\(\int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx\) [100]

Optimal result

Integrand size = 13, antiderivative size = 54 \[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=\frac {\arctan (\sinh (x))}{b}-\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b}} \]

[Out]

arctan(sinh(x))/b-2*a*arctan((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b/(a-b)^(1/2)/(a+b)^(1/2)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {3874, 3855, 3916, 2738, 211} \[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=\frac {\arctan (\sinh (x))}{b}-\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}} \]

[In]

Int[Sech[x]^2/(a + b*Sech[x]),x]

[Out]

ArcTan[Sinh[x]]/b - (2*a*ArcTan[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2738

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[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]

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3874

Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[Csc[e + f*x],
x], x] - Dist[a/b, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]

Rule 3916

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a/b)*Si
n[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \text {sech}(x) \, dx}{b}-\frac {a \int \frac {\text {sech}(x)}{a+b \text {sech}(x)} \, dx}{b} \\ & = \frac {\arctan (\sinh (x))}{b}-\frac {a \int \frac {1}{1+\frac {a \cosh (x)}{b}} \, dx}{b^2} \\ & = \frac {\arctan (\sinh (x))}{b}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{1+\frac {a}{b}-\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^2} \\ & = \frac {\arctan (\sinh (x))}{b}-\frac {2 a \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=\frac {2 \left (\arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {a \arctan \left (\frac {(-a+b) \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}\right )}{b} \]

[In]

Integrate[Sech[x]^2/(a + b*Sech[x]),x]

[Out]

(2*(ArcTan[Tanh[x/2]] + (a*ArcTan[((-a + b)*Tanh[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2]))/b

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {2 a \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{b}\)	\(51\)
risch	\(-\frac {a \ln \left ({\mathrm e}^{x}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, b}+\frac {a \ln \left ({\mathrm e}^{x}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, b}+\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{b}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{b}\)	\(141\)

[In]

int(sech(x)^2/(a+b*sech(x)),x,method=_RETURNVERBOSE)

[Out]

-2*a/b/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))+2/b*arctan(tanh(1/2*x))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 219, normalized size of antiderivative = 4.06 \[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=\left [-\frac {\sqrt {-a^{2} + b^{2}} a \log \left (\frac {a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) - a^{2} + 2 \, b^{2} + 2 \, {\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) + 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \, {\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) + a}\right ) - 2 \, {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}{a^{2} b - b^{3}}, \frac {2 \, {\left (\sqrt {a^{2} - b^{2}} a \arctan \left (-\frac {a \cosh \left (x\right ) + a \sinh \left (x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right ) + {\left (a^{2} - b^{2}\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right )\right )}}{a^{2} b - b^{3}}\right ] \]

[In]

integrate(sech(x)^2/(a+b*sech(x)),x, algorithm="fricas")

[Out]

[-(sqrt(-a^2 + b^2)*a*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^2*cosh(x) + a*b)
*sinh(x) + 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*cosh(x) + 2*(a*cos
h(x) + b)*sinh(x) + a)) - 2*(a^2 - b^2)*arctan(cosh(x) + sinh(x)))/(a^2*b - b^3), 2*(sqrt(a^2 - b^2)*a*arctan(
-(a*cosh(x) + a*sinh(x) + b)/sqrt(a^2 - b^2)) + (a^2 - b^2)*arctan(cosh(x) + sinh(x)))/(a^2*b - b^3)]

Sympy [F]

\[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=\int \frac {\operatorname {sech}^{2}{\left (x \right )}}{a + b \operatorname {sech}{\left (x \right )}}\, dx \]

[In]

integrate(sech(x)**2/(a+b*sech(x)),x)

[Out]

Integral(sech(x)**2/(a + b*sech(x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(sech(x)^2/(a+b*sech(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=-\frac {2 \, a \arctan \left (\frac {a e^{x} + b}{\sqrt {a^{2} - b^{2}}}\right )}{\sqrt {a^{2} - b^{2}} b} + \frac {2 \, \arctan \left (e^{x}\right )}{b} \]

[In]

integrate(sech(x)^2/(a+b*sech(x)),x, algorithm="giac")

[Out]

-2*a*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^2 - b^2)*b) + 2*arctan(e^x)/b

Mupad [B] (verification not implemented)

Time = 4.44 (sec) , antiderivative size = 286, normalized size of antiderivative = 5.30 \[ \int \frac {\text {sech}^2(x)}{a+b \text {sech}(x)} \, dx=\frac {a\,\ln \left (64\,a\,b^4-64\,a^3\,b^2+128\,b^5\,{\mathrm {e}}^x-64\,a\,b^3\,\sqrt {b^2-a^2}+32\,a^3\,b\,\sqrt {b^2-a^2}+32\,a^4\,b\,{\mathrm {e}}^x-128\,b^4\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}-160\,a^2\,b^3\,{\mathrm {e}}^x+96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}\right )}{b\,\sqrt {b^2-a^2}}-\frac {\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{b}-\frac {a\,\ln \left (64\,a\,b^4-64\,a^3\,b^2+128\,b^5\,{\mathrm {e}}^x+64\,a\,b^3\,\sqrt {b^2-a^2}-32\,a^3\,b\,\sqrt {b^2-a^2}+32\,a^4\,b\,{\mathrm {e}}^x+128\,b^4\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}-160\,a^2\,b^3\,{\mathrm {e}}^x-96\,a^2\,b^2\,{\mathrm {e}}^x\,\sqrt {b^2-a^2}\right )}{b\,\sqrt {b^2-a^2}} \]

[In]

int(1/(cosh(x)^2*(a + b/cosh(x))),x)

[Out]

(a*log(64*a*b^4 - 64*a^3*b^2 + 128*b^5*exp(x) - 64*a*b^3*(b^2 - a^2)^(1/2) + 32*a^3*b*(b^2 - a^2)^(1/2) + 32*a
^4*b*exp(x) - 128*b^4*exp(x)*(b^2 - a^2)^(1/2) - 160*a^2*b^3*exp(x) + 96*a^2*b^2*exp(x)*(b^2 - a^2)^(1/2)))/(b
*(b^2 - a^2)^(1/2)) - (log(exp(x) - 1i)*1i - log(exp(x) + 1i)*1i)/b - (a*log(64*a*b^4 - 64*a^3*b^2 + 128*b^5*e
xp(x) + 64*a*b^3*(b^2 - a^2)^(1/2) - 32*a^3*b*(b^2 - a^2)^(1/2) + 32*a^4*b*exp(x) + 128*b^4*exp(x)*(b^2 - a^2)
^(1/2) - 160*a^2*b^3*exp(x) - 96*a^2*b^2*exp(x)*(b^2 - a^2)^(1/2)))/(b*(b^2 - a^2)^(1/2))