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

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Rubi [A]  time = 0.11, antiderivative size = 62, 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 = {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 12

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

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]

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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [B]  time = 0.78, size = 449, normalized size = 7.24 \[ \left [-\frac {a^{2} b - b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} x \cosh \relax (x) - {\left (a^{2} b - b^{3}\right )} \cosh \relax (x)^{2} - {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)^{2} - 2 \, {\left (a^{2} \cosh \relax (x) + a^{2} \sinh \relax (x)\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {b^{2} \cosh \relax (x)^{2} + b^{2} \sinh \relax (x)^{2} + 2 \, a b \cosh \relax (x) + 2 \, a^{2} - b^{2} + 2 \, {\left (b^{2} \cosh \relax (x) + a b\right )} \sinh \relax (x) - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{b \cosh \relax (x)^{2} + b \sinh \relax (x)^{2} + 2 \, a \cosh \relax (x) + 2 \, {\left (b \cosh \relax (x) + a\right )} \sinh \relax (x) + b}\right ) + 2 \, {\left ({\left (a^{3} - a b^{2}\right )} x - {\left (a^{2} b - b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{2 \, {\left ({\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x) + {\left (a^{2} b^{2} - b^{4}\right )} \sinh \relax (x)\right )}}, -\frac {a^{2} b - b^{3} + 2 \, {\left (a^{3} - a b^{2}\right )} x \cosh \relax (x) - {\left (a^{2} b - b^{3}\right )} \cosh \relax (x)^{2} - {\left (a^{2} b - b^{3}\right )} \sinh \relax (x)^{2} + 4 \, {\left (a^{2} \cosh \relax (x) + a^{2} \sinh \relax (x)\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cosh \relax (x) + b \sinh \relax (x) + a\right )}}{a^{2} - b^{2}}\right ) + 2 \, {\left ({\left (a^{3} - a b^{2}\right )} x - {\left (a^{2} b - b^{3}\right )} \cosh \relax (x)\right )} \sinh \relax (x)}{2 \, {\left ({\left (a^{2} b^{2} - b^{4}\right )} \cosh \relax (x) + {\left (a^{2} b^{2} - b^{4}\right )} \sinh \relax (x)\right )}}\right ] \]

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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giac [A]  time = 0.15, size = 62, normalized size = 1.00 \[ \frac {2 \, a^{2} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{2}} - \frac {a x}{b^{2}} - \frac {e^{\left (-x\right )}}{2 \, b} + \frac {e^{x}}{2 \, b} \]

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

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maple [A]  time = 0.06, size = 94, normalized size = 1.52 \[ -\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{2}}-\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{2}}+\frac {2 a^{2} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)+2*a^2/b^2/((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)^2/(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 = 1.04, size = 139, normalized size = 2.24 \[ \frac {{\mathrm {e}}^x}{2\,b}-\frac {{\mathrm {e}}^{-x}}{2\,b}-\frac {a\,x}{b^2}+\frac {a^2\,\ln \left (-\frac {2\,a^2\,{\mathrm {e}}^x}{b^3}-\frac {2\,a^2\,\left (b+a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a^2\,\ln \left (\frac {2\,a^2\,\left (b+a\,{\mathrm {e}}^x\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {2\,a^2\,{\mathrm {e}}^x}{b^3}\right )}{b^2\,\sqrt {a+b}\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(a + b*cosh(x)),x)

[Out]

exp(x)/(2*b) - exp(-x)/(2*b) - (a*x)/b^2 + (a^2*log(- (2*a^2*exp(x))/b^3 - (2*a^2*(b + a*exp(x)))/(b^3*(a + b)
^(1/2)*(a - b)^(1/2))))/(b^2*(a + b)^(1/2)*(a - b)^(1/2)) - (a^2*log((2*a^2*(b + a*exp(x)))/(b^3*(a + b)^(1/2)
*(a - b)^(1/2)) - (2*a^2*exp(x))/b^3))/(b^2*(a + b)^(1/2)*(a - b)^(1/2))

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sympy [A]  time = 93.90, size = 1275, normalized size = 20.56 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*sinh(x), Eq(a, 0) & Eq(b, 0)), (x*tanh(x/2)**3/(b*tanh(x/2)**3 - b*tanh(x/2)) - x*tanh(x/2)/(b*
tanh(x/2)**3 - b*tanh(x/2)) - 3*tanh(x/2)**2/(b*tanh(x/2)**3 - b*tanh(x/2)) + 1/(b*tanh(x/2)**3 - b*tanh(x/2))
, Eq(a, -b)), ((-x*sinh(x)**2/2 + x*cosh(x)**2/2 + sinh(x)*cosh(x)/2)/a, Eq(b, 0)), (-x*tanh(x/2)**2/(b*tanh(x
/2)**2 - b) + x/(b*tanh(x/2)**2 - b) + tanh(x/2)**3/(b*tanh(x/2)**2 - b) - 3*tanh(x/2)/(b*tanh(x/2)**2 - b), E
q(a, b)), (-a**2*x*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2/(a*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 -
a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**3*sqrt(a/(a - b) + b/
(a - b))) + a**2*x*sqrt(a/(a - b) + b/(a - b))/(a*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 - a*b**2*sqrt(
a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**3*sqrt(a/(a - b) + b/(a - b))) - a
**2*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))*tanh(x/2)**2/(a*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**
2 - a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**3*sqrt(a/(a - b) +
b/(a - b))) + a**2*log(-sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2
)**2 - a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**3*sqrt(a/(a - b
) + b/(a - b))) + a**2*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))*tanh(x/2)**2/(a*b**2*sqrt(a/(a - b) + b/(a
- b))*tanh(x/2)**2 - a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**
3*sqrt(a/(a - b) + b/(a - b))) - a**2*log(sqrt(a/(a - b) + b/(a - b)) + tanh(x/2))/(a*b**2*sqrt(a/(a - b) + b/
(a - b))*tanh(x/2)**2 - a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b
**3*sqrt(a/(a - b) + b/(a - b))) + a*b*x*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2/(a*b**2*sqrt(a/(a - b) + b/(
a - b))*tanh(x/2)**2 - a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b*
*3*sqrt(a/(a - b) + b/(a - b))) - a*b*x*sqrt(a/(a - b) + b/(a - b))/(a*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x
/2)**2 - a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**3*sqrt(a/(a -
b) + b/(a - b))) - 2*a*b*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)/(a*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)*
*2 - a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**3*sqrt(a/(a - b)
+ b/(a - b))) + 2*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)/(a*b**2*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2
- a*b**2*sqrt(a/(a - b) + b/(a - b)) - b**3*sqrt(a/(a - b) + b/(a - b))*tanh(x/2)**2 + b**3*sqrt(a/(a - b) + b
/(a - b))), True))

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