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

Optimal. Leaf size=35 \[ \frac{1}{2} \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{-\text{sech}^2(x)}}\right )-\frac{1}{2} \tanh (x) \sqrt{-\text{sech}^2(x)} \]

[Out]

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

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Rubi [A]  time = 0.0235295, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3657, 4122, 195, 217, 206} \[ \frac{1}{2} \tanh ^{-1}\left (\frac{\tanh (x)}{\sqrt{-\text{sech}^2(x)}}\right )-\frac{1}{2} \tanh (x) \sqrt{-\text{sech}^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + Tanh[x]^2)^(3/2),x]

[Out]

ArcTanh[Tanh[x]/Sqrt[-Sech[x]^2]]/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 217

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

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

Mathematica [A]  time = 0.0193486, size = 28, normalized size = 0.8 \[ -\frac{1}{2} \sqrt{-\text{sech}^2(x)} \left (\tanh (x)+2 \cosh (x) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Tanh[x]^2)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.035, size = 28, normalized size = 0.8 \begin{align*} -{\frac{\tanh \left ( x \right ) }{2}\sqrt{-1+ \left ( \tanh \left ( x \right ) \right ) ^{2}}}+{\frac{1}{2}\ln \left ( \tanh \left ( x \right ) +\sqrt{-1+ \left ( \tanh \left ( x \right ) \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C]  time = 1.72371, size = 43, normalized size = 1.23 \begin{align*} \frac{-i \, e^{\left (3 \, x\right )} + i \, e^{x}}{e^{\left (4 \, x\right )} + 2 \, e^{\left (2 \, x\right )} + 1} - i \, \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]

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

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Fricas [A]  time = 2.35453, size = 4, normalized size = 0.11 \begin{align*} 0 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

0

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

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Giac [C]  time = 1.20675, size = 84, normalized size = 2.4 \begin{align*} -\frac{i \, e^{\left (-x\right )} - i \, e^{x}}{{\left (-i \, e^{\left (-x\right )} + i \, e^{x}\right )}^{2} - 4} + \frac{1}{8} \, \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right ) - \frac{1}{4} \, \log \left (-i \, e^{\left (-x\right )} + i \, e^{x} + 2\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]

-(I*e^(-x) - I*e^x)/((-I*e^(-x) + I*e^x)^2 - 4) + 1/8*log((e^(-x) - e^x)^2 + 4) - 1/4*log(-I*e^(-x) + I*e^x +
2)