Integrand size = 11, antiderivative size = 40 \[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=\frac {1}{2 a (1+\cosh (x))}+\frac {\log (1-\cosh (x))}{4 a}+\frac {3 \log (1+\cosh (x))}{4 a} \] Output:
1/2/a/(1+cosh(x))+1/4*ln(1-cosh(x))/a+3/4*ln(1+cosh(x))/a
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.10 \[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=\frac {\left (1+2 \cosh ^2\left (\frac {x}{2}\right ) \left (3 \log \left (\cosh \left (\frac {x}{2}\right )\right )+\log \left (\sinh \left (\frac {x}{2}\right )\right )\right )\right ) \text {sech}(x)}{2 a (1+\text {sech}(x))} \] Input:
Integrate[Coth[x]/(a + a*Sech[x]),x]
Output:
((1 + 2*Cosh[x/2]^2*(3*Log[Cosh[x/2]] + Log[Sinh[x/2]]))*Sech[x])/(2*a*(1 + Sech[x]))
Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 26, 4367, 27, 99, 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.
\(\displaystyle \int \frac {\coth (x)}{a \text {sech}(x)+a} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i}{\cot \left (\frac {\pi }{2}+i x\right ) \left (a+a \csc \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {1}{\cot \left (i x+\frac {\pi }{2}\right ) \left (\csc \left (i x+\frac {\pi }{2}\right ) a+a\right )}dx\) |
\(\Big \downarrow \) 4367 |
\(\displaystyle -a^2 \int \frac {\cosh ^2(x)}{a^3 (1-\cosh (x)) (\cosh (x)+1)^2}d\cosh (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {\cosh ^2(x)}{(1-\cosh (x)) (\cosh (x)+1)^2}d\cosh (x)}{a}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (-\frac {3}{4 (\cosh (x)+1)}+\frac {1}{2 (\cosh (x)+1)^2}-\frac {1}{4 (\cosh (x)-1)}\right )d\cosh (x)}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{2 (\cosh (x)+1)}-\frac {1}{4} \log (1-\cosh (x))-\frac {3}{4} \log (\cosh (x)+1)}{a}\) |
Input:
Int[Coth[x]/(a + a*Sech[x]),x]
Output:
-((-1/2*1/(1 + Cosh[x]) - Log[1 - Cosh[x]]/4 - (3*Log[1 + Cosh[x]])/4)/a)
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_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _.), x_Symbol] :> Simp[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] /; Fr eeQ[{a, b, c, d}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] && Integer Q[n]
Time = 0.15 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95
method | result | size |
default | \(\frac {-\frac {\tanh \left (\frac {x}{2}\right )^{2}}{2}-2 \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )+\ln \left (\tanh \left (\frac {x}{2}\right )\right )-2 \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 a}\) | \(38\) |
risch | \(-\frac {x}{a}+\frac {{\mathrm e}^{x}}{\left (1+{\mathrm e}^{x}\right )^{2} a}+\frac {3 \ln \left (1+{\mathrm e}^{x}\right )}{2 a}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{2 a}\) | \(40\) |
Input:
int(coth(x)/(a+a*sech(x)),x,method=_RETURNVERBOSE)
Output:
1/2/a*(-1/2*tanh(1/2*x)^2-2*ln(tanh(1/2*x)-1)+ln(tanh(1/2*x))-2*ln(tanh(1/ 2*x)+1))
Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (34) = 68\).
Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 3.40 \[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=-\frac {2 \, x \cosh \left (x\right )^{2} + 2 \, x \sinh \left (x\right )^{2} + 2 \, {\left (2 \, x - 1\right )} \cosh \left (x\right ) - 3 \, {\left (\cosh \left (x\right )^{2} + 2 \, {\left (\cosh \left (x\right ) + 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (\cosh \left (x\right )^{2} + 2 \, {\left (\cosh \left (x\right ) + 1\right )} \sinh \left (x\right ) + \sinh \left (x\right )^{2} + 2 \, \cosh \left (x\right ) + 1\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right ) + 2 \, {\left (2 \, x \cosh \left (x\right ) + 2 \, x - 1\right )} \sinh \left (x\right ) + 2 \, x}{2 \, {\left (a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (a \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + a\right )}} \] Input:
integrate(coth(x)/(a+a*sech(x)),x, algorithm="fricas")
Output:
-1/2*(2*x*cosh(x)^2 + 2*x*sinh(x)^2 + 2*(2*x - 1)*cosh(x) - 3*(cosh(x)^2 + 2*(cosh(x) + 1)*sinh(x) + sinh(x)^2 + 2*cosh(x) + 1)*log(cosh(x) + sinh(x ) + 1) - (cosh(x)^2 + 2*(cosh(x) + 1)*sinh(x) + sinh(x)^2 + 2*cosh(x) + 1) *log(cosh(x) + sinh(x) - 1) + 2*(2*x*cosh(x) + 2*x - 1)*sinh(x) + 2*x)/(a* cosh(x)^2 + a*sinh(x)^2 + 2*a*cosh(x) + 2*(a*cosh(x) + a)*sinh(x) + a)
\[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=\frac {\int \frac {\coth {\left (x \right )}}{\operatorname {sech}{\left (x \right )} + 1}\, dx}{a} \] Input:
integrate(coth(x)/(a+a*sech(x)),x)
Output:
Integral(coth(x)/(sech(x) + 1), x)/a
Time = 0.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.30 \[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=\frac {x}{a} + \frac {e^{\left (-x\right )}}{2 \, a e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a} + \frac {3 \, \log \left (e^{\left (-x\right )} + 1\right )}{2 \, a} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{2 \, a} \] Input:
integrate(coth(x)/(a+a*sech(x)),x, algorithm="maxima")
Output:
x/a + e^(-x)/(2*a*e^(-x) + a*e^(-2*x) + a) + 3/2*log(e^(-x) + 1)/a + 1/2*l og(e^(-x) - 1)/a
Time = 0.13 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.40 \[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=\frac {3 \, \log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{4 \, a} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{4 \, a} - \frac {3 \, e^{\left (-x\right )} + 3 \, e^{x} + 2}{4 \, a {\left (e^{\left (-x\right )} + e^{x} + 2\right )}} \] Input:
integrate(coth(x)/(a+a*sech(x)),x, algorithm="giac")
Output:
3/4*log(e^(-x) + e^x + 2)/a + 1/4*log(e^(-x) + e^x - 2)/a - 1/4*(3*e^(-x) + 3*e^x + 2)/(a*(e^(-x) + e^x + 2))
Time = 2.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.62 \[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=\frac {\ln \left ({\mathrm {e}}^{2\,x}-1\right )}{a}-\frac {x}{a}-\frac {1}{a+2\,a\,{\mathrm {e}}^x+a\,{\mathrm {e}}^{2\,x}}+\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^x\,\sqrt {-a^2}}{a}\right )}{\sqrt {-a^2}}+\frac {1}{a+a\,{\mathrm {e}}^x} \] Input:
int(coth(x)/(a + a/cosh(x)),x)
Output:
log(exp(2*x) - 1)/a - x/a - 1/(a + 2*a*exp(x) + a*exp(2*x)) + atan((exp(x) *(-a^2)^(1/2))/a)/(-a^2)^(1/2) + 1/(a + a*exp(x))
Time = 0.20 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.65 \[ \int \frac {\coth (x)}{a+a \text {sech}(x)} \, dx=\frac {e^{2 x} \mathrm {log}\left (e^{x}-1\right )+3 e^{2 x} \mathrm {log}\left (e^{x}+1\right )-2 e^{2 x} x -e^{2 x}+2 e^{x} \mathrm {log}\left (e^{x}-1\right )+6 e^{x} \mathrm {log}\left (e^{x}+1\right )-4 e^{x} x +\mathrm {log}\left (e^{x}-1\right )+3 \,\mathrm {log}\left (e^{x}+1\right )-2 x -1}{2 a \left (e^{2 x}+2 e^{x}+1\right )} \] Input:
int(coth(x)/(a+a*sech(x)),x)
Output:
(e**(2*x)*log(e**x - 1) + 3*e**(2*x)*log(e**x + 1) - 2*e**(2*x)*x - e**(2* x) + 2*e**x*log(e**x - 1) + 6*e**x*log(e**x + 1) - 4*e**x*x + log(e**x - 1 ) + 3*log(e**x + 1) - 2*x - 1)/(2*a*(e**(2*x) + 2*e**x + 1))