3.54 \(\int \frac{1}{\sqrt{-1+\cosh ^2(x)}} \, dx\)

Optimal. Leaf size=15 \[ -\frac{\sinh (x) \tanh ^{-1}(\cosh (x))}{\sqrt{\sinh ^2(x)}} \]

[Out]

-((ArcTanh[Cosh[x]]*Sinh[x])/Sqrt[Sinh[x]^2])

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Rubi [A]  time = 0.0188273, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {3176, 3207, 3770} \[ -\frac{\sinh (x) \tanh ^{-1}(\cosh (x))}{\sqrt{\sinh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-1 + Cosh[x]^2],x]

[Out]

-((ArcTanh[Cosh[x]]*Sinh[x])/Sqrt[Sinh[x]^2])

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-1+\cosh ^2(x)}} \, dx &=\int \frac{1}{\sqrt{\sinh ^2(x)}} \, dx\\ &=\frac{\sinh (x) \int \text{csch}(x) \, dx}{\sqrt{\sinh ^2(x)}}\\ &=-\frac{\tanh ^{-1}(\cosh (x)) \sinh (x)}{\sqrt{\sinh ^2(x)}}\\ \end{align*}

Mathematica [A]  time = 0.008162, size = 18, normalized size = 1.2 \[ \frac{\sinh (x) \log \left (\tanh \left (\frac{x}{2}\right )\right )}{\sqrt{\sinh ^2(x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-1 + Cosh[x]^2],x]

[Out]

(Log[Tanh[x/2]]*Sinh[x])/Sqrt[Sinh[x]^2]

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Maple [A]  time = 0.076, size = 16, normalized size = 1.1 \begin{align*} -{\frac{{\it Artanh} \left ( \cosh \left ( x \right ) \right ) }{\sinh \left ( x \right ) }\sqrt{ \left ( \sinh \left ( x \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sinh(x)^2)^(1/2)*arctanh(cosh(x))/sinh(x)

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Maxima [A]  time = 1.62907, size = 23, normalized size = 1.53 \begin{align*} \log \left (e^{\left (-x\right )} + 1\right ) - \log \left (e^{\left (-x\right )} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(e^(-x) + 1) - log(e^(-x) - 1)

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Fricas [A]  time = 2.03799, size = 78, normalized size = 5.2 \begin{align*} -\log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(cosh(x) + sinh(x) + 1) + log(cosh(x) + sinh(x) - 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(cosh(x)**2 - 1), x)

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Giac [B]  time = 1.29688, size = 53, normalized size = 3.53 \begin{align*} -\frac{\log \left (e^{x} + 1\right )}{\mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} + \frac{\log \left ({\left | e^{x} - 1 \right |}\right )}{\mathrm{sgn}\left (e^{\left (3 \, x\right )} - e^{x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(e^x + 1)/sgn(e^(3*x) - e^x) + log(abs(e^x - 1))/sgn(e^(3*x) - e^x)