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

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Rubi [A]  time = 0.0789683, antiderivative size = 59, normalized size of antiderivative = 1., 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 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]

Rule 12

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

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 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 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])

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*}

Mathematica [A]  time = 0.166832, 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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Maple [A]  time = 0.026, size = 71, normalized size = 1.2 \begin{align*} -2\,{\frac{-a\tanh \left ( x/2 \right ) -b}{ \left ({a}^{2}+{b}^{2} \right ) \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) }}+2\,{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{3/2}}{\it Artanh} \left ( 1/2\,{\frac{2\,a\tanh \left ( x/2 \right ) -2\,b}{\sqrt{{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*sinh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.05738, size = 693, normalized size = 11.75 \begin{align*} -\frac{2 \, a^{3} + 2 \, a b^{2} -{\left (b^{2} \cosh \left (x\right )^{2} + 2 \, b^{2} \cosh \left (x\right ) \sinh \left (x\right ) + b^{2} \sinh \left (x\right )^{2} + b^{2}\right )} \sqrt{a^{2} + b^{2}} \log \left (\frac{b^{2} \cosh \left (x\right )^{2} + b^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) + 2 \, a^{2} + b^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt{a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{b \cosh \left (x\right )^{2} + b \sinh \left (x\right )^{2} + 2 \, a \cosh \left (x\right ) + 2 \,{\left (b \cosh \left (x\right ) + a\right )} \sinh \left (x\right ) - b}\right ) - 2 \,{\left (a^{2} b + b^{3}\right )} \cosh \left (x\right ) - 2 \,{\left (a^{2} b + b^{3}\right )} \sinh \left (x\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right )^{2} + 2 \,{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \cosh \left (x\right ) \sinh \left (x\right ) +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sinh \left (x\right )^{2}} \end{align*}

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

[Out]

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

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Giac [A]  time = 1.32447, size = 117, normalized size = 1.98 \begin{align*} \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 )}} \end{align*}

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))