Integrand size = 11, antiderivative size = 54 \[ \int \frac {\coth (x)}{a+b \cosh (x)} \, dx=\frac {\log (1-\cosh (x))}{2 (a+b)}+\frac {\log (1+\cosh (x))}{2 (a-b)}-\frac {a \log (a+b \cosh (x))}{a^2-b^2} \]
Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {\coth (x)}{a+b \cosh (x)} \, dx=\frac {\log \left (\cosh \left (\frac {x}{2}\right )\right )}{a-b}-\frac {a \log (a+b \cosh (x))}{a^2-b^2}+\frac {\log \left (\sinh \left (\frac {x}{2}\right )\right )}{a+b} \]
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {3042, 26, 3200, 587, 16, 452, 219, 240}
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.
\(\displaystyle \int \frac {\coth (x)}{a+b \cosh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i \tan \left (-\frac {\pi }{2}+i x\right )}{a-b \sin \left (-\frac {\pi }{2}+i x\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {\tan \left (i x-\frac {\pi }{2}\right )}{a-b \sin \left (i x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle -\int \frac {b \cosh (x)}{(a+b \cosh (x)) \left (b^2-b^2 \cosh ^2(x)\right )}d(b \cosh (x))\) |
\(\Big \downarrow \) 587 |
\(\displaystyle \frac {\int \frac {b^2-a b \cosh (x)}{b^2-b^2 \cosh ^2(x)}d(b \cosh (x))}{a^2-b^2}-\frac {a \int \frac {1}{a+b \cosh (x)}d(b \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {\int \frac {b^2-a b \cosh (x)}{b^2-b^2 \cosh ^2(x)}d(b \cosh (x))}{a^2-b^2}-\frac {a \log (a+b \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {b^2 \int \frac {1}{b^2-b^2 \cosh ^2(x)}d(b \cosh (x))-a \int \frac {b \cosh (x)}{b^2-b^2 \cosh ^2(x)}d(b \cosh (x))}{a^2-b^2}-\frac {a \log (a+b \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {b \text {arctanh}(\cosh (x))-a \int \frac {b \cosh (x)}{b^2-b^2 \cosh ^2(x)}d(b \cosh (x))}{a^2-b^2}-\frac {a \log (a+b \cosh (x))}{a^2-b^2}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {\frac {1}{2} a \log \left (b^2-b^2 \cosh ^2(x)\right )+b \text {arctanh}(\cosh (x))}{a^2-b^2}-\frac {a \log (a+b \cosh (x))}{a^2-b^2}\) |
-((a*Log[a + b*Cosh[x]])/(a^2 - b^2)) + (b*ArcTanh[Cosh[x]] + (a*Log[b^2 - b^2*Cosh[x]^2])/2)/(a^2 - b^2)
3.2.83.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, 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_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[(x_.)/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)), x_Symbol] :> Simp[(- c)*(d/(b*c^2 + a*d^2)) Int[1/(c + d*x), x], x] + Simp[1/(b*c^2 + a*d^2) Int[(a*d + b*c*x)/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c ^2 + a*d^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(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] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98
method | result | size |
default | \(\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -\tanh \left (\frac {x}{2}\right )^{2} b -a -b \right )}{\left (a +b \right ) \left (a -b \right )}\) | \(53\) |
risch | \(-\frac {x}{a +b}-\frac {x}{a -b}+\frac {2 x a}{a^{2}-b^{2}}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{a +b}+\frac {\ln \left ({\mathrm e}^{x}+1\right )}{a -b}-\frac {a \ln \left ({\mathrm e}^{2 x}+\frac {2 a \,{\mathrm e}^{x}}{b}+1\right )}{a^{2}-b^{2}}\) | \(88\) |
Time = 0.27 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int \frac {\coth (x)}{a+b \cosh (x)} \, dx=-\frac {a \log \left (\frac {2 \, {\left (b \cosh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - {\left (a + b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (a - b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{2} - b^{2}} \]
-(a*log(2*(b*cosh(x) + a)/(cosh(x) - sinh(x))) - (a + b)*log(cosh(x) + sin h(x) + 1) - (a - b)*log(cosh(x) + sinh(x) - 1))/(a^2 - b^2)
\[ \int \frac {\coth (x)}{a+b \cosh (x)} \, dx=\int \frac {\coth {\left (x \right )}}{a + b \cosh {\left (x \right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.09 \[ \int \frac {\coth (x)}{a+b \cosh (x)} \, dx=-\frac {a \log \left (2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} + b\right )}{a^{2} - b^{2}} + \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \]
-a*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 = 0.25 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.24 \[ \int \frac {\coth (x)}{a+b \cosh (x)} \, dx=-\frac {a b \log \left ({\left | b {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b - b^{3}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{2 \, {\left (a - b\right )}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{2 \, {\left (a + b\right )}} \]
-a*b*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 = 0.45 (sec) , antiderivative size = 148, normalized size of antiderivative = 2.74 \[ \int \frac {\coth (x)}{a+b \cosh (x)} \, dx=\frac {\ln \left (128\,a\,b-128\,a^2-32\,b^2+128\,a^2\,{\mathrm {e}}^x+32\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a+b}+\frac {\ln \left (-128\,a\,b-128\,a^2-32\,b^2-128\,a^2\,{\mathrm {e}}^x-32\,b^2\,{\mathrm {e}}^x-128\,a\,b\,{\mathrm {e}}^x\right )}{a-b}-\frac {a\,\ln \left (16\,a^2\,b-4\,b^3\,{\mathrm {e}}^{2\,x}-4\,b^3+32\,a^3\,{\mathrm {e}}^x-8\,a\,b^2\,{\mathrm {e}}^x+16\,a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2-b^2} \]
log(128*a*b - 128*a^2 - 32*b^2 + 128*a^2*exp(x) + 32*b^2*exp(x) - 128*a*b* exp(x))/(a + b) + log(- 128*a*b - 128*a^2 - 32*b^2 - 128*a^2*exp(x) - 32*b ^2*exp(x) - 128*a*b*exp(x))/(a - b) - (a*log(16*a^2*b - 4*b^3*exp(2*x) - 4 *b^3 + 32*a^3*exp(x) - 8*a*b^2*exp(x) + 16*a^2*b*exp(2*x)))/(a^2 - b^2)