Integrand size = 11, antiderivative size = 54 \[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {\arctan (\sinh (x))}{a}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}} \] Output:
arctan(sinh(x))/a-2*b*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/a/(a-b) ^(1/2)/(a+b)^(1/2)
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {2 \left (\arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {b \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}\right )}{a} \] Input:
Integrate[Sech[x]/(a + b*Cosh[x]),x]
Output:
(2*(ArcTan[Tanh[x/2]] + (b*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/S qrt[-a^2 + b^2]))/a
Time = 0.34 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 3226, 3042, 3138, 221, 4257}
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.
\(\displaystyle \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (\frac {\pi }{2}+i x\right ) \left (a+b \sin \left (\frac {\pi }{2}+i x\right )\right )}dx\) |
\(\Big \downarrow \) 3226 |
\(\displaystyle \frac {\int \text {sech}(x)dx}{a}-\frac {b \int \frac {1}{a+b \cosh (x)}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}-\frac {b \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{a}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle -\frac {2 b \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{a}+\frac {\int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}+\frac {\int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\arctan (\sinh (x))}{a}-\frac {2 b \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}\) |
Input:
Int[Sech[x]/(a + b*Cosh[x]),x]
Output:
ArcTan[Sinh[x]]/a - (2*b*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(a* Sqrt[a - b]*Sqrt[a + b])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[ {a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.51 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94
method | result | size |
default | \(\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a}-\frac {2 b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}\) | \(51\) |
risch | \(\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{a}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{a}+\frac {b \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a}-\frac {b \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a}\) | \(141\) |
Input:
int(sech(x)/(a+b*cosh(x)),x,method=_RETURNVERBOSE)
Output:
2/a*arctan(tanh(1/2*x))-2*b/a/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x )/((a+b)*(a-b))^(1/2))
Time = 0.11 (sec) , antiderivative size = 227, normalized size of antiderivative = 4.20 \[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=\left [\frac {\sqrt {a^{2} - b^{2}} b \log \left (\frac {b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \, {\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) + b}\right ) + 2 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}{a^{3} - a b^{2}}, \frac {2 \, {\left (\sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )\right )}}{a^{3} - a b^{2}}\right ] \] Input:
integrate(sech(x)/(a+b*cosh(x)),x, algorithm="fricas")
Output:
[(sqrt(a^2 - b^2)*b*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)) + 2*(a^2 - b^2)*arctan(cosh(x) + sinh(x)))/(a^3 - a*b^2 ), 2*(sqrt(-a^2 + b^2)*b*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (a^2 - b^2)*arctan(cosh(x) + sinh(x)))/(a^3 - a*b^2)]
\[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=\int \frac {\operatorname {sech}{\left (x \right )}}{a + b \cosh {\left (x \right )}}\, dx \] Input:
integrate(sech(x)/(a+b*cosh(x)),x)
Output:
Integral(sech(x)/(a + b*cosh(x)), x)
Exception generated. \[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sech(x)/(a+b*cosh(x)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.11 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=-\frac {2 \, b \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a} + \frac {2 \, \arctan \left (e^{x}\right )}{a} \] Input:
integrate(sech(x)/(a+b*cosh(x)),x, algorithm="giac")
Output:
-2*b*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*a) + 2*arctan( e^x)/a
Time = 4.82 (sec) , antiderivative size = 286, normalized size of antiderivative = 5.30 \[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {b\,\ln \left (64\,a^4\,b-64\,a^2\,b^3+128\,a^5\,{\mathrm {e}}^x+32\,a\,b^3\,\sqrt {a^2-b^2}-64\,a^3\,b\,\sqrt {a^2-b^2}+32\,a\,b^4\,{\mathrm {e}}^x-128\,a^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^3\,b^2\,{\mathrm {e}}^x+96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a\,\sqrt {a^2-b^2}}-\frac {b\,\ln \left (64\,a^4\,b-64\,a^2\,b^3+128\,a^5\,{\mathrm {e}}^x-32\,a\,b^3\,\sqrt {a^2-b^2}+64\,a^3\,b\,\sqrt {a^2-b^2}+32\,a\,b^4\,{\mathrm {e}}^x+128\,a^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^3\,b^2\,{\mathrm {e}}^x-96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a\,\sqrt {a^2-b^2}}-\frac {\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a} \] Input:
int(1/(cosh(x)*(a + b*cosh(x))),x)
Output:
(b*log(64*a^4*b - 64*a^2*b^3 + 128*a^5*exp(x) + 32*a*b^3*(a^2 - b^2)^(1/2) - 64*a^3*b*(a^2 - b^2)^(1/2) + 32*a*b^4*exp(x) - 128*a^4*exp(x)*(a^2 - b^ 2)^(1/2) - 160*a^3*b^2*exp(x) + 96*a^2*b^2*exp(x)*(a^2 - b^2)^(1/2)))/(a*( a^2 - b^2)^(1/2)) - (b*log(64*a^4*b - 64*a^2*b^3 + 128*a^5*exp(x) - 32*a*b ^3*(a^2 - b^2)^(1/2) + 64*a^3*b*(a^2 - b^2)^(1/2) + 32*a*b^4*exp(x) + 128* a^4*exp(x)*(a^2 - b^2)^(1/2) - 160*a^3*b^2*exp(x) - 96*a^2*b^2*exp(x)*(a^2 - b^2)^(1/2)))/(a*(a^2 - b^2)^(1/2)) - (log(exp(x) - 1i)*1i - log(exp(x) + 1i)*1i)/a
Time = 0.21 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.24 \[ \int \frac {\text {sech}(x)}{a+b \cosh (x)} \, dx=\frac {2 \mathit {atan} \left (e^{x}\right ) a^{2}-2 \mathit {atan} \left (e^{x}\right ) b^{2}+2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b}{a \left (a^{2}-b^{2}\right )} \] Input:
int(sech(x)/(a+b*cosh(x)),x)
Output:
(2*(atan(e**x)*a**2 - atan(e**x)*b**2 + sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt( - a**2 + b**2))*b))/(a*(a**2 - b**2))