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

Optimal. Leaf size=17 \[ -\frac{\sinh (x) \tanh ^{-1}(\cosh (x))}{\sqrt{-\sinh ^2(x)}} \]

[Out]

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

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Rubi [A]  time = 0.0218475, antiderivative size = 17, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, 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 = 0.0072967, size = 20, normalized size = 1.18 \[ \frac{\sinh (x) \log \left (\tanh \left (\frac{x}{2}\right )\right )}{\sqrt{-\sinh ^2(x)}} \]

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 [B]  time = 0.068, size = 34, normalized size = 2. \begin{align*} -{\frac{\sinh \left ( x \right ) }{\cosh \left ( x \right ) }\sqrt{- \left ( \cosh \left ( x \right ) \right ) ^{2}}\arctan \left ({\frac{1}{\sqrt{- \left ( \cosh \left ( x \right ) \right ) ^{2}}}} \right ){\frac{1}{\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)*(-cosh(x)^2)^(1/2)*arctan(1/(-cosh(x)^2)^(1/2))/cosh(x)/(-sinh(x)^2)^(1/2)

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Maxima [C]  time = 1.66846, size = 26, normalized size = 1.53 \begin{align*} -i \, \log \left (e^{\left (-x\right )} + 1\right ) + i \, \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]

-I*log(e^(-x) + 1) + I*log(e^(-x) - 1)

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Fricas [A]  time = 2.21145, size = 4, normalized size = 0.24 \begin{align*} 0 \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]

0

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

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Giac [C]  time = 1.21177, size = 54, normalized size = 3.18 \begin{align*} -\frac{i \, \log \left (e^{x} + 1\right )}{\mathrm{sgn}\left (-e^{\left (3 \, x\right )} + e^{x}\right )} + \frac{i \, \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]

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