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*ln(cosh(x))/(a^2+b^2)+b*ln(a+b*sinh(x))/(a^2+b^2)

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Rubi [A]  time = 0.06, antiderivative size = 48, normalized size of antiderivative = 1.00, 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 31

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

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]

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 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 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]

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*}

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Mathematica [B]  time = 0.10, 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]

-1/2*(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])*Lo
g[Sqrt[-b^2] + b*Sinh[x]]))/(Sqrt[-b^2]*(a^2 + b^2))

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fricas [A]  time = 1.02, size = 57, normalized size = 1.19 \[ \frac {2 \, a \arctan \left (\cosh \relax (x) + \sinh \relax (x)\right ) + b \log \left (\frac {2 \, {\left (b \sinh \relax (x) + a\right )}}{\cosh \relax (x) - \sinh \relax (x)}\right ) - b \log \left (\frac {2 \, \cosh \relax (x)}{\cosh \relax (x) - \sinh \relax (x)}\right )}{a^{2} + b^{2}} \]

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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giac [A]  time = 0.18, size = 89, normalized size = 1.85 \[ \frac {b^{2} \log \left ({\left | -b {\left (e^{\left (-x\right )} - e^{x}\right )} + 2 \, a \right |}\right )}{a^{2} b + b^{3}} + \frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} a}{2 \, {\left (a^{2} + b^{2}\right )}} - \frac {b \log \left ({\left (e^{\left (-x\right )} - e^{x}\right )}^{2} + 4\right )}{2 \, {\left (a^{2} + b^{2}\right )}} \]

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)

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maple [A]  time = 0.04, size = 71, normalized size = 1.48 \[ \frac {b \ln \left (a \left (\tanh ^{2}\left (\frac {x}{2}\right )\right )-2 \tanh \left (\frac {x}{2}\right ) b -a \right )}{a^{2}+b^{2}}-\frac {b \ln \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )}{a^{2}+b^{2}}+\frac {2 a \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{2}+b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.41, size = 66, normalized size = 1.38 \[ -\frac {2 \, a \arctan \left (e^{\left (-x\right )}\right )}{a^{2} + b^{2}} + \frac {b \log \left (-2 \, a e^{\left (-x\right )} + b e^{\left (-2 \, x\right )} - b\right )}{a^{2} + b^{2}} - \frac {b \log \left (e^{\left (-2 \, x\right )} + 1\right )}{a^{2} + b^{2}} \]

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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mupad [B]  time = 1.35, size = 93, normalized size = 1.94 \[ \frac {b\,\ln \left (4\,b^3\,{\mathrm {e}}^{2\,x}-a^2\,b-4\,b^3+2\,a^3\,{\mathrm {e}}^x+8\,a\,b^2\,{\mathrm {e}}^x+a^2\,b\,{\mathrm {e}}^{2\,x}\right )}{a^2+b^2}-\frac {\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )}{b+a\,1{}\mathrm {i}}-\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{a+b\,1{}\mathrm {i}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)*(a + b*sinh(x))),x)

[Out]

(b*log(4*b^3*exp(2*x) - a^2*b - 4*b^3 + 2*a^3*exp(x) + 8*a*b^2*exp(x) + a^2*b*exp(2*x)))/(a^2 + b^2) - log(exp
(x) + 1i)/(a*1i + b) - (log(exp(x)*1i + 1)*1i)/(a + b*1i)

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

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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