3.59 \(\int \frac {\text {sech}^2(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=64 \[ \frac {2 b^2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b}}-\frac {b \tan ^{-1}(\sinh (x))}{a^2}+\frac {\tanh (x)}{a} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2802, 12, 2747, 3770, 2659, 208} \[ \frac {2 b^2 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^2 \sqrt {a-b} \sqrt {a+b}}-\frac {b \tan ^{-1}(\sinh (x))}{a^2}+\frac {\tanh (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(a + b*Cosh[x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 208

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

Rule 2659

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 2747

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[b/(
b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] - Dist[d/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; Fre
eQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 63, normalized size = 0.98 \[ \frac {-\frac {2 b^2 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+a \tanh (x)-2 b \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.65, size = 515, normalized size = 8.05 \[ \left [-\frac {2 \, a^{3} - 2 \, a b^{2} - {\left (b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2} + b^{2}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + 2 \, {\left (a^{2} b - b^{3} + {\left (a^{2} b - b^{3}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )}{a^{4} - a^{2} b^{2} + {\left (a^{4} - a^{2} b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{4} - a^{2} b^{2}\right )} \sinh \relax (x)^{2}}, -\frac {2 \, {\left (a^{3} - a b^{2} + {\left (b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2} + b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (a^{2} b - b^{3} + {\left (a^{2} b - b^{3}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{2} b - b^{3}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)^{2}\right )} \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right )\right )}}{a^{4} - a^{2} b^{2} + {\left (a^{4} - a^{2} b^{2}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} - a^{2} b^{2}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{4} - a^{2} b^{2}\right )} \sinh \relax (x)^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-(2*a^3 - 2*a*b^2 - (b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2 + b^2)*sqrt(a^2 - b^2)*log((b^2*co
sh(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*c
osh(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
- b^3 + (a^2*b - b^3)*cosh(x)^2 + 2*(a^2*b - b^3)*cosh(x)*sinh(x) + (a^2*b - b^3)*sinh(x)^2)*arctan(cosh(x) +
sinh(x)))/(a^4 - a^2*b^2 + (a^4 - a^2*b^2)*cosh(x)^2 + 2*(a^4 - a^2*b^2)*cosh(x)*sinh(x) + (a^4 - a^2*b^2)*si
nh(x)^2), -2*(a^3 - a*b^2 + (b^2*cosh(x)^2 + 2*b^2*cosh(x)*sinh(x) + b^2*sinh(x)^2 + b^2)*sqrt(-a^2 + b^2)*arc
tan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + (a^2*b - b^3 + (a^2*b - b^3)*cosh(x)^2 + 2*(a
^2*b - b^3)*cosh(x)*sinh(x) + (a^2*b - b^3)*sinh(x)^2)*arctan(cosh(x) + sinh(x)))/(a^4 - a^2*b^2 + (a^4 - a^2*
b^2)*cosh(x)^2 + 2*(a^4 - a^2*b^2)*cosh(x)*sinh(x) + (a^4 - a^2*b^2)*sinh(x)^2)]

________________________________________________________________________________________

giac [A]  time = 0.15, size = 61, normalized size = 0.95 \[ \frac {2 \, b^{2} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{2}} - \frac {2 \, b \arctan \left (e^{x}\right )}{a^{2}} - \frac {2}{a {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.09, size = 73, normalized size = 1.14 \[ \frac {2 b^{2} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \tanh \left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}-\frac {2 b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2/(a+b*cosh(x)),x)

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+b*cosh(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*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 3.10, size = 294, normalized size = 4.59 \[ \frac {b^2\,\ln \left (64\,a\,b^3-64\,a^3\,b+32\,b^3\,\sqrt {a^2-b^2}-128\,a^4\,{\mathrm {e}}^x-32\,b^4\,{\mathrm {e}}^x-64\,a^2\,b\,\sqrt {a^2-b^2}-128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}+160\,a^2\,b^2\,{\mathrm {e}}^x+96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^2\,\sqrt {a^2-b^2}}+\frac {b\,\left (\ln \left (32\,{\mathrm {e}}^x-32{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left (32\,{\mathrm {e}}^x+32{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{a^2}-\frac {b^2\,\ln \left (64\,a^3\,b-64\,a\,b^3+32\,b^3\,\sqrt {a^2-b^2}+128\,a^4\,{\mathrm {e}}^x+32\,b^4\,{\mathrm {e}}^x-64\,a^2\,b\,\sqrt {a^2-b^2}-128\,a^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-160\,a^2\,b^2\,{\mathrm {e}}^x+96\,a\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^2\,\sqrt {a^2-b^2}}-\frac {2}{a+a\,{\mathrm {e}}^{2\,x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(a + b*cosh(x))),x)

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{2}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________