3.55 \(\int \frac {\cosh ^3(x)}{a+b \cosh (x)} \, dx\)

Optimal. Leaf size=85 \[ -\frac {2 a^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^3 \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (2 a^2+b^2\right )}{2 b^3}-\frac {a \sinh (x)}{b^2}+\frac {\sinh (x) \cosh (x)}{2 b} \]

[Out]

1/2*(2*a^2+b^2)*x/b^3-a*sinh(x)/b^2+1/2*cosh(x)*sinh(x)/b-2*a^3*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b
^3/(a-b)^(1/2)/(a+b)^(1/2)

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Rubi [A]  time = 0.17, antiderivative size = 85, 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 = {2793, 3023, 2735, 2659, 208} \[ \frac {x \left (2 a^2+b^2\right )}{2 b^3}-\frac {2 a^3 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^3 \sqrt {a-b} \sqrt {a+b}}-\frac {a \sinh (x)}{b^2}+\frac {\sinh (x) \cosh (x)}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Cosh[x]),x]

[Out]

((2*a^2 + b^2)*x)/(2*b^3) - (2*a^3*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^3*Sqrt[a + b])
- (a*Sinh[x])/b^2 + (Cosh[x]*Sinh[x])/(2*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]

Rule 2793

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d
*(m + n)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e + f*x])^n*Simp[a^3*d*(m + n) + b^2*(b*c*(m - 2) + a*d
*(n + 1)) - b*(a*b*c - b^2*d*(m + n - 1) - 3*a^2*d*(m + n))*Sin[e + f*x] - b^2*(b*c*(m - 1) - a*d*(3*m + 2*n -
2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] ||
(EqQ[a, 0] && NeQ[c, 0])))

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 78, normalized size = 0.92 \[ \frac {4 a^2 x+\frac {8 a^3 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}-4 a b \sinh (x)+2 b^2 x+b^2 \sinh (2 x)}{4 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[x]^3/(a + b*Cosh[x]),x]

[Out]

(4*a^2*x + 2*b^2*x + (8*a^3*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 4*a*b*Sinh[x] + b
^2*Sinh[2*x])/(4*b^3)

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fricas [B]  time = 0.89, size = 903, normalized size = 10.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[1/8*((a^2*b^2 - b^4)*cosh(x)^4 + (a^2*b^2 - b^4)*sinh(x)^4 - a^2*b^2 + b^4 + 4*(2*a^4 - a^2*b^2 - b^4)*x*cosh
(x)^2 - 4*(a^3*b - a*b^3)*cosh(x)^3 - 4*(a^3*b - a*b^3 - (a^2*b^2 - b^4)*cosh(x))*sinh(x)^3 + 2*(3*(a^2*b^2 -
b^4)*cosh(x)^2 + 2*(2*a^4 - a^2*b^2 - b^4)*x - 6*(a^3*b - a*b^3)*cosh(x))*sinh(x)^2 + 8*(a^3*cosh(x)^2 + 2*a^3
*cosh(x)*sinh(x) + a^3*sinh(x)^2)*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)) + 4*(a^3*b - a*b^3)*cosh(x) + 4*(a^3*b - a*b^3 + (a^2*b^2 -
b^4)*cosh(x)^3 + 2*(2*a^4 - a^2*b^2 - b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cosh(x)^2)*sinh(x))/((a^2*b^3 - b^5)
*cosh(x)^2 + 2*(a^2*b^3 - b^5)*cosh(x)*sinh(x) + (a^2*b^3 - b^5)*sinh(x)^2), 1/8*((a^2*b^2 - b^4)*cosh(x)^4 +
(a^2*b^2 - b^4)*sinh(x)^4 - a^2*b^2 + b^4 + 4*(2*a^4 - a^2*b^2 - b^4)*x*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x)^
3 - 4*(a^3*b - a*b^3 - (a^2*b^2 - b^4)*cosh(x))*sinh(x)^3 + 2*(3*(a^2*b^2 - b^4)*cosh(x)^2 + 2*(2*a^4 - a^2*b^
2 - b^4)*x - 6*(a^3*b - a*b^3)*cosh(x))*sinh(x)^2 + 16*(a^3*cosh(x)^2 + 2*a^3*cosh(x)*sinh(x) + a^3*sinh(x)^2)
*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 4*(a^3*b - a*b^3)*cosh(x
) + 4*(a^3*b - a*b^3 + (a^2*b^2 - b^4)*cosh(x)^3 + 2*(2*a^4 - a^2*b^2 - b^4)*x*cosh(x) - 3*(a^3*b - a*b^3)*cos
h(x)^2)*sinh(x))/((a^2*b^3 - b^5)*cosh(x)^2 + 2*(a^2*b^3 - b^5)*cosh(x)*sinh(x) + (a^2*b^3 - b^5)*sinh(x)^2)]

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giac [A]  time = 0.12, size = 92, normalized size = 1.08 \[ -\frac {2 \, a^{3} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{3}} + \frac {b e^{\left (2 \, x\right )} - 4 \, a e^{x}}{8 \, b^{2}} + \frac {{\left (2 \, a^{2} + b^{2}\right )} x}{2 \, b^{3}} + \frac {{\left (4 \, a b e^{x} - b^{2}\right )} e^{\left (-2 \, x\right )}}{8 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)^3/(a+b*cosh(x)),x, algorithm="giac")

[Out]

-2*a^3*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*b^3) + 1/8*(b*e^(2*x) - 4*a*e^x)/b^2 + 1/2*(2*a^
2 + b^2)*x/b^3 + 1/8*(4*a*b*e^x - b^2)*e^(-2*x)/b^3

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maple [B]  time = 0.06, size = 174, normalized size = 2.05 \[ \frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}+\frac {a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right ) a^{2}}{b^{3}}-\frac {\ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {a}{b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right ) a^{2}}{b^{3}}+\frac {\ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b}-\frac {2 a^{3} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a+b*cosh(x)),x)

[Out]

1/2/b/(tanh(1/2*x)-1)^2+1/b^2/(tanh(1/2*x)-1)*a+1/2/b/(tanh(1/2*x)-1)-1/b^3*ln(tanh(1/2*x)-1)*a^2-1/2/b*ln(tan
h(1/2*x)-1)-1/2/b/(tanh(1/2*x)+1)^2+1/b^2/(tanh(1/2*x)+1)*a+1/2/b/(tanh(1/2*x)+1)+1/b^3*ln(tanh(1/2*x)+1)*a^2+
1/2/b*ln(tanh(1/2*x)+1)-2*a^3/b^3/((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)^3/(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 = 1.12, size = 167, normalized size = 1.96 \[ \frac {{\mathrm {e}}^{2\,x}}{8\,b}-\frac {{\mathrm {e}}^{-2\,x}}{8\,b}-\frac {a\,{\mathrm {e}}^x}{2\,b^2}+\frac {a\,{\mathrm {e}}^{-x}}{2\,b^2}+\frac {x\,\left (2\,a^2+b^2\right )}{2\,b^3}+\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^x}{b^4}-\frac {2\,a^3\,\left (b+a\,{\mathrm {e}}^x\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a^3\,\ln \left (\frac {2\,a^3\,{\mathrm {e}}^x}{b^4}+\frac {2\,a^3\,\left (b+a\,{\mathrm {e}}^x\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^3\,\sqrt {a+b}\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a + b*cosh(x)),x)

[Out]

exp(2*x)/(8*b) - exp(-2*x)/(8*b) - (a*exp(x))/(2*b^2) + (a*exp(-x))/(2*b^2) + (x*(2*a^2 + b^2))/(2*b^3) + (a^3
*log((2*a^3*exp(x))/b^4 - (2*a^3*(b + a*exp(x)))/(b^4*(a + b)^(1/2)*(a - b)^(1/2))))/(b^3*(a + b)^(1/2)*(a - b
)^(1/2)) - (a^3*log((2*a^3*exp(x))/b^4 + (2*a^3*(b + a*exp(x)))/(b^4*(a + b)^(1/2)*(a - b)^(1/2))))/(b^3*(a +
b)^(1/2)*(a - b)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(x)**3/(a+b*cosh(x)),x)

[Out]

Timed out

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