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

Optimal. Leaf size=25 \[ -\frac{i}{\sinh (x)+i}-\log (\sinh (x))+\log (\sinh (x)+i) \]

[Out]

-Log[Sinh[x]] + Log[I + Sinh[x]] - I/(I + Sinh[x])

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Rubi [A]  time = 0.0400701, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {2707, 44} \[ -\frac{i}{\sinh (x)+i}-\log (\sinh (x))+\log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Int[Coth[x]/(I + Sinh[x])^2,x]

[Out]

-Log[Sinh[x]] + Log[I + Sinh[x]] - I/(I + Sinh[x])

Rule 2707

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^(m - (p + 1)/2))/(a - x)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& EqQ[a^2 - b^2, 0] && IntegerQ[(p + 1)/2]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.026686, size = 25, normalized size = 1. \[ -\frac{i}{\sinh (x)+i}-\log (\sinh (x))+\log (\sinh (x)+i) \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[x]/(I + Sinh[x])^2,x]

[Out]

-Log[Sinh[x]] + Log[I + Sinh[x]] - I/(I + Sinh[x])

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Maple [A]  time = 0.033, size = 23, normalized size = 0.9 \begin{align*} -\ln \left ( \sinh \left ( x \right ) \right ) +\ln \left ( i+\sinh \left ( x \right ) \right ) -{\frac{i}{i+\sinh \left ( x \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-ln(sinh(x))+ln(I+sinh(x))-I/(I+sinh(x))

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Maxima [B]  time = 1.11766, size = 65, normalized size = 2.6 \begin{align*} \frac{2 i \, e^{\left (-x\right )}}{-2 i \, e^{\left (-x\right )} + e^{\left (-2 \, x\right )} - 1} - \log \left (e^{\left (-x\right )} + 1\right ) + 2 \, \log \left (e^{\left (-x\right )} - i\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)/(I+sinh(x))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.0516, size = 162, normalized size = 6.48 \begin{align*} -\frac{{\left (e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )} \log \left (e^{\left (2 \, x\right )} - 1\right ) - 2 \,{\left (e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1\right )} \log \left (e^{x} + i\right ) + 2 i \, e^{x}}{e^{\left (2 \, x\right )} + 2 i \, e^{x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.513079, size = 36, normalized size = 1.44 \begin{align*} 2 \log{\left (e^{x} + i \right )} - \log{\left (e^{2 x} - 1 \right )} - \frac{2 i e^{x}}{e^{2 x} + 2 i e^{x} - 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.12059, size = 45, normalized size = 1.8 \begin{align*} -\frac{2 i \, e^{x}}{{\left (e^{x} + i\right )}^{2}} - \log \left (e^{x} + 1\right ) + 2 \, \log \left (e^{x} + i\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)/(I+sinh(x))^2,x, algorithm="giac")

[Out]

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