\(\int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) [208]

Optimal result

Integrand size = 22, antiderivative size = 41 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\text {arctanh}(\cosh (c+d x))}{a d}+\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))} \]

[Out]

-arctanh(cosh(d*x+c))/a/d+cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2826, 3855, 2727} \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\text {arctanh}(\cosh (c+d x))}{a d}+\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))} \]

[In]

Int[Csch[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

-(ArcTanh[Cosh[c + d*x]]/(a*d)) + Cosh[c + d*x]/(d*(a + I*a*Sinh[c + d*x]))

Rule 2727

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

Rule 2826

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = -\left (i \int \frac {1}{a+i a \sinh (c+d x)} \, dx\right )+\frac {\int \text {csch}(c+d x) \, dx}{a} \\ & = -\frac {\text {arctanh}(\cosh (c+d x))}{a d}+\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.27 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\text {sech}(c+d x) \left (-1+\text {arctanh}\left (\sqrt {\cosh ^2(c+d x)}\right ) \sqrt {\cosh ^2(c+d x)}+i \sinh (c+d x)\right )}{a d} \]

[In]

Integrate[Csch[c + d*x]/(a + I*a*Sinh[c + d*x]),x]

[Out]

-((Sech[c + d*x]*(-1 + ArcTanh[Sqrt[Cosh[c + d*x]^2]]*Sqrt[Cosh[c + d*x]^2] + I*Sinh[c + d*x]))/(a*d))

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 i}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d a}\)	\(36\)
default	\(\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 i}{-i+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}}{d a}\)	\(36\)
risch	\(\frac {2}{d a \left ({\mathrm e}^{d x +c}-i\right )}-\frac {\ln \left ({\mathrm e}^{d x +c}+1\right )}{d a}+\frac {\ln \left ({\mathrm e}^{d x +c}-1\right )}{d a}\)	\(54\)
parallelrisch	\(\frac {\left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\)	\(61\)

[In]

int(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/d/a*(ln(tanh(1/2*d*x+1/2*c))-2*I/(-I+tanh(1/2*d*x+1/2*c)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {{\left (e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) - {\left (e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) - 2}{a d e^{\left (d x + c\right )} - i \, a d} \]

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \int \frac {\operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \]

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)

[Out]

-I*Integral(csch(c + d*x)/(sinh(c + d*x) - I), x)/a

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.51 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} + \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d} \]

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.17 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\frac {\log \left (e^{\left (d x + c\right )} + 1\right )}{a} - \frac {\log \left (e^{\left (d x + c\right )} - 1\right )}{a} - \frac {2}{a {\left (e^{\left (d x + c\right )} - i\right )}}}{d} \]

[In]

integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.64 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.37 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^2\,d^2}}{a\,d}\right )}{\sqrt {-a^2\,d^2}}+\frac {2}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )} \]

[In]

int(1/(sinh(c + d*x)*(a + a*sinh(c + d*x)*1i)),x)

[Out]

2/(a*d*(exp(c + d*x) - 1i)) - (2*atan((exp(d*x)*exp(c)*(-a^2*d^2)^(1/2))/(a*d)))/(-a^2*d^2)^(1/2)