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

Optimal. Leaf size=135 \[ \frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a \sinh (x)}{a^2-b^2}+\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a^2 b \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a^3 b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

[Out]

a^3*b*arctanh((a*cosh(x)+b*sinh(x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)-a^2*b*cosh(x)/(a^2-b^2)^2-1/3*b*cosh(x)^3
/(a^2-b^2)+a*b^2*sinh(x)/(a^2-b^2)^2+a*sinh(x)/(a^2-b^2)+1/3*a*sinh(x)^3/(a^2-b^2)

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Rubi [A]  time = 0.25, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.692, Rules used = {3518, 3109, 2633, 2565, 30, 3100, 2637, 3074, 206} \[ \frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a \sinh (x)}{a^2-b^2}+\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {a^2 b \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {a^3 b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^3/(a + b*Coth[x]),x]

[Out]

(a^3*b*ArcTanh[(a*Cosh[x] + b*Sinh[x])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) - (a^2*b*Cosh[x])/(a^2 - b^2)^2 - (
b*Cosh[x]^3)/(3*(a^2 - b^2)) + (a*b^2*Sinh[x])/(a^2 - b^2)^2 + (a*Sinh[x])/(a^2 - b^2) + (a*Sinh[x]^3)/(3*(a^2
- b^2))

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

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]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3074

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Dist[d^(-1), Subst[Int
[1/(a^2 + b^2 - x^2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b^2,
0]

Rule 3100

Int[cos[(c_.) + (d_.)*(x_)]^(m_)/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(b*Cos[c + d*x]^(m - 1))/(d*(a^2 + b^2)*(m - 1)), x] + (Dist[a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1), x]
, x] + Dist[b^2/(a^2 + b^2), Int[Cos[c + d*x]^(m - 2)/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a,
b, c, d}, x] && NeQ[a^2 + b^2, 0] && GtQ[m, 1]

Rule 3109

Int[(cos[(c_.) + (d_.)*(x_)]^(m_.)*sin[(c_.) + (d_.)*(x_)]^(n_.))/(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(
c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[b/(a^2 + b^2), Int[Cos[c + d*x]^m*Sin[c + d*x]^(n - 1), x], x] + (Dist[
a/(a^2 + b^2), Int[Cos[c + d*x]^(m - 1)*Sin[c + d*x]^n, x], x] - Dist[(a*b)/(a^2 + b^2), Int[(Cos[c + d*x]^(m
- 1)*Sin[c + d*x]^(n - 1))/(a*Cos[c + d*x] + b*Sin[c + d*x]), x], x]) /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 + b
^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3518

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[(Sin[e + f*x]
^m*(a*Cos[e + f*x] + b*Sin[e + f*x])^n)/Cos[e + f*x]^n, x] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] &&
ILtQ[n, 0] && ((LtQ[m, 5] && GtQ[n, -4]) || (EqQ[m, 5] && EqQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {\cosh ^3(x)}{a+b \coth (x)} \, dx &=-\left (i \int \frac {\cosh ^3(x) \sinh (x)}{-i b \cosh (x)-i a \sinh (x)} \, dx\right )\\ &=\frac {a \int \cosh ^3(x) \, dx}{a^2-b^2}-\frac {b \int \cosh ^2(x) \sinh (x) \, dx}{a^2-b^2}+\frac {(i a b) \int \frac {\cosh ^2(x)}{-i b \cosh (x)-i a \sinh (x)} \, dx}{a^2-b^2}\\ &=-\frac {a^2 b \cosh (x)}{\left (a^2-b^2\right )^2}+\frac {\left (i a^3 b\right ) \int \frac {1}{-i b \cosh (x)-i a \sinh (x)} \, dx}{\left (a^2-b^2\right )^2}+\frac {\left (a b^2\right ) \int \cosh (x) \, dx}{\left (a^2-b^2\right )^2}+\frac {(i a) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-i \sinh (x)\right )}{a^2-b^2}-\frac {b \operatorname {Subst}\left (\int x^2 \, dx,x,\cosh (x)\right )}{a^2-b^2}\\ &=-\frac {a^2 b \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \sinh (x)}{a^2-b^2}+\frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )}-\frac {\left (a^3 b\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-b^2-x^2} \, dx,x,-a \cosh (x)-b \sinh (x)\right )}{\left (a^2-b^2\right )^2}\\ &=\frac {a^3 b \tanh ^{-1}\left (\frac {a \cosh (x)+b \sinh (x)}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {a^2 b \cosh (x)}{\left (a^2-b^2\right )^2}-\frac {b \cosh ^3(x)}{3 \left (a^2-b^2\right )}+\frac {a b^2 \sinh (x)}{\left (a^2-b^2\right )^2}+\frac {a \sinh (x)}{a^2-b^2}+\frac {a \sinh ^3(x)}{3 \left (a^2-b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 1.48, size = 167, normalized size = 1.24 \[ \frac {1}{12} \left (\frac {a^3 \sinh (3 x)}{(a-b)^2 (a+b)^2}-\frac {24 a^3 b \tan ^{-1}\left (\frac {a+b \tanh \left (\frac {x}{2}\right )}{\sqrt {b-a} \sqrt {a+b}}\right )}{(b-a)^{5/2} (a+b)^{5/2}}+\frac {3 a \left (3 a^2+b^2\right ) \sinh (x)}{(a-b)^2 (a+b)^2}+\frac {3 b \left (b^2-5 a^2\right ) \cosh (x)}{(a-b)^2 (a+b)^2}-\frac {a b^2 \sinh (3 x)}{(a-b)^2 (a+b)^2}+\frac {b \cosh (3 x)}{(b-a) (a+b)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-24*a^3*b*ArcTan[(a + b*Tanh[x/2])/(Sqrt[-a + b]*Sqrt[a + b])])/((-a + b)^(5/2)*(a + b)^(5/2)) + (3*b*(-5*a^
2 + b^2)*Cosh[x])/((a - b)^2*(a + b)^2) + (b*Cosh[3*x])/((-a + b)*(a + b)) + (3*a*(3*a^2 + b^2)*Sinh[x])/((a -
b)^2*(a + b)^2) + (a^3*Sinh[3*x])/((a - b)^2*(a + b)^2) - (a*b^2*Sinh[3*x])/((a - b)^2*(a + b)^2))/12

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fricas [B]  time = 0.44, size = 1873, normalized size = 13.87 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3
+ a*b^4 - b^5)*cosh(x)*sinh(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 - a^5 - a^4*b
+ 2*a^3*b^2 + 2*a^2*b^3 - a*b^4 - b^5 + 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^4 +
3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5 + 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b
^5)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 + 3*(3*a^5 - 5*a
^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 -
a*b^4 + b^5)*cosh(x)^2 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2
+ 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^4 - 6*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^2)*s
inh(x)^2 + 24*(a^3*b*cosh(x)^3 + 3*a^3*b*cosh(x)^2*sinh(x) + 3*a^3*b*cosh(x)*sinh(x)^2 + a^3*b*sinh(x)^3)*sqrt
(a^2 - b^2)*log(((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 + 2*sqrt(a^2 - b^2)*(cosh(x
) + sinh(x)) + a - b)/((a + b)*cosh(x)^2 + 2*(a + b)*cosh(x)*sinh(x) + (a + b)*sinh(x)^2 - a + b)) + 6*((a^5 -
a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^5 + 2*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 -
b^5)*cosh(x)^3 - (3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2
+ 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2 +
3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*sinh(x)^3), 1/24*((a^5 - a^4*b - 2*a^
3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^6 + 6*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)*sin
h(x)^5 + (a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*sinh(x)^6 - a^5 - a^4*b + 2*a^3*b^2 + 2*a^2*b^3 -
a*b^4 - b^5 + 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^4 + 3*(3*a^5 - 5*a^4*b - 2*a^
3*b^2 + 6*a^2*b^3 - a*b^4 - b^5 + 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^2)*sinh(x)^4 +
4*(5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(x)^3 + 3*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b
^3 - a*b^4 - b^5)*cosh(x))*sinh(x)^3 - 3*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5)*cosh(x)^2 - 3
*(3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*a^2*b^3 - a*b^4 + b^5 - 5*(a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5
)*cosh(x)^4 - 6*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^2)*sinh(x)^2 - 48*(a^3*b*cosh(
x)^3 + 3*a^3*b*cosh(x)^2*sinh(x) + 3*a^3*b*cosh(x)*sinh(x)^2 + a^3*b*sinh(x)^3)*sqrt(-a^2 + b^2)*arctan(sqrt(-
a^2 + b^2)/((a + b)*cosh(x) + (a + b)*sinh(x))) + 6*((a^5 - a^4*b - 2*a^3*b^2 + 2*a^2*b^3 + a*b^4 - b^5)*cosh(
x)^5 + 2*(3*a^5 - 5*a^4*b - 2*a^3*b^2 + 6*a^2*b^3 - a*b^4 - b^5)*cosh(x)^3 - (3*a^5 + 5*a^4*b - 2*a^3*b^2 - 6*
a^2*b^3 - a*b^4 + b^5)*cosh(x))*sinh(x))/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^6 - 3*a^4*b^2 +
3*a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^6 - 3*a^4*b
^2 + 3*a^2*b^4 - b^6)*sinh(x)^3)]

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giac [A]  time = 0.14, size = 164, normalized size = 1.21 \[ \frac {2 \, a^{3} b \arctan \left (-\frac {a e^{x} + b e^{x}}{\sqrt {-a^{2} + b^{2}}}\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (9 \, a e^{\left (2 \, x\right )} - 3 \, b e^{\left (2 \, x\right )} + a - b\right )} e^{\left (-3 \, x\right )}}{24 \, {\left (a^{2} - 2 \, a b + b^{2}\right )}} + \frac {a^{2} e^{\left (3 \, x\right )} + 2 \, a b e^{\left (3 \, x\right )} + b^{2} e^{\left (3 \, x\right )} + 9 \, a^{2} e^{x} + 12 \, a b e^{x} + 3 \, b^{2} e^{x}}{24 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a^3*b*arctan(-(a*e^x + b*e^x)/sqrt(-a^2 + b^2))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) - 1/24*(9*a*e^(2*
x) - 3*b*e^(2*x) + a - b)*e^(-3*x)/(a^2 - 2*a*b + b^2) + 1/24*(a^2*e^(3*x) + 2*a*b*e^(3*x) + b^2*e^(3*x) + 9*a
^2*e^x + 12*a*b*e^x + 3*b^2*e^x)/(a^3 + 3*a^2*b + 3*a*b^2 + b^3)

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maple [A]  time = 0.13, size = 200, normalized size = 1.48 \[ -\frac {2 a^{3} b \arctan \left (\frac {2 \tanh \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2} \sqrt {-a^{2}+b^{2}}}-\frac {4}{3 \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3} \left (4 a +4 b \right )}-\frac {2}{\left (4 a +4 b \right ) \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a}{\left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {b}{2 \left (a +b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {4}{3 \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3} \left (4 a -4 b \right )}+\frac {2}{\left (4 a -4 b \right ) \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a}{\left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}+\frac {b}{2 \left (a -b \right )^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3*b/(a-b)^2/(a+b)^2/(-a^2+b^2)^(1/2)*arctan(1/2*(2*tanh(1/2*x)*b+2*a)/(-a^2+b^2)^(1/2))-4/3/(tanh(1/2*x)-
1)^3/(4*a+4*b)-2/(4*a+4*b)/(tanh(1/2*x)-1)^2-1/(a+b)^2/(tanh(1/2*x)-1)*a-1/2/(a+b)^2/(tanh(1/2*x)-1)*b-4/3/(ta
nh(1/2*x)+1)^3/(4*a-4*b)+2/(4*a-4*b)/(tanh(1/2*x)+1)^2-1/(a-b)^2/(tanh(1/2*x)+1)*a+1/2/(a-b)^2/(tanh(1/2*x)+1)
*b

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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*coth(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 = 2.04, size = 262, normalized size = 1.94 \[ \frac {{\mathrm {e}}^{3\,x}}{24\,a+24\,b}-\frac {{\mathrm {e}}^{-3\,x}}{24\,a-24\,b}+\frac {{\mathrm {e}}^x\,\left (3\,a+b\right )}{8\,{\left (a+b\right )}^2}-\frac {{\mathrm {e}}^{-x}\,\left (3\,a-b\right )}{8\,{\left (a-b\right )}^2}+\frac {2\,\mathrm {atan}\left (\frac {a^3\,b\,{\mathrm {e}}^x\,\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}}{a^5\,\sqrt {a^6\,b^2}-b^5\,\sqrt {a^6\,b^2}+2\,a^2\,b^3\,\sqrt {a^6\,b^2}-2\,a^3\,b^2\,\sqrt {a^6\,b^2}+a\,b^4\,\sqrt {a^6\,b^2}-a^4\,b\,\sqrt {a^6\,b^2}}\right )\,\sqrt {a^6\,b^2}}{\sqrt {-a^{10}+5\,a^8\,b^2-10\,a^6\,b^4+10\,a^4\,b^6-5\,a^2\,b^8+b^{10}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^3/(a + b*coth(x)),x)

[Out]

exp(3*x)/(24*a + 24*b) - exp(-3*x)/(24*a - 24*b) + (exp(x)*(3*a + b))/(8*(a + b)^2) - (exp(-x)*(3*a - b))/(8*(
a - b)^2) + (2*atan((a^3*b*exp(x)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2))/(a^5*
(a^6*b^2)^(1/2) - b^5*(a^6*b^2)^(1/2) + 2*a^2*b^3*(a^6*b^2)^(1/2) - 2*a^3*b^2*(a^6*b^2)^(1/2) + a*b^4*(a^6*b^2
)^(1/2) - a^4*b*(a^6*b^2)^(1/2)))*(a^6*b^2)^(1/2))/(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*
b^2)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cosh ^{3}{\relax (x )}}{a + b \coth {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(x)**3/(a + b*coth(x)), x)

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