 3181

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(-1), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[1/(a + (a + b)*ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{1-\cosh ^2(x)}{1+\cosh ^2(x)} \, dx &=-x+2 \int \frac{1}{1+\cosh ^2(x)} \, dx\\ &=-x+2 \operatorname{Subst}\left (\int \frac{1}{1-2 x^2} \, dx,x,\coth (x)\right )\\ &=-x+\sqrt{2} \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0302714, size = 24, normalized size = 1.26 $-2 \left (\frac{x}{2}-\frac{\tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{2}}\right )}{\sqrt{2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-2*(x/2 - ArcTanh[Tanh[x]/Sqrt[2]]/Sqrt[2])

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Maple [B]  time = 0.014, size = 102, normalized size = 5.4 \begin{align*} -\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) +{\frac{\sqrt{2}}{4}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}}{4}\ln \left ({ \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+\sqrt{2}\tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}} \right ) }+\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((1-cosh(x)^2)/(1+cosh(x)^2),x)

[Out]

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

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Maxima [B]  time = 1.6124, size = 138, normalized size = 7.26 \begin{align*} \frac{3}{16} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (2 \, x\right )} + 3}\right ) - \frac{5}{16} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (-2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (-2 \, x\right )} + 3}\right ) - 2 \, x + \frac{1}{4} \, \log \left (e^{\left (4 \, x\right )} + 6 \, e^{\left (2 \, x\right )} + 1\right ) - \frac{1}{4} \, \log \left (6 \, e^{\left (-2 \, x\right )} + e^{\left (-4 \, x\right )} + 1\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/16*sqrt(2)*log(-(2*sqrt(2) - e^(2*x) - 3)/(2*sqrt(2) + e^(2*x) + 3)) - 5/16*sqrt(2)*log(-(2*sqrt(2) - e^(-2*
x) - 3)/(2*sqrt(2) + e^(-2*x) + 3)) - 2*x + 1/4*log(e^(4*x) + 6*e^(2*x) + 1) - 1/4*log(6*e^(-2*x) + e^(-4*x) +
1)

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Fricas [B]  time = 2.27978, size = 220, normalized size = 11.58 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-\frac{3 \,{\left (2 \, \sqrt{2} - 3\right )} \cosh \left (x\right )^{2} - 4 \,{\left (3 \, \sqrt{2} - 4\right )} \cosh \left (x\right ) \sinh \left (x\right ) + 3 \,{\left (2 \, \sqrt{2} - 3\right )} \sinh \left (x\right )^{2} + 2 \, \sqrt{2} - 3}{\cosh \left (x\right )^{2} + \sinh \left (x\right )^{2} + 3}\right ) - x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log(-(3*(2*sqrt(2) - 3)*cosh(x)^2 - 4*(3*sqrt(2) - 4)*cosh(x)*sinh(x) + 3*(2*sqrt(2) - 3)*sinh(x)^
2 + 2*sqrt(2) - 3)/(cosh(x)^2 + sinh(x)^2 + 3)) - x

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Sympy [B]  time = 5.58405, size = 61, normalized size = 3.21 \begin{align*} - x - \frac{\sqrt{2} \log{\left (4 \tanh ^{2}{\left (\frac{x}{2} \right )} - 4 \sqrt{2} \tanh{\left (\frac{x}{2} \right )} + 4 \right )}}{2} + \frac{\sqrt{2} \log{\left (4 \tanh ^{2}{\left (\frac{x}{2} \right )} + 4 \sqrt{2} \tanh{\left (\frac{x}{2} \right )} + 4 \right )}}{2} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x - sqrt(2)*log(4*tanh(x/2)**2 - 4*sqrt(2)*tanh(x/2) + 4)/2 + sqrt(2)*log(4*tanh(x/2)**2 + 4*sqrt(2)*tanh(x/2
) + 4)/2

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Giac [B]  time = 1.15681, size = 51, normalized size = 2.68 \begin{align*} \frac{1}{2} \, \sqrt{2} \log \left (-\frac{2 \, \sqrt{2} - e^{\left (2 \, x\right )} - 3}{2 \, \sqrt{2} + e^{\left (2 \, x\right )} + 3}\right ) - x \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log(-(2*sqrt(2) - e^(2*x) - 3)/(2*sqrt(2) + e^(2*x) + 3)) - x