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)/(-sinh(x)^2)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.49, size = 1, normalized size = 0.06 \[ 0 \]

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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giac [C]  time = 0.15, size = 40, normalized size = 2.35 \[ -\frac {i \, \log \left (e^{x} + 1\right )}{\mathrm {sgn}\left (-e^{\left (3 \, x\right )} + e^{x}\right )} + \frac {i \, \log \left ({\left | e^{x} - 1 \right |}\right )}{\mathrm {sgn}\left (-e^{\left (3 \, x\right )} + e^{x}\right )} \]

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)

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maple [B]  time = 0.14, size = 34, normalized size = 2.00 \[ -\frac {\sinh \relax (x ) \sqrt {-\left (\cosh ^{2}\relax (x )\right )}\, \arctan \left (\frac {1}{\sqrt {-\left (\cosh ^{2}\relax (x )\right )}}\right )}{\cosh \relax (x ) \sqrt {-\left (\sinh ^{2}\relax (x )\right )}} \]

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 = 0.82, size = 19, normalized size = 1.12 \[ -i \, \log \left (e^{\left (-x\right )} + 1\right ) + i \, \log \left (e^{\left (-x\right )} - 1\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.06 \[ \int \frac {1}{\sqrt {1-{\mathrm {cosh}\relax (x)}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1 - cosh(x)^2)^(1/2),x)

[Out]

int(1/(1 - cosh(x)^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {1 - \cosh ^{2}{\relax (x )}}}\, dx \]

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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