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

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

[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)

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Rubi [A]  time = 0.168612, antiderivative size = 85, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [A]  time = 0.126834, size = 78, normalized size = 0.92 \[ \frac{\frac{8 a^3 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+4 a^2 x-4 a b \sinh (x)+2 b^2 x+b^2 \sinh (2 x)}{4 b^3} \]

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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Maple [B]  time = 0.023, size = 174, normalized size = 2.1 \begin{align*} -2\,{\frac{{a}^{3}}{{b}^{3}\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tanh \left ( x/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-{\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 ) ^{-1}}+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}+{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\frac{1}{2\,b}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) +1 \right ) }+{\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 ) ^{-1}}+{\frac{1}{2\,b} \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}-{\frac{{a}^{2}}{{b}^{3}}\ln \left ( \tanh \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{1}{2\,b}\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)^3/(a+b*cosh(x)),x)

[Out]

-2*a^3/b^3/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-1/2/b/(tanh(1/2*x)+1)^2+1/b^2/(t
anh(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)+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)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [B]  time = 2.39567, size = 2083, normalized size = 24.51 \begin{align*} \text{result too large to display} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.23676, size = 124, normalized size = 1.46 \begin{align*} -\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}} \end{align*}

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