Integrand size = 15, antiderivative size = 60 \[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=-\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {B \log (\sinh (x))}{a}-\frac {B \log (a+b \sinh (x))}{a} \] Output:
-2*A*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2)+B*ln(sinh( x))/a-B*ln(a+b*sinh(x))/a
Time = 1.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.08 \[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=\frac {2 A \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+\frac {B (\log (\sinh (x))-\log (a+b \sinh (x)))}{a} \] Input:
Integrate[(A + B*Coth[x])/(a + b*Sinh[x]),x]
Output:
(2*A*ArcTan[(b - a*Tanh[x/2])/Sqrt[-a^2 - b^2]])/Sqrt[-a^2 - b^2] + (B*(Lo g[Sinh[x]] - Log[a + b*Sinh[x]]))/a
Time = 0.38 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4901, 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.
\(\displaystyle \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+i B \cot (i x)}{a-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 4901 |
\(\displaystyle \int \left (\frac {A}{a+b \sinh (x)}+\frac {B \coth (x)}{a+b \sinh (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {B \log (a+b \sinh (x))}{a}+\frac {B \log (\sinh (x))}{a}\) |
Input:
Int[(A + B*Coth[x])/(a + b*Sinh[x]),x]
Output:
(-2*A*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/Sqrt[a^2 + b^2] + (B*Log [Sinh[x]])/a - (B*Log[a + b*Sinh[x]])/a
Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; !InertTrigFreeQ[u]
Time = 0.65 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22
method | result | size |
parts | \(\frac {2 A \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}-\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}{a}+\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a}\) | \(73\) |
default | \(\frac {B \ln \left (\tanh \left (\frac {x}{2}\right )\right )}{a}+\frac {-B \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )+\frac {2 A a \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{a}\) | \(76\) |
risch | \(-\frac {2 x B}{a}-\frac {2 x \,a^{3} B}{-a^{4}-a^{2} b^{2}}-\frac {2 x B a \,b^{2}}{-a^{4}-a^{2} b^{2}}+\frac {B \ln \left ({\mathrm e}^{2 x}-1\right )}{a}-\frac {a \ln \left ({\mathrm e}^{x}+\frac {a^{2} A -\sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a^{2} A -\sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B \,b^{2}}{\left (a^{2}+b^{2}\right ) a}+\frac {\ln \left ({\mathrm e}^{x}+\frac {a^{2} A -\sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) \sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{\left (a^{2}+b^{2}\right ) a}-\frac {a \ln \left ({\mathrm e}^{x}+\frac {a^{2} A +\sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a^{2} A +\sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) B \,b^{2}}{\left (a^{2}+b^{2}\right ) a}-\frac {\ln \left ({\mathrm e}^{x}+\frac {a^{2} A +\sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{A a b}\right ) \sqrt {A^{2} a^{4}+A^{2} a^{2} b^{2}}}{\left (a^{2}+b^{2}\right ) a}\) | \(443\) |
Input:
int((A+B*coth(x))/(a+b*sinh(x)),x,method=_RETURNVERBOSE)
Output:
2*A/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))-B/a *ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)+B/a*ln(tanh(1/2*x))
Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (56) = 112\).
Time = 0.11 (sec) , antiderivative size = 183, normalized size of antiderivative = 3.05 \[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=\frac {\sqrt {a^{2} + b^{2}} A a \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {2 \, \sinh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{3} + a b^{2}} \] Input:
integrate((A+B*coth(x))/(a+b*sinh(x)),x, algorithm="fricas")
Output:
(sqrt(a^2 + b^2)*A*a*log((b^2*cosh(x)^2 + b^2*sinh(x)^2 + 2*a*b*cosh(x) + 2*a^2 + b^2 + 2*(b^2*cosh(x) + a*b)*sinh(x) - 2*sqrt(a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a))/(b*cosh(x)^2 + b*sinh(x)^2 + 2*a*cosh(x) + 2*(b*cosh(x) + a)*sinh(x) - b)) - (B*a^2 + B*b^2)*log(2*(b*sinh(x) + a)/(cosh(x) - sin h(x))) + (B*a^2 + B*b^2)*log(2*sinh(x)/(cosh(x) - sinh(x))))/(a^3 + a*b^2)
\[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=\int \frac {A + B \coth {\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:
integrate((A+B*coth(x))/(a+b*sinh(x)),x)
Output:
Integral((A + B*coth(x))/(a + b*sinh(x)), x)
Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.77 \[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=-B {\left (\frac {\log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a} - \frac {\log \left (e^{\left (-x\right )} + 1\right )}{a} - \frac {\log \left (e^{\left (-x\right )} - 1\right )}{a}\right )} + \frac {A \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} \] Input:
integrate((A+B*coth(x))/(a+b*sinh(x)),x, algorithm="maxima")
Output:
-B*(log(-2*a*e^(-x) + b*e^(-2*x) - b)/a - log(e^(-x) + 1)/a - log(e^(-x) - 1)/a) + A*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2)))/sqrt(a^2 + b^2)
Time = 0.14 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.70 \[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=\frac {A \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} + \frac {B \log \left (e^{x} + 1\right )}{a} - \frac {B \log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{a} + \frac {B \log \left ({\left | e^{x} - 1 \right |}\right )}{a} \] Input:
integrate((A+B*coth(x))/(a+b*sinh(x)),x, algorithm="giac")
Output:
A*log(abs(2*b*e^x + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^ 2 + b^2)))/sqrt(a^2 + b^2) + B*log(e^x + 1)/a - B*log(abs(b*e^(2*x) + 2*a* e^x - b))/a + B*log(abs(e^x - 1))/a
Time = 11.95 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.73 \[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=\frac {B\,\ln \left (16\,B^2\,a^2+16\,B^2\,b^2-16\,B^2\,a^2\,{\mathrm {e}}^{2\,x}-16\,B^2\,b^2\,{\mathrm {e}}^{2\,x}\right )}{a}-\frac {2\,\mathrm {atan}\left (\frac {A^2\,b^2\,{\mathrm {e}}^x\,\sqrt {-a^2-b^2}+A^2\,a\,b\,\sqrt {-a^2-b^2}}{A\,b\,\sqrt {A^2}\,\left (a^2+b^2\right )}\right )\,\sqrt {A^2}}{\sqrt {-a^2-b^2}}-\frac {B\,\ln \left (32\,B^2\,a\,{\mathrm {e}}^x-16\,B^2\,b+16\,B^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a} \] Input:
int((A + B*coth(x))/(a + b*sinh(x)),x)
Output:
(B*log(16*B^2*a^2 + 16*B^2*b^2 - 16*B^2*a^2*exp(2*x) - 16*B^2*b^2*exp(2*x) ))/a - (2*atan((A^2*b^2*exp(x)*(- a^2 - b^2)^(1/2) + A^2*a*b*(- a^2 - b^2) ^(1/2))/(A*b*(A^2)^(1/2)*(a^2 + b^2)))*(A^2)^(1/2))/(- a^2 - b^2)^(1/2) - (B*log(32*B^2*a*exp(x) - 16*B^2*b + 16*B^2*b*exp(2*x)))/a
Time = 0.16 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.32 \[ \int \frac {A+B \coth (x)}{a+b \sinh (x)} \, dx=\frac {2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a^{2} i +\mathrm {log}\left (e^{x}-1\right ) a^{2} b +\mathrm {log}\left (e^{x}-1\right ) b^{3}+\mathrm {log}\left (e^{x}+1\right ) a^{2} b +\mathrm {log}\left (e^{x}+1\right ) b^{3}-\mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) a^{2} b -\mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) b^{3}}{a \left (a^{2}+b^{2}\right )} \] Input:
int((A+B*coth(x))/(a+b*sinh(x)),x)
Output:
(2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a**2*i + log (e**x - 1)*a**2*b + log(e**x - 1)*b**3 + log(e**x + 1)*a**2*b + log(e**x + 1)*b**3 - log(e**(2*x)*b + 2*e**x*a - b)*a**2*b - log(e**(2*x)*b + 2*e**x *a - b)*b**3)/(a*(a**2 + b**2))