3.56 \(\int \frac{\cosh ^2(x)}{a+b \cosh (x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.106229, antiderivative size = 62, 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 = {2746, 12, 2735, 2659, 208} \[ \frac{2 a^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{b^2 \sqrt{a-b} \sqrt{a+b}}-\frac{a x}{b^2}+\frac{\sinh (x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(a + b*Cosh[x]),x]

[Out]

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

Rule 2746

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

Rule 12

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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 208

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

Rubi steps

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

Mathematica [A]  time = 0.105418, size = 57, normalized size = 0.92 \[ \frac{a \left (-\frac{2 a \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-x\right )+b \sinh (x)}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 94, normalized size = 1.5 \begin{align*} 2\,{\frac{{a}^{2}}{{b}^{2}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{{b}^{2}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2/(a+b*cosh(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.25264, size = 1080, normalized size = 17.42 \begin{align*} \left [-\frac{a^{2} b - b^{3} + 2 \,{\left (a^{3} - a b^{2}\right )} x \cosh \left (x\right ) -{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )^{2} -{\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )^{2} - 2 \,{\left (a^{2} \cosh \left (x\right ) + a^{2} \sinh \left (x\right )\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 ({\left (a^{3} - a b^{2}\right )} x -{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left ({\left (a^{2} b^{2} - b^{4}\right )} \cosh \left (x\right ) +{\left (a^{2} b^{2} - b^{4}\right )} \sinh \left (x\right )\right )}}, -\frac{a^{2} b - b^{3} + 2 \,{\left (a^{3} - a b^{2}\right )} x \cosh \left (x\right ) -{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )^{2} -{\left (a^{2} b - b^{3}\right )} \sinh \left (x\right )^{2} + 4 \,{\left (a^{2} \cosh \left (x\right ) + a^{2} \sinh \left (x\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cosh \left (x\right ) + b \sinh \left (x\right ) + a\right )}}{a^{2} - b^{2}}\right ) + 2 \,{\left ({\left (a^{3} - a b^{2}\right )} x -{\left (a^{2} b - b^{3}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )}{2 \,{\left ({\left (a^{2} b^{2} - b^{4}\right )} \cosh \left (x\right ) +{\left (a^{2} b^{2} - b^{4}\right )} \sinh \left (x\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^2/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**2/(a+b*cosh(x)),x)

[Out]

Timed out

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Giac [A]  time = 1.20146, size = 84, normalized size = 1.35 \begin{align*} \frac{2 \, a^{2} \arctan \left (\frac{b e^{x} + a}{\sqrt{-a^{2} + b^{2}}}\right )}{\sqrt{-a^{2} + b^{2}} b^{2}} - \frac{a x}{b^{2}} - \frac{e^{\left (-x\right )}}{2 \, b} + \frac{e^{x}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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