3.425 \(\int \frac {\coth (x)}{1-\sinh ^2(x)} \, dx\)

Optimal. Leaf size=17 \[ \log (\sinh (x))-\frac {1}{2} \log \left (1-\sinh ^2(x)\right ) \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3194, 36, 31, 29} \[ \log (\sinh (x))-\frac {1}{2} \log \left (1-\sinh ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(1 - Sinh[x]^2),x]

[Out]

Log[Sinh[x]] - Log[1 - Sinh[x]^2]/2

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\coth (x)}{1-\sinh ^2(x)} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1-x) x} \, dx,x,\sinh ^2(x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sinh ^2(x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\sinh ^2(x)\right )\\ &=\log (\sinh (x))-\frac {1}{2} \log \left (1-\sinh ^2(x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 1.35 \[ -2 \left (\frac {1}{4} \log \left (1-\sinh ^2(x)\right )-\frac {1}{2} \log (\sinh (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(1 - Sinh[x]^2),x]

[Out]

-2*(-1/2*Log[Sinh[x]] + Log[1 - Sinh[x]^2]/4)

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fricas [B]  time = 1.03, size = 47, normalized size = 2.76 \[ -\frac {1}{2} \, \log \left (\frac {2 \, {\left (\cosh \relax (x)^{2} + \sinh \relax (x)^{2} - 3\right )}}{\cosh \relax (x)^{2} - 2 \, \cosh \relax (x) \sinh \relax (x) + \sinh \relax (x)^{2}}\right ) + \log \left (\frac {2 \, \sinh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(2*(cosh(x)^2 + sinh(x)^2 - 3)/(cosh(x)^2 - 2*cosh(x)*sinh(x) + sinh(x)^2)) + log(2*sinh(x)/(cosh(x) -
sinh(x)))

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giac [A]  time = 0.13, size = 25, normalized size = 1.47 \[ -\frac {1}{2} \, \log \left ({\left | e^{\left (4 \, x\right )} - 6 \, e^{\left (2 \, x\right )} + 1 \right |}\right ) + \log \left ({\left | e^{\left (2 \, x\right )} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(abs(e^(4*x) - 6*e^(2*x) + 1)) + log(abs(e^(2*x) - 1))

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maple [B]  time = 0.09, size = 41, normalized size = 2.41 \[ -\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )-2 \tanh \left (\frac {x}{2}\right )-1\right )}{2}-\frac {\ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+2 \tanh \left (\frac {x}{2}\right )-1\right )}{2}+\ln \left (\tanh \left (\frac {x}{2}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)/(1-sinh(x)^2),x)

[Out]

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

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maxima [B]  time = 0.33, size = 45, normalized size = 2.65 \[ -\frac {1}{2} \, \log \left (2 \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 1\right ) + \log \left (e^{\left (-x\right )} + 1\right ) + \log \left (e^{\left (-x\right )} - 1\right ) - \frac {1}{2} \, \log \left (-2 \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 27, normalized size = 1.59 \[ \ln \left (5184\,{\mathrm {e}}^{2\,x}-5184\right )-\frac {\ln \left (9\,{\mathrm {e}}^{4\,x}-54\,{\mathrm {e}}^{2\,x}+9\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-coth(x)/(sinh(x)^2 - 1),x)

[Out]

log(5184*exp(2*x) - 5184) - log(9*exp(4*x) - 54*exp(2*x) + 9)/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)/(1-sinh(x)**2),x)

[Out]

-Integral(coth(x)/(sinh(x)**2 - 1), x)

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