3.90 \(\int \frac{\text{csch}(x)}{i+\tanh (x)} \, dx\)

Optimal. Leaf size=33 \[ i \tanh ^{-1}(\cosh (x))-\frac{i \tanh ^{-1}\left (\frac{\cosh (x)+i \sinh (x)}{\sqrt{2}}\right )}{\sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.102202, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {3518, 3110, 3770, 3074, 206} \[ i \tanh ^{-1}(\cosh (x))-\frac{i \tanh ^{-1}\left (\frac{\cosh (x)+i \sinh (x)}{\sqrt{2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[Csch[x]/(I + Tanh[x]),x]

[Out]

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

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rule 3110

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Int[ExpandTrig[(cos[c + d*x]^m*sin[c + d*x]^n)/(a*cos[c + d*x] + b*sin[c + d
*x]), x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IntegersQ[m, n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0676641, size = 46, normalized size = 1.39 \[ -i \left (\sqrt{2} \tanh ^{-1}\left (\frac{1+i \tanh \left (\frac{x}{2}\right )}{\sqrt{2}}\right )+\log \left (\sinh \left (\frac{x}{2}\right )\right )-\log \left (\cosh \left (\frac{x}{2}\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[x]/(I + Tanh[x]),x]

[Out]

(-I)*(Sqrt[2]*ArcTanh[(1 + I*Tanh[x/2])/Sqrt[2]] - Log[Cosh[x/2]] + Log[Sinh[x/2]])

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Maple [A]  time = 0.036, size = 29, normalized size = 0.9 \begin{align*} \sqrt{2}\arctan \left ({\frac{\sqrt{2}}{4} \left ( 2\,\tanh \left ( x/2 \right ) -2\,i \right ) } \right ) -i\ln \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(x)/(I+tanh(x)),x)

[Out]

2^(1/2)*arctan(1/4*(2*tanh(1/2*x)-2*I)*2^(1/2))-I*ln(tanh(1/2*x))

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Maxima [A]  time = 1.56259, size = 46, normalized size = 1.39 \begin{align*} -\sqrt{2} \arctan \left (\left (\frac{1}{2} i + \frac{1}{2}\right ) \, \sqrt{2} e^{\left (-x\right )}\right ) + 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(csch(x)/(I+tanh(x)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.32153, size = 180, normalized size = 5.45 \begin{align*} -\frac{1}{2} i \, \sqrt{2} \log \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} + e^{x}\right ) + \frac{1}{2} i \, \sqrt{2} \log \left (\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} + e^{x}\right ) + i \, \log \left (e^{x} + 1\right ) - i \, \log \left (e^{x} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(I+tanh(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{csch}{\left (x \right )}}{\tanh{\left (x \right )} + i}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(I+tanh(x)),x)

[Out]

Integral(csch(x)/(tanh(x) + I), x)

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Giac [A]  time = 1.22636, size = 38, normalized size = 1.15 \begin{align*} \sqrt{2} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} e^{x}\right ) + i \, \log \left (e^{x} + 1\right ) - i \, \log \left ({\left | e^{x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(x)/(I+tanh(x)),x, algorithm="giac")

[Out]

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