\(\int \frac {\coth (x)}{i+\text {csch}(x)} \, dx\) [107]

Optimal result

Integrand size = 11, antiderivative size = 13 \[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=-i \log (i-\sinh (x)) \]

[Out]

-I*ln(I-sinh(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3964, 31} \[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=-i \log (-\sinh (x)+i) \]

[In]

Int[Coth[x]/(I + Csch[x]),x]

[Out]

(-I)*Log[I - Sinh[x]]

Rule 31

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

Rule 3964

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[1/(a^(m - n
- 1)*b^n*d), Subst[Int[(a - b*x)^((m - 1)/2)*((a + b*x)^((m - 1)/2 + n)/x^(m + n)), x], x, Sin[c + d*x]], x] /
; FreeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{i+i x} \, dx,x,i \sinh (x)\right ) \\ & = -i \log (i-\sinh (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=-i \log (i-\sinh (x)) \]

[In]

Integrate[Coth[x]/(I + Csch[x]),x]

[Out]

(-I)*Log[I - Sinh[x]]

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15

method	result	size
risch	\(i x -2 i \ln \left ({\mathrm e}^{x}-i\right )\)	\(15\)
derivativedivides	\(i \ln \left (\operatorname {csch}\left (x \right )\right )-\frac {i \ln \left (1+\operatorname {csch}\left (x \right )^{2}\right )}{2}-\arctan \left (\operatorname {csch}\left (x \right )\right )\)	\(23\)
default	\(i \ln \left (\operatorname {csch}\left (x \right )\right )-\frac {i \ln \left (1+\operatorname {csch}\left (x \right )^{2}\right )}{2}-\arctan \left (\operatorname {csch}\left (x \right )\right )\)	\(23\)

[In]

int(coth(x)/(I+csch(x)),x,method=_RETURNVERBOSE)

[Out]

I*x-2*I*ln(exp(x)-I)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=i \, x - 2 i \, \log \left (e^{x} - i\right ) \]

[In]

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

[Out]

I*x - 2*I*log(e^x - I)

Sympy [F]

\[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=\int \frac {\coth {\left (x \right )}}{\operatorname {csch}{\left (x \right )} + i}\, dx \]

[In]

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

[Out]

Integral(coth(x)/(csch(x) + I), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.15 \[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=-i \, x - 2 i \, \log \left (i \, e^{\left (-x\right )} - 1\right ) \]

[In]

integrate(coth(x)/(I+csch(x)),x, algorithm="maxima")

[Out]

-I*x - 2*I*log(I*e^(-x) - 1)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=i \, x - 2 i \, \log \left (e^{x} - i\right ) \]

[In]

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

[Out]

I*x - 2*I*log(e^x - I)

Mupad [B] (verification not implemented)

Time = 2.22 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {\coth (x)}{i+\text {csch}(x)} \, dx=x\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,2{}\mathrm {i} \]

[In]

int(coth(x)/(1/sinh(x) + 1i),x)

[Out]

x*1i - log(exp(x) - 1i)*2i