3.194 \(\int \frac{\text{sech}(x)}{a+b \sinh (x)} \, dx\)

Optimal. Leaf size=48 \[ \frac{b \log (a+b \sinh (x))}{a^2+b^2}+\frac{a \tan ^{-1}(\sinh (x))}{a^2+b^2}-\frac{b \log (\cosh (x))}{a^2+b^2} \]

[Out]

(a*ArcTan[Sinh[x]])/(a^2 + b^2) - (b*Log[Cosh[x]])/(a^2 + b^2) + (b*Log[a + b*Sinh[x]])/(a^2 + b^2)

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Rubi [A]  time = 0.060295, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {2668, 706, 31, 635, 204, 260} \[ \frac{b \log (a+b \sinh (x))}{a^2+b^2}+\frac{a \tan ^{-1}(\sinh (x))}{a^2+b^2}-\frac{b \log (\cosh (x))}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]/(a + b*Sinh[x]),x]

[Out]

(a*ArcTan[Sinh[x]])/(a^2 + b^2) - (b*Log[Cosh[x]])/(a^2 + b^2) + (b*Log[a + b*Sinh[x]])/(a^2 + b^2)

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 706

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[e^2/(c*d^2 + a*e^2), Int[1/(d + e*x), x],
x] + Dist[1/(c*d^2 + a*e^2), Int[(c*d - c*e*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a
*e^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

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

Mathematica [B]  time = 0.0953978, size = 99, normalized size = 2.06 \[ -\frac{b \left (\left (\sqrt{-b^2}-a\right ) \log \left (\sqrt{-b^2}-b \sinh (x)\right )-2 \sqrt{-b^2} \log (a+b \sinh (x))+\left (a+\sqrt{-b^2}\right ) \log \left (\sqrt{-b^2}+b \sinh (x)\right )\right )}{2 \sqrt{-b^2} \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]/(a + b*Sinh[x]),x]

[Out]

-(b*((-a + Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Sinh[x]] - 2*Sqrt[-b^2]*Log[a + b*Sinh[x]] + (a + Sqrt[-b^2])*Log[Sq
rt[-b^2] + b*Sinh[x]]))/(2*Sqrt[-b^2]*(a^2 + b^2))

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Maple [A]  time = 0.021, size = 71, normalized size = 1.5 \begin{align*} -{\frac{b}{{a}^{2}+{b}^{2}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{a\arctan \left ( \tanh \left ( x/2 \right ) \right ) }{{a}^{2}+{b}^{2}}}+{\frac{b}{{a}^{2}+{b}^{2}}\ln \left ( a \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}-2\,\tanh \left ( x/2 \right ) b-a \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)/(a+b*sinh(x)),x)

[Out]

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

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Maxima [A]  time = 1.84252, size = 89, normalized size = 1.85 \begin{align*} -\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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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Fricas [A]  time = 2.15307, size = 177, normalized size = 3.69 \begin{align*} \frac{2 \, a \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) + b \log \left (\frac{2 \,{\left (b \sinh \left (x\right ) + a\right )}}{\cosh \left (x\right ) - \sinh \left (x\right )}\right ) - b \log \left (\frac{2 \, \cosh \left (x\right )}{\cosh \left (x\right ) - \sinh \left (x\right )}\right )}{a^{2} + b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a*arctan(cosh(x) + sinh(x)) + b*log(2*(b*sinh(x) + a)/(cosh(x) - sinh(x))) - b*log(2*cosh(x)/(cosh(x) - sin
h(x))))/(a^2 + b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sech(x)/(a + b*sinh(x)), x)

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Giac [A]  time = 1.2414, size = 120, normalized size = 2.5 \begin{align*} \frac{b^{2} \log \left ({\left | -b{\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} a}{2 \,{\left (a^{2} + b^{2}\right )}} - \frac{b \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \,{\left (a^{2} + b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*log(abs(-b*(e^(-x) - e^x) + 2*a))/(a^2*b + b^3) + 1/2*(pi + 2*arctan(1/2*(e^(2*x) - 1)*e^(-x)))*a/(a^2 + b
^2) - 1/2*b*log((e^(-x) - e^x)^2 + 4)/(a^2 + b^2)