3.57 \(\int \frac {\cosh (x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=52 \[ \frac {x}{b}-\frac {2 a \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2735, 2659, 208} \[ \frac {x}{b}-\frac {2 a \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b \sqrt {a-b} \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]/(a + b*Cosh[x]),x]

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 48, normalized size = 0.92 \[ \frac {\frac {2 a \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+x}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]/(a + b*Cosh[x]),x]

[Out]

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

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fricas [A]  time = 0.48, size = 218, normalized size = 4.19 \[ \left [\frac {\sqrt {a^{2} - b^{2}} a \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + {\left (a^{2} - b^{2}\right )} x}{a^{2} b - b^{3}}, \frac {2 \, \sqrt {-a^{2} + b^{2}} a \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} x}{a^{2} b - b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(sqrt(a^2 - b^2)*a*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)*s
inh(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)) + (a^2 - b^2)*x)/(a^2*b - b^3), (2*sqrt(-a^2 + b^2)*a*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x)
+ b*sinh(x) + a)/(a^2 - b^2)) + (a^2 - b^2)*x)/(a^2*b - b^3)]

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giac [A]  time = 0.12, size = 42, normalized size = 0.81 \[ -\frac {2 \, a \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b} + \frac {x}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 64, normalized size = 1.23 \[ -\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b}-\frac {2 a \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 0.22, size = 109, normalized size = 2.10 \[ \frac {x}{b}+\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^2}-\frac {2\,a\,\left (b+a\,{\mathrm {e}}^x\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a\,\ln \left (\frac {2\,a\,{\mathrm {e}}^x}{b^2}+\frac {2\,a\,\left (b+a\,{\mathrm {e}}^x\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b\,\sqrt {a+b}\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)/(a + b*cosh(x)),x)

[Out]

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

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sympy [A]  time = 25.02, size = 241, normalized size = 4.63 \[ \begin {cases} \tilde {\infty } x & \text {for}\: a = 0 \wedge b = 0 \\\frac {x}{b} - \frac {1}{b \tanh {\left (\frac {x}{2} \right )}} & \text {for}\: a = - b \\\frac {\sinh {\relax (x )}}{a} & \text {for}\: b = 0 \\\frac {x}{b} - \frac {\tanh {\left (\frac {x}{2} \right )}}{b} & \text {for}\: a = b \\\frac {a x \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} + \frac {a \log {\left (- \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} - \frac {a \log {\left (\sqrt {\frac {a}{a - b} + \frac {b}{a - b}} + \tanh {\left (\frac {x}{2} \right )} \right )}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} - \frac {b x \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}}{a b \sqrt {\frac {a}{a - b} + \frac {b}{a - b}} - b^{2} \sqrt {\frac {a}{a - b} + \frac {b}{a - b}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x, Eq(a, 0) & Eq(b, 0)), (x/b - 1/(b*tanh(x/2)), Eq(a, -b)), (sinh(x)/a, Eq(b, 0)), (x/b - tanh
(x/2)/b, Eq(a, b)), (a*x*sqrt(a/(a - b) + b/(a - b))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/(a - b) +
b/(a - b))) + a*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/(
a - b) + b/(a - b))) - a*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*
sqrt(a/(a - b) + b/(a - b))) - b*x*sqrt(a/(a - b) + b/(a - b))/(a*b*sqrt(a/(a - b) + b/(a - b)) - b**2*sqrt(a/
(a - b) + b/(a - b))), True))

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