3.98 \(\int \frac{\cosh (x)}{a+b \text{sech}(x)} \, dx\)

Optimal. Leaf size=62 \[ \frac{2 b^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^2 \sqrt{a-b} \sqrt{a+b}}-\frac{b x}{a^2}+\frac{\sinh (x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.0924905, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {3853, 12, 3783, 2659, 205} \[ \frac{2 b^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^2 \sqrt{a-b} \sqrt{a+b}}-\frac{b x}{a^2}+\frac{\sinh (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(a + b*Sech[x]),x]

[Out]

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

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 12

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

Rule 3783

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

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 205

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

Rubi steps

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

Mathematica [A]  time = 0.112964, size = 57, normalized size = 0.92 \[ \frac{b \left (-\frac{2 b \tan ^{-1}\left (\frac{(b-a) \tanh \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}-x\right )+a \sinh (x)}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/(a + b*Sech[x]),x]

[Out]

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

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Maple [A]  time = 0.031, size = 94, normalized size = 1.5 \begin{align*} -{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }+2\,{\frac{{b}^{2}}{{a}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}\arctan \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/(tanh(1/2*x)+1)-b/a^2*ln(tanh(1/2*x)+1)-1/a/(tanh(1/2*x)-1)+b/a^2*ln(tanh(1/2*x)-1)+2/a^2*b^2/((a+b)*(a-b
))^(1/2)*arctan((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(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(cosh(x)/(a+b*sech(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.6303, size = 1064, normalized size = 17.16 \begin{align*} \left [-\frac{a^{3} - a b^{2} + 2 \,{\left (a^{2} b - b^{3}\right )} x \cosh \left (x\right ) -{\left (a^{3} - a b^{2}\right )} \cosh \left (x\right )^{2} -{\left (a^{3} - a b^{2}\right )} \sinh \left (x\right )^{2} + 2 \,{\left (b^{2} \cosh \left (x\right ) + b^{2} \sinh \left (x\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{a^{2} \cosh \left (x\right )^{2} + a^{2} \sinh \left (x\right )^{2} + 2 \, a b \cosh \left (x\right ) - a^{2} + 2 \, b^{2} + 2 \,{\left (a^{2} \cosh \left (x\right ) + a b\right )} \sinh \left (x\right ) - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + b\right )}}{a \cosh \left (x\right )^{2} + a \sinh \left (x\right )^{2} + 2 \, b \cosh \left (x\right ) + 2 \,{\left (a \cosh \left (x\right ) + b\right )} \sinh \left (x\right ) + a}\right ) + 2 \,{\left ({\left (a^{2} b - b^{3}\right )} x -{\left (a^{3} - a b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left ({\left (a^{4} - a^{2} b^{2}\right )} \cosh \left (x\right ) +{\left (a^{4} - a^{2} b^{2}\right )} \sinh \left (x\right )\right )}}, -\frac{a^{3} - a b^{2} + 2 \,{\left (a^{2} b - b^{3}\right )} x \cosh \left (x\right ) -{\left (a^{3} - a b^{2}\right )} \cosh \left (x\right )^{2} -{\left (a^{3} - a b^{2}\right )} \sinh \left (x\right )^{2} + 4 \,{\left (b^{2} \cosh \left (x\right ) + b^{2} \sinh \left (x\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cosh \left (x\right ) + a \sinh \left (x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right ) + 2 \,{\left ({\left (a^{2} b - b^{3}\right )} x -{\left (a^{3} - a b^{2}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left ({\left (a^{4} - a^{2} b^{2}\right )} \cosh \left (x\right ) +{\left (a^{4} - a^{2} b^{2}\right )} \sinh \left (x\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a^3 - a*b^2 + 2*(a^2*b - b^3)*x*cosh(x) - (a^3 - a*b^2)*cosh(x)^2 - (a^3 - a*b^2)*sinh(x)^2 + 2*(b^2*co
sh(x) + b^2*sinh(x))*sqrt(-a^2 + b^2)*log((a^2*cosh(x)^2 + a^2*sinh(x)^2 + 2*a*b*cosh(x) - a^2 + 2*b^2 + 2*(a^
2*cosh(x) + a*b)*sinh(x) - 2*sqrt(-a^2 + b^2)*(a*cosh(x) + a*sinh(x) + b))/(a*cosh(x)^2 + a*sinh(x)^2 + 2*b*co
sh(x) + 2*(a*cosh(x) + b)*sinh(x) + a)) + 2*((a^2*b - b^3)*x - (a^3 - a*b^2)*cosh(x))*sinh(x))/((a^4 - a^2*b^2
)*cosh(x) + (a^4 - a^2*b^2)*sinh(x)), -1/2*(a^3 - a*b^2 + 2*(a^2*b - b^3)*x*cosh(x) - (a^3 - a*b^2)*cosh(x)^2
- (a^3 - a*b^2)*sinh(x)^2 + 4*(b^2*cosh(x) + b^2*sinh(x))*sqrt(a^2 - b^2)*arctan(-(a*cosh(x) + a*sinh(x) + b)/
sqrt(a^2 - b^2)) + 2*((a^2*b - b^3)*x - (a^3 - a*b^2)*cosh(x))*sinh(x))/((a^4 - a^2*b^2)*cosh(x) + (a^4 - a^2*
b^2)*sinh(x))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(x)/(a + b*sech(x)), x)

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Giac [A]  time = 1.16149, size = 84, normalized size = 1.35 \begin{align*} \frac{2 \, b^{2} \arctan \left (\frac{a e^{x} + b}{\sqrt{a^{2} - b^{2}}}\right )}{\sqrt{a^{2} - b^{2}} a^{2}} - \frac{b x}{a^{2}} - \frac{e^{\left (-x\right )}}{2 \, a} + \frac{e^{x}}{2 \, a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2*arctan((a*e^x + b)/sqrt(a^2 - b^2))/(sqrt(a^2 - b^2)*a^2) - b*x/a^2 - 1/2*e^(-x)/a + 1/2*e^x/a