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

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Rubi [A]  time = 0.116694, antiderivative size = 64, normalized size of antiderivative = 1., 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 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 12

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

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 3770

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

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 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]

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 = 0.102917, 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

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Maple [A]  time = 0.03, size = 73, normalized size = 1.1 \begin{align*} 2\,{\frac{{b}^{2}}{{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{\tanh \left ( x/2 \right ) }{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}-2\,{\frac{b\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{{a}^{2}}} \end{align*}

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))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.56925, size = 1301, normalized size = 20.33 \begin{align*} \text{result too large to display} \end{align*}

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)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{2}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}

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)

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Giac [A]  time = 1.16106, size = 82, normalized size = 1.28 \begin{align*} \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 )}} \end{align*}

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))