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

Optimal. Leaf size=31 \[ \frac{1}{2} \tanh ^{-1}\left (\frac{\coth (x)}{\sqrt{\text{csch}^2(x)}}\right )-\frac{1}{2} \coth (x) \sqrt{\text{csch}^2(x)} \]

[Out]

ArcTanh[Coth[x]/Sqrt[Csch[x]^2]]/2 - (Coth[x]*Sqrt[Csch[x]^2])/2

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Rubi [A]  time = 0.0228202, antiderivative size = 31, 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} \[ \frac{1}{2} \tanh ^{-1}\left (\frac{\coth (x)}{\sqrt{\text{csch}^2(x)}}\right )-\frac{1}{2} \coth (x) \sqrt{\text{csch}^2(x)} \]

Antiderivative was successfully verified.

[In]

Int[(-1 + Coth[x]^2)^(3/2),x]

[Out]

ArcTanh[Coth[x]/Sqrt[Csch[x]^2]]/2 - (Coth[x]*Sqrt[Csch[x]^2])/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+\coth ^2(x)\right )^{3/2} \, dx &=\int \text{csch}^2(x)^{3/2} \, dx\\ &=-\operatorname{Subst}\left (\int \sqrt{-1+x^2} \, dx,x,\coth (x)\right )\\ &=-\frac{1}{2} \coth (x) \sqrt{\text{csch}^2(x)}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1+x^2}} \, dx,x,\coth (x)\right )\\ &=-\frac{1}{2} \coth (x) \sqrt{\text{csch}^2(x)}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\frac{\coth (x)}{\sqrt{\text{csch}^2(x)}}\right )\\ &=\frac{1}{2} \tanh ^{-1}\left (\frac{\coth (x)}{\sqrt{\text{csch}^2(x)}}\right )-\frac{1}{2} \coth (x) \sqrt{\text{csch}^2(x)}\\ \end{align*}

Mathematica [A]  time = 0.0384073, size = 40, normalized size = 1.29 \[ -\frac{1}{8} \sinh (x) \sqrt{\text{csch}^2(x)} \left (\text{csch}^2\left (\frac{x}{2}\right )+\text{sech}^2\left (\frac{x}{2}\right )+4 \log \left (\tanh \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + Coth[x]^2)^(3/2),x]

[Out]

-(Sqrt[Csch[x]^2]*(Csch[x/2]^2 + 4*Log[Tanh[x/2]] + Sech[x/2]^2)*Sinh[x])/8

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Maple [A]  time = 0.042, size = 28, normalized size = 0.9 \begin{align*} -{\frac{{\rm coth} \left (x\right )}{2}\sqrt{-1+ \left ({\rm coth} \left (x\right ) \right ) ^{2}}}+{\frac{1}{2}\ln \left ({\rm coth} \left (x\right )+\sqrt{-1+ \left ({\rm coth} \left (x\right ) \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.91738, size = 62, normalized size = 2. \begin{align*} -\frac{e^{\left (-x\right )} + e^{\left (-3 \, x\right )}}{2 \, e^{\left (-2 \, x\right )} - e^{\left (-4 \, x\right )} - 1} - \frac{1}{2} \, \log \left (e^{\left (-x\right )} + 1\right ) + \frac{1}{2} \, \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.96156, size = 737, normalized size = 23.77 \begin{align*} -\frac{2 \, \cosh \left (x\right )^{3} + 6 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} + 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 )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) +{\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 )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 2 \,{\left (3 \, \cosh \left (x\right )^{2} + 1\right )} \sinh \left (x\right ) + 2 \, \cosh \left (x\right )}{2 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*cosh(x)^3 + 6*cosh(x)*sinh(x)^2 + 2*sinh(x)^3 - (cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*co
sh(x)^2 - 1)*sinh(x)^2 - 2*cosh(x)^2 + 4*(cosh(x)^3 - cosh(x))*sinh(x) + 1)*log(cosh(x) + sinh(x) + 1) + (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)*log(cosh(x) + sinh(x) - 1) + 2*(3*cosh(x)^2 + 1)*sinh(x) + 2*cosh(x))/(cosh(x)^4 + 4*cosh(x)*si
nh(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 (\coth ^{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+coth(x)**2)**(3/2),x)

[Out]

Integral((coth(x)**2 - 1)**(3/2), x)

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Giac [B]  time = 1.17869, size = 70, normalized size = 2.26 \begin{align*} -\frac{1}{4} \,{\left (\frac{4 \,{\left (e^{\left (-x\right )} + e^{x}\right )}}{{\left (e^{\left (-x\right )} + e^{x}\right )}^{2} - 4} - \log \left (e^{\left (-x\right )} + e^{x} + 2\right ) + \log \left (e^{\left (-x\right )} + e^{x} - 2\right )\right )} \mathrm{sgn}\left (e^{\left (2 \, x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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