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

Optimal. Leaf size=59 \[ \frac {\text {sech}(x) (a \sinh (x)+b)}{a^2+b^2}-\frac {2 b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

[Out]

-2*b^2*arctanh((b-a*tanh(1/2*x))/(a^2+b^2)^(1/2))/(a^2+b^2)^(3/2)+sech(x)*(b+a*sinh(x))/(a^2+b^2)

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Rubi [A]  time = 0.08, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2696, 12, 2660, 618, 206} \[ \frac {\text {sech}(x) (a \sinh (x)+b)}{a^2+b^2}-\frac {2 b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^2/(a + b*Sinh[x]),x]

[Out]

(-2*b^2*ArcTanh[(b - a*Tanh[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(3/2) + (Sech[x]*(b + a*Sinh[x]))/(a^2 + b^2)

Rule 12

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

Rule 206

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {\text {sech}^2(x)}{a+b \sinh (x)} \, dx &=\frac {\text {sech}(x) (b+a \sinh (x))}{a^2+b^2}+\frac {\int \frac {b^2}{a+b \sinh (x)} \, dx}{a^2+b^2}\\ &=\frac {\text {sech}(x) (b+a \sinh (x))}{a^2+b^2}+\frac {b^2 \int \frac {1}{a+b \sinh (x)} \, dx}{a^2+b^2}\\ &=\frac {\text {sech}(x) (b+a \sinh (x))}{a^2+b^2}+\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x-a x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^2+b^2}\\ &=\frac {\text {sech}(x) (b+a \sinh (x))}{a^2+b^2}-\frac {\left (4 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{4 \left (a^2+b^2\right )-x^2} \, dx,x,2 b-2 a \tanh \left (\frac {x}{2}\right )\right )}{a^2+b^2}\\ &=-\frac {2 b^2 \tanh ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2}}+\frac {\text {sech}(x) (b+a \sinh (x))}{a^2+b^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 67, normalized size = 1.14 \[ \frac {\frac {2 b^2 \tan ^{-1}\left (\frac {b-a \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2-b^2}}\right )}{\sqrt {-a^2-b^2}}+a \tanh (x)+b \text {sech}(x)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 2.36, size = 259, normalized size = 4.39 \[ -\frac {2 \, a^{3} + 2 \, a b^{2} - {\left (b^{2} \cosh \relax (x)^{2} + 2 \, b^{2} \cosh \relax (x) \sinh \relax (x) + b^{2} \sinh \relax (x)^{2} + b^{2}\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) - b}\right ) - 2 \, {\left (a^{2} b + b^{3}\right )} \cosh \relax (x) - 2 \, {\left (a^{2} b + b^{3}\right )} \sinh \relax (x)}{a^{4} + 2 \, a^{2} b^{2} + b^{4} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x)^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \relax (x) \sinh \relax (x) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \relax (x)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2/(a+b*sinh(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*cos
h(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*co
sh(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)*cosh(x) - 2*(a^2*b + b^3)*sinh(x))/(a^4 + 2*a^2*b^2 + b^4 + (a^4 + 2*a^2*b^2 + b^4)*cosh(x)^2 + 2*(a^4
+ 2*a^2*b^2 + b^4)*cosh(x)*sinh(x) + (a^4 + 2*a^2*b^2 + b^4)*sinh(x)^2)

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giac [A]  time = 0.18, size = 87, normalized size = 1.47 \[ \frac {b^{2} \log \left (\frac {{\left | 2 \, b e^{x} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{x} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b e^{x} - a\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, x\right )} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 71, normalized size = 1.20 \[ \frac {2 b^{2} \arctanh \left (\frac {2 a \tanh \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}}}-\frac {2 \left (-a \tanh \left (\frac {x}{2}\right )-b \right )}{\left (a^{2}+b^{2}\right ) \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2/(a^2+b^2)^(3/2)*arctanh(1/2*(2*a*tanh(1/2*x)-2*b)/(a^2+b^2)^(1/2))-2/(a^2+b^2)*(-a*tanh(1/2*x)-b)/(tanh(
1/2*x)^2+1)

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maxima [A]  time = 0.42, size = 89, normalized size = 1.51 \[ \frac {b^{2} \log \left (\frac {b e^{\left (-x\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-x\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (b e^{\left (-x\right )} + a\right )}}{a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*log((b*e^(-x) - a - sqrt(a^2 + b^2))/(b*e^(-x) - a + sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) + 2*(b*e^(-x) + a
)/(a^2 + b^2 + (a^2 + b^2)*e^(-2*x))

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mupad [B]  time = 1.04, size = 321, normalized size = 5.44 \[ -\frac {\frac {2\,a}{a^2+b^2}-\frac {2\,b\,{\mathrm {e}}^x}{a^2+b^2}}{{\mathrm {e}}^{2\,x}+1}-\frac {2\,\mathrm {atan}\left (\left ({\mathrm {e}}^x\,\left (\frac {2}{\sqrt {b^4}\,{\left (a^2+b^2\right )}^2}+\frac {2\,a\,\left (a^3\,\sqrt {b^4}+a\,b^2\,\sqrt {b^4}\right )}{b^4\,\sqrt {-{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}\right )-\frac {2\,a\,\left (b^3\,\sqrt {b^4}+a^2\,b\,\sqrt {b^4}\right )}{b^4\,\sqrt {-{\left (a^2+b^2\right )}^3}\,\left (a^2+b^2\right )\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}\right )\,\left (\frac {b^3\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}{2}+\frac {a^2\,b\,\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}}{2}\right )\right )\,\sqrt {b^4}}{\sqrt {-a^6-3\,a^4\,b^2-3\,a^2\,b^4-b^6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^2*(a + b*sinh(x))),x)

[Out]

- ((2*a)/(a^2 + b^2) - (2*b*exp(x))/(a^2 + b^2))/(exp(2*x) + 1) - (2*atan((exp(x)*(2/((b^4)^(1/2)*(a^2 + b^2)^
2) + (2*a*(a^3*(b^4)^(1/2) + a*b^2*(b^4)^(1/2)))/(b^4*(-(a^2 + b^2)^3)^(1/2)*(a^2 + b^2)*(- a^6 - b^6 - 3*a^2*
b^4 - 3*a^4*b^2)^(1/2))) - (2*a*(b^3*(b^4)^(1/2) + a^2*b*(b^4)^(1/2)))/(b^4*(-(a^2 + b^2)^3)^(1/2)*(a^2 + b^2)
*(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2)))*((b^3*(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/2 + (a^2*b*(
- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2))/2))*(b^4)^(1/2))/(- a^6 - b^6 - 3*a^2*b^4 - 3*a^4*b^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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