Integrand size = 15, antiderivative size = 89 \[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {a B \arctan (\sinh (x))}{a^2+b^2}-\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}-\frac {b B \log (\cosh (x))}{a^2+b^2}+\frac {b B \log (a+b \sinh (x))}{a^2+b^2} \] Output:
a*B*arctan(sinh(x))/(a^2+b^2)-2*A*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2 ))/(a^2+b^2)^(1/2)-b*B*ln(cosh(x))/(a^2+b^2)+b*B*ln(a+b*sinh(x))/(a^2+b^2)
Time = 0.25 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.04 \[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {2 a B \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^2+b^2}+\frac {2 A \arctan \left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}-\frac {b B (\log (\cosh (x))-\log (a+b \sinh (x)))}{a^2+b^2} \] Input:
Integrate[(A + B*Sech[x])/(a + b*Sinh[x]),x]
Output:
(2*a*B*ArcTan[Tanh[x/2]])/(a^2 + b^2) + (2*A*ArcTan[(b - a*Tanh[x/2])/Sqrt [-a^2 - b^2]])/Sqrt[-a^2 - b^2] - (b*B*(Log[Cosh[x]] - Log[a + b*Sinh[x]]) )/(a^2 + b^2)
Time = 0.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4714, 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.
\(\displaystyle \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+B \sec (i x)}{a-i b \sin (i x)}dx\) |
\(\Big \downarrow \) 4714 |
\(\displaystyle \int \frac {\text {sech}(x) (A \cosh (x)+B)}{a+b \sinh (x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {B+A \cos (i x)}{\cos (i x) (a-i b \sin (i x))}dx\) |
\(\Big \downarrow \) 4901 |
\(\displaystyle \int \left (\frac {A}{a+b \sinh (x)}+\frac {B \text {sech}(x)}{a+b \sinh (x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 A \text {arctanh}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\sqrt {a^2+b^2}}+\frac {a B \arctan (\sinh (x))}{a^2+b^2}+\frac {b B \log (a+b \sinh (x))}{a^2+b^2}-\frac {b B \log (\cosh (x))}{a^2+b^2}\) |
Input:
Int[(A + B*Sech[x])/(a + b*Sinh[x]),x]
Output:
(a*B*ArcTan[Sinh[x]])/(a^2 + b^2) - (2*A*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^ 2 + b^2]])/Sqrt[a^2 + b^2] - (b*B*Log[Cosh[x]])/(a^2 + b^2) + (b*B*Log[a + b*Sinh[x]])/(a^2 + b^2)
Int[(u_)*((A_) + (B_.)*sec[(a_.) + (b_.)*(x_)]), x_Symbol] :> Int[ActivateT rig[u]*((B + A*Cos[a + b*x])/Cos[a + b*x]), x] /; FreeQ[{a, b, A, B}, x] && KnownSineIntegrandQ[u, x]
Int[u_, x_Symbol] :> With[{v = ExpandTrig[u, x]}, Int[v, x] /; SumQ[v]] /; !InertTrigFreeQ[u]
Time = 2.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.15
method | result | size |
parts | \(\frac {2 A \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}+B \left (\frac {b \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )}{a^{2}+b^{2}}+\frac {-b \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )+2 a \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{2}+b^{2}}\right )\) | \(102\) |
default | \(\frac {B b \ln \left (\tanh \left (\frac {x}{2}\right )^{2} a -2 b \tanh \left (\frac {x}{2}\right )-a \right )-\frac {2 \left (-a^{2} A -A \,b^{2}\right ) \operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\sqrt {a^{2}+b^{2}}}}{a^{2}+b^{2}}+\frac {2 B \left (-\frac {b \ln \left (1+\tanh \left (\frac {x}{2}\right )^{2}\right )}{2}+a \arctan \left (\tanh \left (\frac {x}{2}\right )\right )\right )}{a^{2}+b^{2}}\) | \(117\) |
risch | \(\frac {2 x B b}{a^{2}+b^{2}}+\frac {2 x B \,a^{2} b}{-a^{4}-2 a^{2} b^{2}-b^{4}}+\frac {2 x B \,b^{3}}{-a^{4}-2 a^{2} b^{2}-b^{4}}+\frac {i B \ln \left ({\mathrm e}^{x}+i\right ) a}{a^{2}+b^{2}}-\frac {B \ln \left ({\mathrm e}^{x}+i\right ) b}{a^{2}+b^{2}}-\frac {i B \ln \left ({\mathrm e}^{x}-i\right ) a}{a^{2}+b^{2}}-\frac {B \ln \left ({\mathrm e}^{x}-i\right ) b}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {A a -\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) b B}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {A a -\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {A^{2} a^{2}+A^{2} b^{2}}}{a^{2}+b^{2}}+\frac {\ln \left ({\mathrm e}^{x}+\frac {A a +\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) b B}{a^{2}+b^{2}}-\frac {\ln \left ({\mathrm e}^{x}+\frac {A a +\sqrt {A^{2} a^{2}+A^{2} b^{2}}}{A b}\right ) \sqrt {A^{2} a^{2}+A^{2} b^{2}}}{a^{2}+b^{2}}\) | \(362\) |
Input:
int((A+B*sech(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*( b/(a^2+b^2)*ln(tanh(1/2*x)^2*a-2*b*tanh(1/2*x)-a)+2/(a^2+b^2)*(-1/2*b*ln(1 +tanh(1/2*x)^2)+a*arctan(tanh(1/2*x))))
Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (85) = 170\).
Time = 1.76 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {2 \, B a \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + B b \log \left (\frac {2 \, {\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - B b \log \left (\frac {2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) + \sqrt {a^{2} + b^{2}} A \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right )}{a^{2} + b^{2}} \] Input:
integrate((A+B*sech(x))/(a+b*sinh(x)),x, algorithm="fricas")
Output:
(2*B*a*arctan(cosh(x) + sinh(x)) + B*b*log(2*(b*sinh(x) + a)/(cosh(x) - si nh(x))) - B*b*log(2*cosh(x)/(cosh(x) - sinh(x))) + sqrt(a^2 + b^2)*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)))/(a^2 + b^2)
\[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx=\int \frac {A + B \operatorname {sech}{\left (x \right )}}{a + b \sinh {\left (x \right )}}\, dx \] Input:
integrate((A+B*sech(x))/(a+b*sinh(x)),x)
Output:
Integral((A + B*sech(x))/(a + b*sinh(x)), x)
Time = 0.13 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.40 \[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx=-B {\left (\frac {2 \, a \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} - \frac {b \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} + \frac {b \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}}\right )} + \frac {A \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{\sqrt {a^{2} + b^{2}}} \] Input:
integrate((A+B*sech(x))/(a+b*sinh(x)),x, algorithm="maxima")
Output:
-B*(2*a*arctan(e^(-x))/(a^2 + b^2) - b*log(-2*a*e^(-x) + b*e^(-2*x) - b)/( a^2 + b^2) + b*log(e^(-2*x) + 1)/(a^2 + b^2)) + 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 = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.38 \[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {2 \, B a \arctan \left (e^{x}\right )}{a^{2} + b^{2}} - \frac {B b \log \left (e^{\left (2 \, x\right )} + 1\right )}{a^{2} + b^{2}} + \frac {B b \log \left ({\left | b e^{\left (2 \, x\right )} + 2 \, a e^{x} - b \right |}\right )}{a^{2} + b^{2}} + \frac {A \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{\sqrt {a^{2} + b^{2}}} \] Input:
integrate((A+B*sech(x))/(a+b*sinh(x)),x, algorithm="giac")
Output:
2*B*a*arctan(e^x)/(a^2 + b^2) - B*b*log(e^(2*x) + 1)/(a^2 + b^2) + B*b*log (abs(b*e^(2*x) + 2*a*e^x - b))/(a^2 + b^2) + A*log(abs(2*b*e^x + 2*a - 2*s qrt(a^2 + b^2))/abs(2*b*e^x + 2*a + 2*sqrt(a^2 + b^2)))/sqrt(a^2 + b^2)
Time = 11.60 (sec) , antiderivative size = 864, normalized size of antiderivative = 9.71 \[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx =\text {Too large to display} \] Input:
int((A + B/cosh(x))/(a + b*sinh(x)),x)
Output:
(log(((A*((a^2 + b^2)^3)^(1/2) + B*b^3 + B*a^2*b)*(b^3*(32*A^2 + 64*B^2 - 96*A*B*exp(x)) - 128*A^2*exp(x)*((a^2 + b^2)^3)^(1/2) - a*b^2*(96*A^2*exp( x) + 128*B^2*exp(x) - 192*A*B) - 128*a^3*exp(x)*(A^2 + B^2) + a^2*b*(64*A^ 2 + 64*B^2 - 384*A*B*exp(x)) + (32*A*b^6*(2*B + 3*A*exp(x)))/((a^2 + b^2)^ 3)^(1/2) + (32*A*a^4*b^2*(5*B + 3*A*exp(x)))/((a^2 + b^2)^3)^(1/2) + (32*A *a^2*b^4*(7*B + 6*A*exp(x)))/((a^2 + b^2)^3)^(1/2) + (32*A*a^3*b^3*(4*A - 19*B*exp(x)))/((a^2 + b^2)^3)^(1/2) + (64*A*a*b^5*(A - 4*B*exp(x)))/((a^2 + b^2)^3)^(1/2) + (32*A*a^5*b*(2*A - 11*B*exp(x)))/((a^2 + b^2)^3)^(1/2))) /(b^5*(a^2 + b^2)^2) - (32*B*(2*B^2*b^2 - A^2*b^2 + 4*A*B*a^2*exp(x) + A*B *b^2*exp(x) + A^2*a*b*exp(x) - 4*B^2*a*b*exp(x) - 2*A*B*a*b))/b^5)*(A*((a^ 2 + b^2)^3)^(1/2) + B*b^3 + B*a^2*b))/(a^4 + b^4 + 2*a^2*b^2) - (B*log(exp (x) + 1i))/(a*1i + b) - (B*log(exp(x) - 1i)*1i)/(a + b*1i) + (log(- (32*B* (2*B^2*b^2 - A^2*b^2 + 4*A*B*a^2*exp(x) + A*B*b^2*exp(x) + A^2*a*b*exp(x) - 4*B^2*a*b*exp(x) - 2*A*B*a*b))/b^5 - ((B*b^3 - A*((a^2 + b^2)^3)^(1/2) + B*a^2*b)*(a*b^2*(96*A^2*exp(x) + 128*B^2*exp(x) - 192*A*B) - 128*A^2*exp( x)*((a^2 + b^2)^3)^(1/2) - b^3*(32*A^2 + 64*B^2 - 96*A*B*exp(x)) + 128*a^3 *exp(x)*(A^2 + B^2) - a^2*b*(64*A^2 + 64*B^2 - 384*A*B*exp(x)) + (32*A*b^6 *(2*B + 3*A*exp(x)))/((a^2 + b^2)^3)^(1/2) + (32*A*a^4*b^2*(5*B + 3*A*exp( x)))/((a^2 + b^2)^3)^(1/2) + (32*A*a^2*b^4*(7*B + 6*A*exp(x)))/((a^2 + b^2 )^3)^(1/2) + (32*A*a^3*b^3*(4*A - 19*B*exp(x)))/((a^2 + b^2)^3)^(1/2) +...
Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \text {sech}(x)}{a+b \sinh (x)} \, dx=\frac {2 \mathit {atan} \left (e^{x}\right ) a b +2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a i -\mathrm {log}\left (e^{2 x}+1\right ) b^{2}+\mathrm {log}\left (e^{2 x} b +2 e^{x} a -b \right ) b^{2}}{a^{2}+b^{2}} \] Input:
int((A+B*sech(x))/(a+b*sinh(x)),x)
Output:
(2*atan(e**x)*a*b + 2*sqrt(a**2 + b**2)*atan((e**x*b*i + a*i)/sqrt(a**2 + b**2))*a*i - log(e**(2*x) + 1)*b**2 + log(e**(2*x)*b + 2*e**x*a - b)*b**2) /(a**2 + b**2)