3.204 \(\int (1-\tanh ^2(x))^{3/2} \, dx\)

Optimal. Leaf size=22 \[ \frac{1}{2} \sin ^{-1}(\tanh (x))+\frac{1}{2} \tanh (x) \sqrt{\text{sech}^2(x)} \]

[Out]

ArcSin[Tanh[x]]/2 + (Sqrt[Sech[x]^2]*Tanh[x])/2

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Rubi [A]  time = 0.0199402, antiderivative size = 22, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3657, 4122, 195, 216} \[ \frac{1}{2} \sin ^{-1}(\tanh (x))+\frac{1}{2} \tanh (x) \sqrt{\text{sech}^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[(1 - Tanh[x]^2)^(3/2),x]

[Out]

ArcSin[Tanh[x]]/2 + (Sqrt[Sech[x]^2]*Tanh[x])/2

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 4122

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 216

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

Rubi steps

\begin{align*} \int \left (1-\tanh ^2(x)\right )^{3/2} \, dx &=\int \text{sech}^2(x)^{3/2} \, dx\\ &=\operatorname{Subst}\left (\int \sqrt{1-x^2} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{2} \sqrt{\text{sech}^2(x)} \tanh (x)+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2}} \, dx,x,\tanh (x)\right )\\ &=\frac{1}{2} \sin ^{-1}(\tanh (x))+\frac{1}{2} \sqrt{\text{sech}^2(x)} \tanh (x)\\ \end{align*}

Mathematica [A]  time = 0.0181831, size = 29, normalized size = 1.32 \[ \frac{\text{sech}(x) \left (2 \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+\tanh (x) \text{sech}(x)\right )}{2 \sqrt{\text{sech}^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - Tanh[x]^2)^(3/2),x]

[Out]

(Sech[x]*(2*ArcTan[Tanh[x/2]] + Sech[x]*Tanh[x]))/(2*Sqrt[Sech[x]^2])

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Maple [A]  time = 0.03, size = 21, normalized size = 1. \begin{align*}{\frac{\tanh \left ( x \right ) }{2}\sqrt{1- \left ( \tanh \left ( x \right ) \right ) ^{2}}}+{\frac{\arcsin \left ( \tanh \left ( x \right ) \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-tanh(x)^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 1.68189, size = 38, normalized size = 1.73 \begin{align*} \frac{e^{\left (3 \, x\right )} - e^{x}}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1} + \arctan \left (e^{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tanh(x)^2)^(3/2),x, algorithm="maxima")

[Out]

(e^(3*x) - e^x)/(e^(4*x) + 2*e^(2*x) + 1) + arctan(e^x)

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Fricas [B]  time = 2.14712, size = 504, normalized size = 22.91 \begin{align*} \frac{\cosh \left (x\right )^{3} + 3 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + \sinh \left (x\right )^{3} +{\left (\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) +{\left (3 \, \cosh \left (x\right )^{2} - 1\right )} \sinh \left (x\right ) - \cosh \left (x\right )}{\cosh \left (x\right )^{4} + 4 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right )^{2} + 4 \,{\left (\cosh \left (x\right )^{3} + \cosh \left (x\right )\right )} \sinh \left (x\right ) + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tanh(x)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (1 - \tanh ^{2}{\left (x \right )}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tanh(x)**2)**(3/2),x)

[Out]

Integral((1 - tanh(x)**2)**(3/2), x)

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Giac [B]  time = 1.29679, size = 61, normalized size = 2.77 \begin{align*} \frac{1}{4} \, \pi - \frac{e^{\left (-x\right )} - e^{x}}{{\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4} + \frac{1}{2} \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1-tanh(x)^2)^(3/2),x, algorithm="giac")

[Out]

1/4*pi - (e^(-x) - e^x)/((e^(-x) - e^x)^2 + 4) + 1/2*arctan(1/2*(e^(2*x) - 1)*e^(-x))