Integrand size = 22, antiderivative size = 41 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\text {arctanh}(\cosh (c+d x))}{a d}+\frac {\cosh (c+d x)}{d (a+i a \sinh (c+d x))} \] Output:
-arctanh(cosh(d*x+c))/a/d+cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))
Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.27 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\text {sech}(c+d x) \left (-1+\text {arctanh}\left (\sqrt {\cosh ^2(c+d x)}\right ) \sqrt {\cosh ^2(c+d x)}+i \sinh (c+d x)\right )}{a d} \] Input:
Integrate[Csch[c + d*x]/(a + I*a*Sinh[c + d*x]),x]
Output:
-((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))
Time = 0.33 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 26, 3226, 26, 3042, 26, 3127, 4257}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\sin (i c+i d x) (a+a \sin (i c+i d x))}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\sin (i c+i d x) (\sin (i c+i d x) a+a)}dx\) |
\(\Big \downarrow \) 3226 |
\(\displaystyle i \left (\frac {\int -i \text {csch}(c+d x)dx}{a}-\int \frac {1}{i \sinh (c+d x) a+a}dx\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (-\int \frac {1}{i \sinh (c+d x) a+a}dx-\frac {i \int \text {csch}(c+d x)dx}{a}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle i \left (-\int \frac {1}{\sin (i c+i d x) a+a}dx-\frac {i \int i \csc (i c+i d x)dx}{a}\right )\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \left (\frac {\int \csc (i c+i d x)dx}{a}-\int \frac {1}{\sin (i c+i d x) a+a}dx\right )\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle i \left (\frac {\int \csc (i c+i d x)dx}{a}-\frac {i \cosh (c+d x)}{d (a+i a \sinh (c+d x))}\right )\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle i \left (\frac {i \text {arctanh}(\cosh (c+d x))}{a d}-\frac {i \cosh (c+d x)}{d (a+i a \sinh (c+d x))}\right )\) |
Input:
Int[Csch[c + d*x]/(a + I*a*Sinh[c + d*x]),x]
Output:
I*((I*ArcTanh[Cosh[c + d*x]])/(a*d) - (I*Cosh[c + d*x])/(d*(a + I*a*Sinh[c + d*x])))
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
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]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[ {a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.27 (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 | \(\frac {\left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (i-\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\) | \(61\) |
Input:
int(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d/a*(ln(tanh(1/2*d*x+1/2*c))-2*I/(-I+tanh(1/2*d*x+1/2*c)))
Time = 0.10 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.39 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {{\left (e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} + 1\right ) - {\left (e^{\left (d x + c\right )} - i\right )} \log \left (e^{\left (d x + c\right )} - 1\right ) - 2}{a d e^{\left (d x + c\right )} - i \, a d} \] Input:
integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")
Output:
-((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)
\[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=- \frac {i \int \frac {\operatorname {csch}{\left (c + d x \right )}}{\sinh {\left (c + d x \right )} - i}\, dx}{a} \] Input:
integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)
Output:
-I*Integral(csch(c + d*x)/(sinh(c + d*x) - I), x)/a
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.51 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} + \frac {2}{{\left (a e^{\left (-d x - c\right )} + i \, a\right )} d} \] Input:
integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")
Output:
-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)
Time = 0.12 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.17 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {\frac {\log \left (e^{\left (d x + c\right )} + 1\right )}{a} - \frac {\log \left (e^{\left (d x + c\right )} - 1\right )}{a} - \frac {2}{a {\left (e^{\left (d x + c\right )} - i\right )}}}{d} \] Input:
integrate(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x, algorithm="giac")
Output:
-(log(e^(d*x + c) + 1)/a - log(e^(d*x + c) - 1)/a - 2/(a*(e^(d*x + c) - I) ))/d
Time = 1.84 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.37 \[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\sqrt {-a^2\,d^2}}{a\,d}\right )}{\sqrt {-a^2\,d^2}}+\frac {2}{a\,d\,\left ({\mathrm {e}}^{c+d\,x}-\mathrm {i}\right )} \] Input:
int(1/(sinh(c + d*x)*(a + a*sinh(c + d*x)*1i)),x)
Output:
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)
\[ \int \frac {\text {csch}(c+d x)}{a+i a \sinh (c+d x)} \, dx=\frac {\int \frac {\mathrm {csch}\left (d x +c \right )}{\sinh \left (d x +c \right ) i +1}d x}{a} \] Input:
int(csch(d*x+c)/(a+I*a*sinh(d*x+c)),x)
Output:
int(csch(c + d*x)/(sinh(c + d*x)*i + 1),x)/a