3.23 \(\int \sqrt{-1+\text{csch}^2(x)} \, dx\)

Optimal. Leaf size=33 \[ -\tan ^{-1}\left (\frac{\coth (x)}{\sqrt{\coth ^2(x)-2}}\right )-\tanh ^{-1}\left (\frac{\coth (x)}{\sqrt{\coth ^2(x)-2}}\right ) \]

[Out]

-ArcTan[Coth[x]/Sqrt[-2 + Coth[x]^2]] - ArcTanh[Coth[x]/Sqrt[-2 + Coth[x]^2]]

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Rubi [A]  time = 0.0241303, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {4128, 402, 217, 206, 377, 203} \[ -\tan ^{-1}\left (\frac{\coth (x)}{\sqrt{\coth ^2(x)-2}}\right )-\tanh ^{-1}\left (\frac{\coth (x)}{\sqrt{\coth ^2(x)-2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-1 + Csch[x]^2],x]

[Out]

-ArcTan[Coth[x]/Sqrt[-2 + Coth[x]^2]] - ArcTanh[Coth[x]/Sqrt[-2 + Coth[x]^2]]

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff/f, Subst[Int[(a + b + b*ff^2*x^2)^p/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p},
x] && NeQ[a + b, 0] && NeQ[p, -1]

Rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Dist[b/d, Int[(a + b*x^2)^(p - 1), x], x]
- Dist[(b*c - a*d)/d, Int[(a + b*x^2)^(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d,
0] && GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4])

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])

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

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

Rubi steps

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

Mathematica [B]  time = 0.0375777, size = 68, normalized size = 2.06 \[ \frac{\sqrt{2} \sinh (x) \sqrt{\text{csch}^2(x)-1} \left (\log \left (\sqrt{2} \cosh (x)+\sqrt{\cosh (2 x)-3}\right )+\tan ^{-1}\left (\frac{\sqrt{2} \cosh (x)}{\sqrt{\cosh (2 x)-3}}\right )\right )}{\sqrt{\cosh (2 x)-3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-1 + Csch[x]^2],x]

[Out]

(Sqrt[2]*Sqrt[-1 + Csch[x]^2]*(ArcTan[(Sqrt[2]*Cosh[x])/Sqrt[-3 + Cosh[2*x]]] + Log[Sqrt[2]*Cosh[x] + Sqrt[-3
+ Cosh[2*x]]])*Sinh[x])/Sqrt[-3 + Cosh[2*x]]

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Maple [F]  time = 0.115, size = 0, normalized size = 0. \begin{align*} \int \sqrt{-1+ \left ({\rm csch} \left (x\right ) \right ) ^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\operatorname{csch}\left (x\right )^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(csch(x)^2 - 1), x)

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Fricas [B]  time = 2.25153, size = 1273, normalized size = 38.58 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 - 1)*sqrt(-(cosh(x)^2 + sinh(x)^2 - 3)/(cosh(x)^
2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(cosh(x)^4 + 4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 2*(3*cosh(x)^2 + 2)*sinh(x)
^2 + 4*cosh(x)^2 + 4*(cosh(x)^3 + 2*cosh(x))*sinh(x) - 1)) + 1/2*arctan(sqrt(2)*(cosh(x)^2 + 2*cosh(x)*sinh(x)
+ sinh(x)^2 - 1)*sqrt(-(cosh(x)^2 + sinh(x)^2 - 3)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2))/(cosh(x)^4 +
4*cosh(x)*sinh(x)^3 + sinh(x)^4 + 6*(cosh(x)^2 - 1)*sinh(x)^2 - 6*cosh(x)^2 + 4*(cosh(x)^3 - 3*cosh(x))*sinh(x
) + 1)) - 1/2*log((cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2 + sqrt(2)*sqrt(-(cosh(x)^2 + sinh(x)^2 - 3)/(cosh
(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2)) + 1/2*log((cosh(x)^2
+ 2*cosh(x)*sinh(x) + sinh(x)^2 - sqrt(2)*sqrt(-(cosh(x)^2 + sinh(x)^2 - 3)/(cosh(x)^2 - 2*cosh(x)*sinh(x) +
sinh(x)^2)) + 1)/(cosh(x)^2 + 2*cosh(x)*sinh(x) + sinh(x)^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(csch(x)**2 - 1), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{\operatorname{csch}\left (x\right )^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(csch(x)^2 - 1), x)