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

Optimal. Leaf size=25 \[ -\frac {x}{a}+\frac {\sinh (x)}{a}+\frac {\sinh (x)}{a (\cosh (x)+1)} \]

[Out]

-x/a+sinh(x)/a+sinh(x)/a/(1+cosh(x))

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Rubi [A]  time = 0.07, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {2746, 12, 2735, 2648} \[ -\frac {x}{a}+\frac {\sinh (x)}{a}+\frac {\sinh (x)}{a (\cosh (x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^2/(a + a*Cosh[x]),x]

[Out]

-(x/a) + Sinh[x]/a + Sinh[x]/(a*(1 + Cosh[x]))

Rule 12

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[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+a \cosh (x)} \, dx &=\frac {\sinh (x)}{a}-\frac {\int \frac {a \cosh (x)}{a+a \cosh (x)} \, dx}{a}\\ &=\frac {\sinh (x)}{a}-\int \frac {\cosh (x)}{a+a \cosh (x)} \, dx\\ &=-\frac {x}{a}+\frac {\sinh (x)}{a}+\int \frac {1}{a+a \cosh (x)} \, dx\\ &=-\frac {x}{a}+\frac {\sinh (x)}{a}+\frac {\sinh (x)}{a+a \cosh (x)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 32, normalized size = 1.28 \[ \frac {-2 x+3 \tanh \left (\frac {x}{2}\right )+\sinh \left (\frac {3 x}{2}\right ) \text {sech}\left (\frac {x}{2}\right )}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x + Sech[x/2]*Sinh[(3*x)/2] + 3*Tanh[x/2])/(2*a)

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fricas [A]  time = 1.46, size = 47, normalized size = 1.88 \[ -\frac {2 \, x \cosh \relax (x) - \cosh \relax (x)^{2} + 2 \, {\left (x - \cosh \relax (x) - 1\right )} \sinh \relax (x) - \sinh \relax (x)^{2} + 2 \, x + 5}{2 \, {\left (a \cosh \relax (x) + a \sinh \relax (x) + a\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*x*cosh(x) - cosh(x)^2 + 2*(x - cosh(x) - 1)*sinh(x) - sinh(x)^2 + 2*x + 5)/(a*cosh(x) + a*sinh(x) + a)

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giac [A]  time = 0.12, size = 35, normalized size = 1.40 \[ -\frac {x}{a} - \frac {{\left (5 \, e^{x} + 1\right )} e^{\left (-x\right )}}{2 \, a {\left (e^{x} + 1\right )}} + \frac {e^{x}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/a - 1/2*(5*e^x + 1)*e^(-x)/(a*(e^x + 1)) + 1/2*e^x/a

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maple [B]  time = 0.07, size = 59, normalized size = 2.36 \[ \frac {\tanh \left (\frac {x}{2}\right )}{a}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{a}-\frac {1}{a \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 41, normalized size = 1.64 \[ -\frac {x}{a} + \frac {5 \, e^{\left (-x\right )} + 1}{2 \, {\left (a e^{\left (-x\right )} + a e^{\left (-2 \, x\right )}\right )}} - \frac {e^{\left (-x\right )}}{2 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/a + 1/2*(5*e^(-x) + 1)/(a*e^(-x) + a*e^(-2*x)) - 1/2*e^(-x)/a

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mupad [B]  time = 0.90, size = 34, normalized size = 1.36 \[ \frac {{\mathrm {e}}^x}{2\,a}-\frac {x}{a}-\frac {2}{a\,\left ({\mathrm {e}}^x+1\right )}-\frac {{\mathrm {e}}^{-x}}{2\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^2/(a + a*cosh(x)),x)

[Out]

exp(x)/(2*a) - x/a - 2/(a*(exp(x) + 1)) - exp(-x)/(2*a)

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sympy [B]  time = 0.60, size = 63, normalized size = 2.52 \[ - \frac {x \tanh ^{2}{\left (\frac {x}{2} \right )}}{a \tanh ^{2}{\left (\frac {x}{2} \right )} - a} + \frac {x}{a \tanh ^{2}{\left (\frac {x}{2} \right )} - a} + \frac {\tanh ^{3}{\left (\frac {x}{2} \right )}}{a \tanh ^{2}{\left (\frac {x}{2} \right )} - a} - \frac {3 \tanh {\left (\frac {x}{2} \right )}}{a \tanh ^{2}{\left (\frac {x}{2} \right )} - a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*tanh(x/2)**2/(a*tanh(x/2)**2 - a) + x/(a*tanh(x/2)**2 - a) + tanh(x/2)**3/(a*tanh(x/2)**2 - a) - 3*tanh(x/2
)/(a*tanh(x/2)**2 - a)

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