3.213 \(\int \frac{\coth ^2(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=12 \[ -\tanh ^{-1}(\cosh (x))+i \coth (x) \]

[Out]

-ArcTanh[Cosh[x]] + I*Coth[x]

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Rubi [A]  time = 0.0444647, antiderivative size = 12, normalized size of antiderivative = 1., 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 = {2706, 3767, 8, 3770} \[ -\tanh ^{-1}(\cosh (x))+i \coth (x) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]^2/(I + Sinh[x]),x]

[Out]

-ArcTanh[Cosh[x]] + I*Coth[x]

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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{\coth ^2(x)}{i+\sinh (x)} \, dx &=-\left (i \int \text{csch}^2(x) \, dx\right )+\int \text{csch}(x) \, dx\\ &=-\tanh ^{-1}(\cosh (x))-\operatorname{Subst}(\int 1 \, dx,x,-i \coth (x))\\ &=-\tanh ^{-1}(\cosh (x))+i \coth (x)\\ \end{align*}

Mathematica [B]  time = 0.0346959, size = 32, normalized size = 2.67 \[ \frac{1}{2} i \tanh \left (\frac{x}{2}\right )+\frac{1}{2} i \coth \left (\frac{x}{2}\right )+\log \left (\tanh \left (\frac{x}{2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]^2/(I + Sinh[x]),x]

[Out]

(I/2)*Coth[x/2] + Log[Tanh[x/2]] + (I/2)*Tanh[x/2]

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Maple [A]  time = 0.026, size = 23, normalized size = 1.9 \begin{align*}{\frac{i}{2}}\tanh \left ({\frac{x}{2}} \right ) +{{\frac{i}{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(x)^2/(I+sinh(x)),x)

[Out]

1/2*I*tanh(1/2*x)+1/2*I/tanh(1/2*x)+ln(tanh(1/2*x))

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Maxima [B]  time = 1.07307, size = 36, normalized size = 3. \begin{align*} -\frac{2 i}{e^{\left (-2 \, x\right )} - 1} - \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(coth(x)^2/(I+sinh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.0131, size = 108, normalized size = 9. \begin{align*} -\frac{{\left (e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} + 1\right ) -{\left (e^{\left (2 \, x\right )} - 1\right )} \log \left (e^{x} - 1\right ) - 2 i}{e^{\left (2 \, x\right )} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^2/(I+sinh(x)),x, algorithm="fricas")

[Out]

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

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Sympy [B]  time = 0.217057, size = 22, normalized size = 1.83 \begin{align*} \log{\left (e^{x} - 1 \right )} - \log{\left (e^{x} + 1 \right )} + \frac{2 i}{e^{2 x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)**2/(I+sinh(x)),x)

[Out]

log(exp(x) - 1) - log(exp(x) + 1) + 2*I/(exp(2*x) - 1)

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Giac [B]  time = 1.12792, size = 32, normalized size = 2.67 \begin{align*} \frac{2 i}{e^{\left (2 \, x\right )} - 1} - \log \left (e^{x} + 1\right ) + \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(x)^2/(I+sinh(x)),x, algorithm="giac")

[Out]

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