3.45 \(\int \frac{\text{csch}^2(x)}{i+\sinh (x)} \, dx\)

Optimal. Leaf size=23 \[ 2 i \coth (x)-\tanh ^{-1}(\cosh (x))+\frac{\coth (x)}{\sinh (x)+i} \]

[Out]

-ArcTanh[Cosh[x]] + (2*I)*Coth[x] + Coth[x]/(I + Sinh[x])

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Rubi [A]  time = 0.0584309, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2768, 2748, 3767, 8, 3770} \[ 2 i \coth (x)-\tanh ^{-1}(\cosh (x))+\frac{\coth (x)}{\sinh (x)+i} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]^2/(I + Sinh[x]),x]

[Out]

-ArcTanh[Cosh[x]] + (2*I)*Coth[x] + Coth[x]/(I + Sinh[x])

Rule 2768

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(b
^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(b*c - a*d)*(a + b*Sin[e + f*x])), x] + Dist[d/(a*(b*c - a*
d)), Int[(c + d*Sin[e + f*x])^n*(a*n - b*(n + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, 0] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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

Mathematica [A]  time = 0.039471, size = 36, normalized size = 1.57 \[ \text{sech}(x) \left (2 i \sinh (x)+i \text{csch}(x)-\sqrt{\cosh ^2(x)} \tanh ^{-1}\left (\sqrt{\cosh ^2(x)}\right )+1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]^2/(I + Sinh[x]),x]

[Out]

Sech[x]*(1 - ArcTanh[Sqrt[Cosh[x]^2]]*Sqrt[Cosh[x]^2] + I*Csch[x] + (2*I)*Sinh[x])

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Maple [A]  time = 0.028, size = 35, normalized size = 1.5 \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 ) +{2\,i \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(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))+2*I/(tanh(1/2*x)+I)

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Maxima [B]  time = 1.23491, size = 72, normalized size = 3.13 \begin{align*} -\frac{4 \,{\left (-i \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 2\right )}}{2 \, e^{\left (-x\right )} + 2 i \, e^{\left (-2 \, x\right )} - 2 \, e^{\left (-3 \, x\right )} - 2 i} - \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(csch(x)^2/(I+sinh(x)),x, algorithm="maxima")

[Out]

-4*(-I*e^(-x) + e^(-2*x) - 2)/(2*e^(-x) + 2*I*e^(-2*x) - 2*e^(-3*x) - 2*I) - log(e^(-x) + 1) + log(e^(-x) - 1)

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Fricas [B]  time = 2.10151, size = 208, normalized size = 9.04 \begin{align*} -\frac{{\left (e^{\left (3 \, x\right )} + i \, e^{\left (2 \, x\right )} - e^{x} - i\right )} \log \left (e^{x} + 1\right ) -{\left (e^{\left (3 \, x\right )} + i \, e^{\left (2 \, x\right )} - e^{x} - i\right )} \log \left (e^{x} - 1\right ) - 2 \, e^{\left (2 \, x\right )} - 2 i \, e^{x} + 4}{e^{\left (3 \, x\right )} + i \, e^{\left (2 \, x\right )} - e^{x} - i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((e^(3*x) + I*e^(2*x) - e^x - I)*log(e^x + 1) - (e^(3*x) + I*e^(2*x) - e^x - I)*log(e^x - 1) - 2*e^(2*x) - 2*
I*e^x + 4)/(e^(3*x) + I*e^(2*x) - e^x - I)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.50181, size = 59, normalized size = 2.57 \begin{align*} \frac{2 \,{\left (e^{\left (2 \, x\right )} + i \, e^{x} - 2\right )}}{e^{\left (3 \, x\right )} + i \, e^{\left (2 \, x\right )} - e^{x} - i} - \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(csch(x)^2/(I+sinh(x)),x, algorithm="giac")

[Out]

2*(e^(2*x) + I*e^x - 2)/(e^(3*x) + I*e^(2*x) - e^x - I) - log(e^x + 1) + log(abs(e^x - 1))