Integrand size = 11, antiderivative size = 53 \[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\frac {\log (1-\cosh (x))}{2 (a+b)}-\frac {\log (1+\cosh (x))}{2 (a-b)}+\frac {b \log (a+b \cosh (x))}{a^2-b^2} \] Output:
ln(1-cosh(x))/(2*a+2*b)-ln(1+cosh(x))/(2*a-2*b)+b*ln(a+b*cosh(x))/(a^2-b^2 )
Time = 0.07 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92 \[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\frac {\log \left (\cosh \left (\frac {x}{2}\right )\right )}{-a+b}+\frac {b \log (a+b \cosh (x))}{a^2-b^2}+\frac {\log \left (\sinh \left (\frac {x}{2}\right )\right )}{a+b} \] Input:
Integrate[Csch[x]/(a + b*Cosh[x]),x]
Output:
Log[Cosh[x/2]]/(-a + b) + (b*Log[a + b*Cosh[x]])/(a^2 - b^2) + Log[Sinh[x/ 2]]/(a + b)
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 26, 3147, 477, 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.
\(\displaystyle \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {i}{\cos \left (-\frac {\pi }{2}+i x\right ) \left (a-b \sin \left (-\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle i \int \frac {1}{\cos \left (i x-\frac {\pi }{2}\right ) \left (a-b \sin \left (i x-\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle -b \int \frac {1}{(a+b \cosh (x)) \left (b^2-b^2 \cosh ^2(x)\right )}d(b \cosh (x))\) |
\(\Big \downarrow \) 477 |
\(\displaystyle -\frac {\int \left (-\frac {b^2}{\left (a^2-b^2\right ) (a+b \cosh (x))}+\frac {b}{2 (a+b) (b-b \cosh (x))}+\frac {b}{2 (a-b) (\cosh (x) b+b)}\right )d(b \cosh (x))}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {b^2 \log (a+b \cosh (x))}{a^2-b^2}-\frac {b \log (b-b \cosh (x))}{2 (a+b)}+\frac {b \log (b \cosh (x)+b)}{2 (a-b)}}{b}\) |
Input:
Int[Csch[x]/(a + b*Cosh[x]),x]
Output:
-((-1/2*(b*Log[b - b*Cosh[x]])/(a + b) - (b^2*Log[a + b*Cosh[x]])/(a^2 - b ^2) + (b*Log[b + b*Cosh[x]])/(2*(a - b)))/b)
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[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ a^p Int[ExpandIntegrand[(c + d*x)^n*(1 - Rt[-b/a, 2]*x)^p*(1 + Rt[-b/a, 2 ]*x)^p, x], x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[p, 0] && IntegerQ[n] & & NiceSqrtQ[-b/a] && !FractionalPowerFactorQ[Rt[-b/a, 2]]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.68 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b}+\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -b \tanh \left (\frac {x}{2}\right )^{2}-a -b \right )}{\left (a +b \right ) \left (a -b \right )}\) | \(52\) |
risch | \(-\frac {x}{a +b}+\frac {x}{a -b}-\frac {2 x b}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{a +b}-\frac {\ln \left ({\mathrm e}^{x}+1\right )}{a -b}+\frac {b \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right )}{a^{2}-b^{2}}\) | \(87\) |
Input:
int(csch(x)/(a+b*cosh(x)),x,method=_RETURNVERBOSE)
Output:
1/(a+b)*ln(tanh(1/2*x))+b/(a+b)/(a-b)*ln(tanh(1/2*x)^2*a-b*tanh(1/2*x)^2-a -b)
Time = 0.10 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09 \[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\frac {b \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a + b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) + {\left (a - b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} - b^{2}} \] Input:
integrate(csch(x)/(a+b*cosh(x)),x, algorithm="fricas")
Output:
(b*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - (a + b)*log(cosh(x) + sinh (x) + 1) + (a - b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2)
\[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\int \frac {\operatorname {csch}{\left (x \right )}}{a + b \cosh {\left (x \right )}}\, dx \] Input:
integrate(csch(x)/(a+b*cosh(x)),x)
Output:
Integral(csch(x)/(a + b*cosh(x)), x)
Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11 \[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\frac {b \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{2} - b^{2}} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \] Input:
integrate(csch(x)/(a+b*cosh(x)),x, algorithm="maxima")
Output:
b*log(2*a*e^(-x) + b*e^(-2*x) + b)/(a^2 - b^2) - log(e^(-x) + 1)/(a - b) + log(e^(-x) - 1)/(a + b)
Time = 0.11 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26 \[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\frac {b^{2} \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{2 \, {\left (a - b\right )}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{2 \, {\left (a + b\right )}} \] Input:
integrate(csch(x)/(a+b*cosh(x)),x, algorithm="giac")
Output:
b^2*log(abs(b*(e^(-x) + e^x) + 2*a))/(a^2*b - b^3) - 1/2*log(e^(-x) + e^x + 2)/(a - b) + 1/2*log(e^(-x) + e^x - 2)/(a + b)
Time = 2.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 3.02 \[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\frac {\ln \left (128\,a\,b^2-128\,a^2\,b+32\,a^3-32\,a^3\,{\mathrm {e}}^x-128\,a\,b^2\,{\mathrm {e}}^x+128\,a^2\,b\,{\mathrm {e}}^x\right )}{a+b}-\frac {\ln \left (128\,a\,b^2+128\,a^2\,b+32\,a^3+32\,a^3\,{\mathrm {e}}^x+128\,a\,b^2\,{\mathrm {e}}^x+128\,a^2\,b\,{\mathrm {e}}^x\right )}{a-b}+\frac {b\,\ln \left (16\,b^3\,{\mathrm {e}}^{2\,x}-4\,a^2\,b+16\,b^3-8\,a^3\,{\mathrm {e}}^x+32\,a\,b^2\,{\mathrm {e}}^x-4\,a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2} \] Input:
int(1/(sinh(x)*(a + b*cosh(x))),x)
Output:
log(128*a*b^2 - 128*a^2*b + 32*a^3 - 32*a^3*exp(x) - 128*a*b^2*exp(x) + 12 8*a^2*b*exp(x))/(a + b) - log(128*a*b^2 + 128*a^2*b + 32*a^3 + 32*a^3*exp( x) + 128*a*b^2*exp(x) + 128*a^2*b*exp(x))/(a - b) + (b*log(16*b^3*exp(2*x) - 4*a^2*b + 16*b^3 - 8*a^3*exp(x) + 32*a*b^2*exp(x) - 4*a^2*b*exp(2*x)))/ (a^2 - b^2)
Time = 0.25 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.25 \[ \int \frac {\text {csch}(x)}{a+b \cosh (x)} \, dx=\frac {\mathrm {log}\left (e^{x}-1\right ) a -\mathrm {log}\left (e^{x}-1\right ) b -\mathrm {log}\left (e^{x}+1\right ) a -\mathrm {log}\left (e^{x}+1\right ) b +\mathrm {log}\left (e^{2 x} b +2 e^{x} a +b \right ) b}{a^{2}-b^{2}} \] Input:
int(csch(x)/(a+b*cosh(x)),x)
Output:
(log(e**x - 1)*a - log(e**x - 1)*b - log(e**x + 1)*a - log(e**x + 1)*b + l og(e**(2*x)*b + 2*e**x*a + b)*b)/(a**2 - b**2)