3.24 \(\int \frac{\cosh ^4(x)}{a+a \cosh (x)} \, dx\)

Optimal. Leaf size=54 \[ -\frac{3 x}{2 a}+\frac{4 \sinh ^3(x)}{3 a}+\frac{4 \sinh (x)}{a}-\frac{\sinh (x) \cosh ^3(x)}{a \cosh (x)+a}-\frac{3 \sinh (x) \cosh (x)}{2 a} \]

[Out]

(-3*x)/(2*a) + (4*Sinh[x])/a - (3*Cosh[x]*Sinh[x])/(2*a) - (Cosh[x]^3*Sinh[x])/(a + a*Cosh[x]) + (4*Sinh[x]^3)
/(3*a)

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Rubi [A]  time = 0.0758203, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {2767, 2748, 2635, 8, 2633} \[ -\frac{3 x}{2 a}+\frac{4 \sinh ^3(x)}{3 a}+\frac{4 \sinh (x)}{a}-\frac{\sinh (x) \cosh ^3(x)}{a \cosh (x)+a}-\frac{3 \sinh (x) \cosh (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(a + a*Cosh[x]),x]

[Out]

(-3*x)/(2*a) + (4*Sinh[x])/a - (3*Cosh[x]*Sinh[x])/(2*a) - (Cosh[x]^3*Sinh[x])/(a + a*Cosh[x]) + (4*Sinh[x]^3)
/(3*a)

Rule 2767

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int \frac{\cosh ^4(x)}{a+a \cosh (x)} \, dx &=-\frac{\cosh ^3(x) \sinh (x)}{a+a \cosh (x)}-\frac{\int \cosh ^2(x) (3 a-4 a \cosh (x)) \, dx}{a^2}\\ &=-\frac{\cosh ^3(x) \sinh (x)}{a+a \cosh (x)}-\frac{3 \int \cosh ^2(x) \, dx}{a}+\frac{4 \int \cosh ^3(x) \, dx}{a}\\ &=-\frac{3 \cosh (x) \sinh (x)}{2 a}-\frac{\cosh ^3(x) \sinh (x)}{a+a \cosh (x)}+\frac{(4 i) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (x)\right )}{a}-\frac{3 \int 1 \, dx}{2 a}\\ &=-\frac{3 x}{2 a}+\frac{4 \sinh (x)}{a}-\frac{3 \cosh (x) \sinh (x)}{2 a}-\frac{\cosh ^3(x) \sinh (x)}{a+a \cosh (x)}+\frac{4 \sinh ^3(x)}{3 a}\\ \end{align*}

Mathematica [A]  time = 0.0752781, size = 53, normalized size = 0.98 \[ \frac{\text{sech}\left (\frac{x}{2}\right ) \left (45 \sinh \left (\frac{x}{2}\right )+18 \sinh \left (\frac{3 x}{2}\right )-2 \sinh \left (\frac{5 x}{2}\right )+\sinh \left (\frac{7 x}{2}\right )-36 x \cosh \left (\frac{x}{2}\right )\right )}{24 a} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^4/(a + a*Cosh[x]),x]

[Out]

(Sech[x/2]*(-36*x*Cosh[x/2] + 45*Sinh[x/2] + 18*Sinh[(3*x)/2] - 2*Sinh[(5*x)/2] + Sinh[(7*x)/2]))/(24*a)

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Maple [B]  time = 0.024, size = 111, normalized size = 2.1 \begin{align*}{\frac{1}{a}\tanh \left ({\frac{x}{2}} \right ) }-{\frac{1}{3\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-3}}+{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-2}}-{\frac{5}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{3}{2\,a}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }-{\frac{1}{3\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-2}}-{\frac{5}{2\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3}{2\,a}\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)^4/(a+a*cosh(x)),x)

[Out]

1/a*tanh(1/2*x)-1/3/a/(tanh(1/2*x)+1)^3+1/a/(tanh(1/2*x)+1)^2-5/2/a/(tanh(1/2*x)+1)-3/2/a*ln(tanh(1/2*x)+1)-1/
3/a/(tanh(1/2*x)-1)^3-1/a/(tanh(1/2*x)-1)^2-5/2/a/(tanh(1/2*x)-1)+3/2/a*ln(tanh(1/2*x)-1)

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Maxima [A]  time = 1.06566, size = 89, normalized size = 1.65 \begin{align*} -\frac{3 \, x}{2 \, a} - \frac{21 \, e^{\left (-x\right )} - 3 \, e^{\left (-2 \, x\right )} + e^{\left (-3 \, x\right )}}{24 \, a} - \frac{2 \, e^{\left (-x\right )} - 18 \, e^{\left (-2 \, x\right )} - 69 \, e^{\left (-3 \, x\right )} - 1}{24 \,{\left (a e^{\left (-3 \, x\right )} + a e^{\left (-4 \, x\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+a*cosh(x)),x, algorithm="maxima")

[Out]

-3/2*x/a - 1/24*(21*e^(-x) - 3*e^(-2*x) + e^(-3*x))/a - 1/24*(2*e^(-x) - 18*e^(-2*x) - 69*e^(-3*x) - 1)/(a*e^(
-3*x) + a*e^(-4*x))

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Fricas [B]  time = 1.91861, size = 346, normalized size = 6.41 \begin{align*} \frac{\cosh \left (x\right )^{4} +{\left (4 \, \cosh \left (x\right ) - 1\right )} \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 3 \, \cosh \left (x\right )^{3} +{\left (6 \, \cosh \left (x\right )^{2} - 9 \, \cosh \left (x\right ) + 20\right )} \sinh \left (x\right )^{2} - 3 \,{\left (12 \, x - 1\right )} \cosh \left (x\right ) + 20 \, \cosh \left (x\right )^{2} +{\left (4 \, \cosh \left (x\right )^{3} - 3 \, \cosh \left (x\right )^{2} - 36 \, x + 32 \, \cosh \left (x\right ) + 39\right )} \sinh \left (x\right ) - 36 \, x - 69}{24 \,{\left (a \cosh \left (x\right ) + a \sinh \left (x\right ) + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+a*cosh(x)),x, algorithm="fricas")

[Out]

1/24*(cosh(x)^4 + (4*cosh(x) - 1)*sinh(x)^3 + sinh(x)^4 - 3*cosh(x)^3 + (6*cosh(x)^2 - 9*cosh(x) + 20)*sinh(x)
^2 - 3*(12*x - 1)*cosh(x) + 20*cosh(x)^2 + (4*cosh(x)^3 - 3*cosh(x)^2 - 36*x + 32*cosh(x) + 39)*sinh(x) - 36*x
- 69)/(a*cosh(x) + a*sinh(x) + a)

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Sympy [B]  time = 3.19746, size = 337, normalized size = 6.24 \begin{align*} - \frac{9 x \tanh ^{6}{\left (\frac{x}{2} \right )}}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} + \frac{27 x \tanh ^{4}{\left (\frac{x}{2} \right )}}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} - \frac{27 x \tanh ^{2}{\left (\frac{x}{2} \right )}}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} + \frac{9 x}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} + \frac{6 \tanh ^{7}{\left (\frac{x}{2} \right )}}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} - \frac{48 \tanh ^{5}{\left (\frac{x}{2} \right )}}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} + \frac{50 \tanh ^{3}{\left (\frac{x}{2} \right )}}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} - \frac{24 \tanh{\left (\frac{x}{2} \right )}}{6 a \tanh ^{6}{\left (\frac{x}{2} \right )} - 18 a \tanh ^{4}{\left (\frac{x}{2} \right )} + 18 a \tanh ^{2}{\left (\frac{x}{2} \right )} - 6 a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**4/(a+a*cosh(x)),x)

[Out]

-9*x*tanh(x/2)**6/(6*a*tanh(x/2)**6 - 18*a*tanh(x/2)**4 + 18*a*tanh(x/2)**2 - 6*a) + 27*x*tanh(x/2)**4/(6*a*ta
nh(x/2)**6 - 18*a*tanh(x/2)**4 + 18*a*tanh(x/2)**2 - 6*a) - 27*x*tanh(x/2)**2/(6*a*tanh(x/2)**6 - 18*a*tanh(x/
2)**4 + 18*a*tanh(x/2)**2 - 6*a) + 9*x/(6*a*tanh(x/2)**6 - 18*a*tanh(x/2)**4 + 18*a*tanh(x/2)**2 - 6*a) + 6*ta
nh(x/2)**7/(6*a*tanh(x/2)**6 - 18*a*tanh(x/2)**4 + 18*a*tanh(x/2)**2 - 6*a) - 48*tanh(x/2)**5/(6*a*tanh(x/2)**
6 - 18*a*tanh(x/2)**4 + 18*a*tanh(x/2)**2 - 6*a) + 50*tanh(x/2)**3/(6*a*tanh(x/2)**6 - 18*a*tanh(x/2)**4 + 18*
a*tanh(x/2)**2 - 6*a) - 24*tanh(x/2)/(6*a*tanh(x/2)**6 - 18*a*tanh(x/2)**4 + 18*a*tanh(x/2)**2 - 6*a)

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Giac [A]  time = 1.12639, size = 95, normalized size = 1.76 \begin{align*} -\frac{3 \, x}{2 \, a} - \frac{{\left (69 \, e^{\left (3 \, x\right )} + 18 \, e^{\left (2 \, x\right )} - 2 \, e^{x} + 1\right )} e^{\left (-3 \, x\right )}}{24 \, a{\left (e^{x} + 1\right )}} + \frac{a^{2} e^{\left (3 \, x\right )} - 3 \, a^{2} e^{\left (2 \, x\right )} + 21 \, a^{2} e^{x}}{24 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^4/(a+a*cosh(x)),x, algorithm="giac")

[Out]

-3/2*x/a - 1/24*(69*e^(3*x) + 18*e^(2*x) - 2*e^x + 1)*e^(-3*x)/(a*(e^x + 1)) + 1/24*(a^2*e^(3*x) - 3*a^2*e^(2*
x) + 21*a^2*e^x)/a^3