Integrand size = 11, antiderivative size = 33 \[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=i \text {arctanh}(\cosh (x))-\frac {i \text {arctanh}\left (\frac {\cosh (x)+i \sinh (x)}{\sqrt {2}}\right )}{\sqrt {2}} \] Output:
I*arctanh(cosh(x))-1/2*I*arctanh(1/2*(cosh(x)+I*sinh(x))*2^(1/2))*2^(1/2)
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.39 \[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=-i \left (\sqrt {2} \text {arctanh}\left (\frac {1+i \tanh \left (\frac {x}{2}\right )}{\sqrt {2}}\right )-\log \left (\cosh \left (\frac {x}{2}\right )\right )+\log \left (\sinh \left (\frac {x}{2}\right )\right )\right ) \] Input:
Integrate[Csch[x]/(I + Tanh[x]),x]
Output:
(-I)*(Sqrt[2]*ArcTanh[(1 + I*Tanh[x/2])/Sqrt[2]] - Log[Cosh[x/2]] + Log[Si nh[x/2]])
Time = 0.35 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.818, Rules used = {3042, 26, 26, 4001, 26, 3042, 26, 3589, 2009}
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.
\(\displaystyle \int \frac {\text {csch}(x)}{\tanh (x)+i} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\sin (i x) (i-i \tan (i x))}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int -\frac {i}{\sin (i x) (1-\tan (i x))}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {1}{\sin (i x) (1-\tan (i x))}dx\) |
\(\Big \downarrow \) 4001 |
\(\displaystyle \int -\frac {i \coth (x)}{\cosh (x)-i \sinh (x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\coth (x)}{\cosh (x)-i \sinh (x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -i \int \frac {i \cos (i x)}{(\cos (i x)-\sin (i x)) \sin (i x)}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle \int \frac {\cos (i x)}{\sin (i x) (\cos (i x)-\sin (i x))}dx\) |
\(\Big \downarrow \) 3589 |
\(\displaystyle \int \left (\frac {1}{\cosh (x)-i \sinh (x)}-i \text {csch}(x)\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle i \text {arctanh}(\cosh (x))-\frac {i \text {arctanh}\left (\frac {\cosh (x)+i \sinh (x)}{\sqrt {2}}\right )}{\sqrt {2}}\) |
Input:
Int[Csch[x]/(I + Tanh[x]),x]
Output:
I*ArcTanh[Cosh[x]] - (I*ArcTanh[(Cosh[x] + I*Sinh[x])/Sqrt[2]])/Sqrt[2]
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[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_. ) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[Ex pandTrig[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]
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/C os[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]))
Time = 0.89 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88
method | result | size |
default | \(-i \ln \left (\tanh \left (\frac {x}{2}\right )\right )+\sqrt {2}\, \arctan \left (\frac {\left (2 \tanh \left (\frac {x}{2}\right )-2 i\right ) \sqrt {2}}{4}\right )\) | \(29\) |
risch | \(-i \ln \left ({\mathrm e}^{x}-1\right )+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{x}-\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right )}{2}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{x}+\frac {\sqrt {2}}{2}-\frac {i \sqrt {2}}{2}\right )}{2}+i \ln \left ({\mathrm e}^{x}+1\right )\) | \(60\) |
Input:
int(csch(x)/(I+tanh(x)),x,method=_RETURNVERBOSE)
Output:
-I*ln(tanh(1/2*x))+2^(1/2)*arctan(1/4*(2*tanh(1/2*x)-2*I)*2^(1/2))
Time = 0.10 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.30 \[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=-\frac {1}{2} i \, \sqrt {2} \log \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} + e^{x}\right ) + \frac {1}{2} i \, \sqrt {2} \log \left (\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} + e^{x}\right ) + i \, \log \left (e^{x} + 1\right ) - i \, \log \left (e^{x} - 1\right ) \] Input:
integrate(csch(x)/(I+tanh(x)),x, algorithm="fricas")
Output:
-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)
\[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=\int \frac {\operatorname {csch}{\left (x \right )}}{\tanh {\left (x \right )} + i}\, dx \] Input:
integrate(csch(x)/(I+tanh(x)),x)
Output:
Integral(csch(x)/(tanh(x) + I), x)
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=-\sqrt {2} \arctan \left (\left (\frac {1}{2} i + \frac {1}{2}\right ) \, \sqrt {2} e^{\left (-x\right )}\right ) + i \, \log \left (e^{\left (-x\right )} + 1\right ) - i \, \log \left (e^{\left (-x\right )} - 1\right ) \] Input:
integrate(csch(x)/(I+tanh(x)),x, algorithm="maxima")
Output:
-sqrt(2)*arctan((1/2*I + 1/2)*sqrt(2)*e^(-x)) + I*log(e^(-x) + 1) - I*log( e^(-x) - 1)
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=\sqrt {2} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} e^{x}\right ) + i \, \log \left (e^{x} + 1\right ) - i \, \log \left ({\left | e^{x} - 1 \right |}\right ) \] Input:
integrate(csch(x)/(I+tanh(x)),x, algorithm="giac")
Output:
sqrt(2)*arctan(-(1/2*I - 1/2)*sqrt(2)*e^x) + I*log(e^x + 1) - I*log(abs(e^ x - 1))
Time = 0.43 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.85 \[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=\ln \left (-8\,{\mathrm {e}}^x-8\right )\,1{}\mathrm {i}-\ln \left (8-8\,{\mathrm {e}}^x\right )\,1{}\mathrm {i}-\frac {\sqrt {2}\,\ln \left ({\mathrm {e}}^x\,\left (4-4{}\mathrm {i}\right )-\sqrt {2}\,4{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {\sqrt {2}\,\ln \left ({\mathrm {e}}^x\,\left (4-4{}\mathrm {i}\right )+\sqrt {2}\,4{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} \] Input:
int(1/(sinh(x)*(tanh(x) + 1i)),x)
Output:
log(- 8*exp(x) - 8)*1i - log(8 - 8*exp(x))*1i - (2^(1/2)*log(exp(x)*(4 - 4 i) - 2^(1/2)*4i)*1i)/2 + (2^(1/2)*log(exp(x)*(4 - 4i) + 2^(1/2)*4i)*1i)/2
Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 3.36 \[ \int \frac {\text {csch}(x)}{i+\tanh (x)} \, dx=\frac {-\sqrt {2}\, \mathrm {log}\left (e^{x} i +e^{x}-\sqrt {2}\right ) i -\sqrt {2}\, \mathrm {log}\left (e^{x} i +e^{x}-\sqrt {2}\right )+\sqrt {2}\, \mathrm {log}\left (e^{x} i +e^{x}+\sqrt {2}\right ) i +\sqrt {2}\, \mathrm {log}\left (e^{x} i +e^{x}+\sqrt {2}\right )+2 \,\mathrm {log}\left (e^{x}-1\right ) i +2 \,\mathrm {log}\left (e^{x}-1\right )-2 \,\mathrm {log}\left (e^{x}+1\right ) i -2 \,\mathrm {log}\left (e^{x}+1\right )}{2 i -2} \] Input:
int(csch(x)/(I+tanh(x)),x)
Output:
( - sqrt(2)*log(e**x*i + e**x - sqrt(2))*i - sqrt(2)*log(e**x*i + e**x - s qrt(2)) + sqrt(2)*log(e**x*i + e**x + sqrt(2))*i + sqrt(2)*log(e**x*i + e* *x + sqrt(2)) + 2*log(e**x - 1)*i + 2*log(e**x - 1) - 2*log(e**x + 1)*i - 2*log(e**x + 1))/(2*(i - 1))