Integrand size = 11, antiderivative size = 66 \[ \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx=\frac {\log (\cosh (x))}{a}+\frac {\log (1-\text {sech}(x))}{2 (a+b)}+\frac {\log (1+\text {sech}(x))}{2 (a-b)}-\frac {b^2 \log (a+b \text {sech}(x))}{a \left (a^2-b^2\right )} \]
ln(cosh(x))/a+1/2*ln(1-sech(x))/(a+b)+1/2*ln(1+sech(x))/(a-b)-b^2*ln(a+b*s ech(x))/a/(a^2-b^2)
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.95 \[ \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx=\frac {1}{2} \left (\frac {2 \log (\cosh (x))}{a}+\frac {\log (1-\text {sech}(x))}{a+b}+\frac {\log (1+\text {sech}(x))}{a-b}-\frac {2 b^2 \log (a+b \text {sech}(x))}{a^3-a b^2}\right ) \]
((2*Log[Cosh[x]])/a + Log[1 - Sech[x]]/(a + b) + Log[1 + Sech[x]]/(a - b) - (2*b^2*Log[a + b*Sech[x]])/(a^3 - a*b^2))/2
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 26, 4373, 615, 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.
\(\displaystyle \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {i}{\cot \left (\frac {\pi }{2}+i x\right ) \left (a+b \csc \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 26 |
\(\displaystyle -i \int \frac {1}{\cot \left (i x+\frac {\pi }{2}\right ) \left (a+b \csc \left (i x+\frac {\pi }{2}\right )\right )}dx\) |
\(\Big \downarrow \) 4373 |
\(\displaystyle -b^2 \int \frac {\cosh (x)}{b (a+b \text {sech}(x)) \left (b^2-b^2 \text {sech}^2(x)\right )}d(b \text {sech}(x))\) |
\(\Big \downarrow \) 615 |
\(\displaystyle -b^2 \int \left (\frac {\cosh (x)}{a b^3}+\frac {1}{2 b^2 (a+b) (b-b \text {sech}(x))}+\frac {1}{a (a-b) (a+b) (a+b \text {sech}(x))}-\frac {1}{2 (a-b) b^2 (\text {sech}(x) b+b)}\right )d(b \text {sech}(x))\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -b^2 \left (\frac {\log (a+b \text {sech}(x))}{a \left (a^2-b^2\right )}+\frac {\log (b \text {sech}(x))}{a b^2}-\frac {\log (b-b \text {sech}(x))}{2 b^2 (a+b)}-\frac {\log (b \text {sech}(x)+b)}{2 b^2 (a-b)}\right )\) |
-(b^2*(Log[b*Sech[x]]/(a*b^2) - Log[b - b*Sech[x]]/(2*b^2*(a + b)) + Log[a + b*Sech[x]]/(a*(a^2 - b^2)) - Log[b + b*Sech[x]]/(2*(a - b)*b^2)))
3.2.20.3.1 Defintions of rubi rules used
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[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && ILtQ[p, 0]
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n _), x_Symbol] :> Simp[-(-1)^((m - 1)/2)/(d*b^(m - 1)) Subst[Int[(b^2 - x^ 2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Time = 0.33 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.18
method | result | size |
default | \(-\frac {b^{2} \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -\tanh \left (\frac {x}{2}\right )^{2} b +a +b \right )}{a \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a +b}\) | \(78\) |
risch | \(\frac {x}{a}-\frac {x}{a +b}-\frac {x}{a -b}+\frac {2 b^{2} x}{a \left (a^{2}-b^{2}\right )}+\frac {\ln \left ({\mathrm e}^{x}-1\right )}{a +b}+\frac {\ln \left ({\mathrm e}^{x}+1\right )}{a -b}-\frac {b^{2} \ln \left ({\mathrm e}^{2 x}+\frac {2 b \,{\mathrm e}^{x}}{a}+1\right )}{a \left (a^{2}-b^{2}\right )}\) | \(103\) |
-b^2/a/(a+b)/(a-b)*ln(tanh(1/2*x)^2*a-tanh(1/2*x)^2*b+a+b)-1/a*ln(tanh(1/2 *x)-1)-1/a*ln(tanh(1/2*x)+1)+1/(a+b)*ln(tanh(1/2*x))
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23 \[ \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx=-\frac {b^{2} \log \left (\frac {2 \, {\left (a \cosh \left (x\right ) + b\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + {\left (a^{2} - b^{2}\right )} x - {\left (a^{2} + a b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) + 1\right ) - {\left (a^{2} - a b\right )} \log \left (\cosh \left (x\right ) + \sinh \left (x\right ) - 1\right )}{a^{3} - a b^{2}} \]
-(b^2*log(2*(a*cosh(x) + b)/(cosh(x) - sinh(x))) + (a^2 - b^2)*x - (a^2 + a*b)*log(cosh(x) + sinh(x) + 1) - (a^2 - a*b)*log(cosh(x) + sinh(x) - 1))/ (a^3 - a*b^2)
\[ \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx=\int \frac {\coth {\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx=-\frac {b^{2} \log \left (2 \, b e^{\left (-x\right )} + a e^{\left (-2 \, x\right )} + a\right )}{a^{3} - a b^{2}} + \frac {x}{a} + \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a - b} + \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a + b} \]
-b^2*log(2*b*e^(-x) + a*e^(-2*x) + a)/(a^3 - a*b^2) + x/a + log(e^(-x) + 1 )/(a - b) + log(e^(-x) - 1)/(a + b)
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx=-\frac {b^{2} \log \left ({\left | a {\left (e^{\left (-x\right )} + e^{x}\right )} + 2 \, b \right |}\right )}{a^{3} - a b^{2}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} + 2\right )}{2 \, {\left (a - b\right )}} + \frac {\log \left (e^{\left (-x\right )} + e^{x} - 2\right )}{2 \, {\left (a + b\right )}} \]
-b^2*log(abs(a*(e^(-x) + e^x) + 2*b))/(a^3 - a*b^2) + 1/2*log(e^(-x) + e^x + 2)/(a - b) + 1/2*log(e^(-x) + e^x - 2)/(a + b)
Time = 0.41 (sec) , antiderivative size = 271, normalized size of antiderivative = 4.11 \[ \int \frac {\coth (x)}{a+b \text {sech}(x)} \, dx=\frac {\ln \left (64\,a\,b^4+32\,a^4\,b+32\,b^5+96\,a^2\,b^3+64\,a^3\,b^2+32\,b^5\,{\mathrm {e}}^x+64\,a\,b^4\,{\mathrm {e}}^x+32\,a^4\,b\,{\mathrm {e}}^x+96\,a^2\,b^3\,{\mathrm {e}}^x+64\,a^3\,b^2\,{\mathrm {e}}^x\right )}{a-b}-\frac {x}{a}+\frac {\ln \left (64\,a\,b^4-32\,a^4\,b-32\,b^5-96\,a^2\,b^3+64\,a^3\,b^2+32\,b^5\,{\mathrm {e}}^x-64\,a\,b^4\,{\mathrm {e}}^x+32\,a^4\,b\,{\mathrm {e}}^x+96\,a^2\,b^3\,{\mathrm {e}}^x-64\,a^3\,b^2\,{\mathrm {e}}^x\right )}{a+b}+\frac {b^2\,\ln \left (4\,a^5\,{\mathrm {e}}^{2\,x}+4\,a\,b^4+4\,a^5+4\,a^3\,b^2+8\,b^5\,{\mathrm {e}}^x+4\,a^3\,b^2\,{\mathrm {e}}^{2\,x}+8\,a^4\,b\,{\mathrm {e}}^x+4\,a\,b^4\,{\mathrm {e}}^{2\,x}+8\,a^2\,b^3\,{\mathrm {e}}^x\right )}{a\,b^2-a^3} \]
log(64*a*b^4 + 32*a^4*b + 32*b^5 + 96*a^2*b^3 + 64*a^3*b^2 + 32*b^5*exp(x) + 64*a*b^4*exp(x) + 32*a^4*b*exp(x) + 96*a^2*b^3*exp(x) + 64*a^3*b^2*exp( x))/(a - b) - x/a + log(64*a*b^4 - 32*a^4*b - 32*b^5 - 96*a^2*b^3 + 64*a^3 *b^2 + 32*b^5*exp(x) - 64*a*b^4*exp(x) + 32*a^4*b*exp(x) + 96*a^2*b^3*exp( x) - 64*a^3*b^2*exp(x))/(a + b) + (b^2*log(4*a^5*exp(2*x) + 4*a*b^4 + 4*a^ 5 + 4*a^3*b^2 + 8*b^5*exp(x) + 4*a^3*b^2*exp(2*x) + 8*a^4*b*exp(x) + 4*a*b ^4*exp(2*x) + 8*a^2*b^3*exp(x)))/(a*b^2 - a^3)