\(\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx\) [54]
Optimal result
Integrand size = 13, antiderivative size = 112 \[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=-\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {2 a^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b}}+\frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh ^2(x) \sinh (x)}{3 b}
\]
[Out]
-1/2*a*(2*a^2+b^2)*x/b^4+1/3*(3*a^2+2*b^2)*sinh(x)/b^3-1/2*a*cosh(x)*sinh(x)/b^2+1/3*cosh(x)^2*sinh(x)/b+2*a^4
*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/b^4/(a-b)^(1/2)/(a+b)^(1/2)
Rubi [A] (verified)
Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of
steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2872, 3128, 3102, 2814, 2738,
214} \[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=\frac {2 a^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4 \sqrt {a-b} \sqrt {a+b}}-\frac {a x \left (2 a^2+b^2\right )}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}-\frac {a \sinh (x) \cosh (x)}{2 b^2}+\frac {\sinh (x) \cosh ^2(x)}{3 b}
\]
[In]
Int[Cosh[x]^4/(a + b*Cosh[x]),x]
[Out]
-1/2*(a*(2*a^2 + b^2)*x)/b^4 + (2*a^4*ArcTanh[(Sqrt[a - b]*Tanh[x/2])/Sqrt[a + b]])/(Sqrt[a - b]*b^4*Sqrt[a +
b]) + ((3*a^2 + 2*b^2)*Sinh[x])/(3*b^3) - (a*Cosh[x]*Sinh[x])/(2*b^2) + (Cosh[x]^2*Sinh[x])/(3*b)
Rule 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 2738
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 2814
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 2872
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-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 3102
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 3128
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d
^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Rubi steps \begin{align*}
\text {integral}& = \frac {\cosh ^2(x) \sinh (x)}{3 b}+\frac {\int \frac {\cosh (x) \left (2 a+2 b \cosh (x)-3 a \cosh ^2(x)\right )}{a+b \cosh (x)} \, dx}{3 b} \\ & = -\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh ^2(x) \sinh (x)}{3 b}+\frac {\int \frac {-3 a^2+a b \cosh (x)+2 \left (3 a^2+2 b^2\right ) \cosh ^2(x)}{a+b \cosh (x)} \, dx}{6 b^2} \\ & = \frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh ^2(x) \sinh (x)}{3 b}+\frac {\int \frac {-3 a^2 b-3 a \left (2 a^2+b^2\right ) \cosh (x)}{a+b \cosh (x)} \, dx}{6 b^3} \\ & = -\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh ^2(x) \sinh (x)}{3 b}+\frac {a^4 \int \frac {1}{a+b \cosh (x)} \, dx}{b^4} \\ & = -\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh ^2(x) \sinh (x)}{3 b}+\frac {\left (2 a^4\right ) \text {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{b^4} \\ & = -\frac {a \left (2 a^2+b^2\right ) x}{2 b^4}+\frac {2 a^4 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^4 \sqrt {a+b}}+\frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}-\frac {a \cosh (x) \sinh (x)}{2 b^2}+\frac {\cosh ^2(x) \sinh (x)}{3 b} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88
\[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=\frac {-6 a \left (2 a^2+b^2\right ) x-\frac {24 a^4 \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+3 b \left (4 a^2+3 b^2\right ) \sinh (x)-3 a b^2 \sinh (2 x)+b^3 \sinh (3 x)}{12 b^4}
\]
[In]
Integrate[Cosh[x]^4/(a + b*Cosh[x]),x]
[Out]
(-6*a*(2*a^2 + b^2)*x - (24*a^4*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] + 3*b*(4*a^2 +
3*b^2)*Sinh[x] - 3*a*b^2*Sinh[2*x] + b^3*Sinh[3*x])/(12*b^4)
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(202\) vs. \(2(94)=188\).
Time = 0.31 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.81
| | |
| method | result | size |
| | |
| default |
\(\frac {2 a^{4} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a +b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a \left (2 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{4}}-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}-\frac {-a -b}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 a^{2}+a b +2 b^{2}}{2 b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a \left (2 a^{2}+b^{2}\right ) \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{4}}\) |
\(203\) |
| risch |
\(-\frac {a^{3} x}{b^{4}}-\frac {a x}{2 b^{2}}+\frac {{\mathrm e}^{3 x}}{24 b}-\frac {a \,{\mathrm e}^{2 x}}{8 b^{2}}+\frac {{\mathrm e}^{x} a^{2}}{2 b^{3}}+\frac {3 \,{\mathrm e}^{x}}{8 b}-\frac {{\mathrm e}^{-x} a^{2}}{2 b^{3}}-\frac {3 \,{\mathrm e}^{-x}}{8 b}+\frac {a \,{\mathrm e}^{-2 x}}{8 b^{2}}-\frac {{\mathrm e}^{-3 x}}{24 b}+\frac {a^{4} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, b^{4}}-\frac {a^{4} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, b^{4}}\) |
\(213\) |
| | |
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[In]
int(cosh(x)^4/(a+b*cosh(x)),x,method=_RETURNVERBOSE)
[Out]
2*a^4/b^4/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))-1/3/b/(tanh(1/2*x)-1)^3-1/2*(a+b)
/b^2/(tanh(1/2*x)-1)^2-1/2*(2*a^2+a*b+2*b^2)/b^3/(tanh(1/2*x)-1)+1/2*a*(2*a^2+b^2)/b^4*ln(tanh(1/2*x)-1)-1/3/b
/(tanh(1/2*x)+1)^3-1/2*(-a-b)/b^2/(tanh(1/2*x)+1)^2-1/2*(2*a^2+a*b+2*b^2)/b^3/(tanh(1/2*x)+1)-1/2*a*(2*a^2+b^2
)/b^4*ln(tanh(1/2*x)+1)
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 778 vs. \(2 (94) = 188\).
Time = 0.29 (sec) , antiderivative size = 1625, normalized size of antiderivative = 14.51
\[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=\text {Too large to display}
\]
[In]
integrate(cosh(x)^4/(a+b*cosh(x)),x, algorithm="fricas")
[Out]
[1/24*((a^2*b^3 - b^5)*cosh(x)^6 + (a^2*b^3 - b^5)*sinh(x)^6 - 3*(a^3*b^2 - a*b^4)*cosh(x)^5 - 3*(a^3*b^2 - a*
b^4 - 2*(a^2*b^3 - b^5)*cosh(x))*sinh(x)^5 - a^2*b^3 + b^5 - 12*(2*a^5 - a^3*b^2 - a*b^4)*x*cosh(x)^3 + 3*(4*a
^4*b - a^2*b^3 - 3*b^5)*cosh(x)^4 + 3*(4*a^4*b - a^2*b^3 - 3*b^5 + 5*(a^2*b^3 - b^5)*cosh(x)^2 - 5*(a^3*b^2 -
a*b^4)*cosh(x))*sinh(x)^4 + 2*(10*(a^2*b^3 - b^5)*cosh(x)^3 - 15*(a^3*b^2 - a*b^4)*cosh(x)^2 - 6*(2*a^5 - a^3*
b^2 - a*b^4)*x + 6*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x))*sinh(x)^3 - 3*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x)^2 -
3*(4*a^4*b - a^2*b^3 - 3*b^5 - 5*(a^2*b^3 - b^5)*cosh(x)^4 + 10*(a^3*b^2 - a*b^4)*cosh(x)^3 + 12*(2*a^5 - a^3*
b^2 - a*b^4)*x*cosh(x) - 6*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x)^2)*sinh(x)^2 + 24*(a^4*cosh(x)^3 + 3*a^4*cosh(x
)^2*sinh(x) + 3*a^4*cosh(x)*sinh(x)^2 + a^4*sinh(x)^3)*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)) + 3*(a^3*b^2 - a*b^4)*cosh(x) + 3*(2*(
a^2*b^3 - b^5)*cosh(x)^5 + a^3*b^2 - a*b^4 - 5*(a^3*b^2 - a*b^4)*cosh(x)^4 - 12*(2*a^5 - a^3*b^2 - a*b^4)*x*co
sh(x)^2 + 4*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x)^3 - 2*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x))*sinh(x))/((a^2*b^4
- b^6)*cosh(x)^3 + 3*(a^2*b^4 - b^6)*cosh(x)^2*sinh(x) + 3*(a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^2*b^4 - b^6)
*sinh(x)^3), 1/24*((a^2*b^3 - b^5)*cosh(x)^6 + (a^2*b^3 - b^5)*sinh(x)^6 - 3*(a^3*b^2 - a*b^4)*cosh(x)^5 - 3*(
a^3*b^2 - a*b^4 - 2*(a^2*b^3 - b^5)*cosh(x))*sinh(x)^5 - a^2*b^3 + b^5 - 12*(2*a^5 - a^3*b^2 - a*b^4)*x*cosh(x
)^3 + 3*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x)^4 + 3*(4*a^4*b - a^2*b^3 - 3*b^5 + 5*(a^2*b^3 - b^5)*cosh(x)^2 - 5
*(a^3*b^2 - a*b^4)*cosh(x))*sinh(x)^4 + 2*(10*(a^2*b^3 - b^5)*cosh(x)^3 - 15*(a^3*b^2 - a*b^4)*cosh(x)^2 - 6*(
2*a^5 - a^3*b^2 - a*b^4)*x + 6*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x))*sinh(x)^3 - 3*(4*a^4*b - a^2*b^3 - 3*b^5)*
cosh(x)^2 - 3*(4*a^4*b - a^2*b^3 - 3*b^5 - 5*(a^2*b^3 - b^5)*cosh(x)^4 + 10*(a^3*b^2 - a*b^4)*cosh(x)^3 + 12*(
2*a^5 - a^3*b^2 - a*b^4)*x*cosh(x) - 6*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x)^2)*sinh(x)^2 - 48*(a^4*cosh(x)^3 +
3*a^4*cosh(x)^2*sinh(x) + 3*a^4*cosh(x)*sinh(x)^2 + a^4*sinh(x)^3)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(
b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 3*(a^3*b^2 - a*b^4)*cosh(x) + 3*(2*(a^2*b^3 - b^5)*cosh(x)^5 + a^3*b
^2 - a*b^4 - 5*(a^3*b^2 - a*b^4)*cosh(x)^4 - 12*(2*a^5 - a^3*b^2 - a*b^4)*x*cosh(x)^2 + 4*(4*a^4*b - a^2*b^3 -
3*b^5)*cosh(x)^3 - 2*(4*a^4*b - a^2*b^3 - 3*b^5)*cosh(x))*sinh(x))/((a^2*b^4 - b^6)*cosh(x)^3 + 3*(a^2*b^4 -
b^6)*cosh(x)^2*sinh(x) + 3*(a^2*b^4 - b^6)*cosh(x)*sinh(x)^2 + (a^2*b^4 - b^6)*sinh(x)^3)]
Sympy [F(-1)]
Timed out. \[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=\text {Timed out}
\]
[In]
integrate(cosh(x)**4/(a+b*cosh(x)),x)
[Out]
Timed out
Maxima [F(-2)]
Exception generated. \[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError}
\]
[In]
integrate(cosh(x)^4/(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 de
Giac [A] (verification not implemented)
none
Time = 0.25 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.19
\[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=\frac {2 \, a^{4} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{4}} + \frac {b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} + 9 \, b^{2} e^{x}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} + a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (3 \, a b^{2} e^{x} - b^{3} - 3 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}}
\]
[In]
integrate(cosh(x)^4/(a+b*cosh(x)),x, algorithm="giac")
[Out]
2*a^4*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*b^4) + 1/24*(b^2*e^(3*x) - 3*a*b*e^(2*x) + 12*a^2
*e^x + 9*b^2*e^x)/b^3 - 1/2*(2*a^3 + a*b^2)*x/b^4 + 1/24*(3*a*b^2*e^x - b^3 - 3*(4*a^2*b + 3*b^3)*e^(2*x))*e^(
-3*x)/b^4
Mupad [B] (verification not implemented)
Time = 2.08 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.87
\[
\int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx=\frac {{\mathrm {e}}^{3\,x}}{24\,b}-\frac {{\mathrm {e}}^{-3\,x}}{24\,b}-\frac {x\,\left (2\,a^3+a\,b^2\right )}{2\,b^4}+\frac {{\mathrm {e}}^x\,\left (4\,a^2+3\,b^2\right )}{8\,b^3}+\frac {a\,{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {a\,{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-x}\,\left (4\,a^2+3\,b^2\right )}{8\,b^3}+\frac {a^4\,\ln \left (-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}-\frac {2\,a^4\,\left (b+a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a^4\,\ln \left (\frac {2\,a^4\,\left (b+a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}}
\]
[In]
int(cosh(x)^4/(a + b*cosh(x)),x)
[Out]
exp(3*x)/(24*b) - exp(-3*x)/(24*b) - (x*(a*b^2 + 2*a^3))/(2*b^4) + (exp(x)*(4*a^2 + 3*b^2))/(8*b^3) + (a*exp(-
2*x))/(8*b^2) - (a*exp(2*x))/(8*b^2) - (exp(-x)*(4*a^2 + 3*b^2))/(8*b^3) + (a^4*log(- (2*a^4*exp(x))/b^5 - (2*
a^4*(b + a*exp(x)))/(b^5*(a + b)^(1/2)*(a - b)^(1/2))))/(b^4*(a + b)^(1/2)*(a - b)^(1/2)) - (a^4*log((2*a^4*(b
+ a*exp(x)))/(b^5*(a + b)^(1/2)*(a - b)^(1/2)) - (2*a^4*exp(x))/b^5))/(b^4*(a + b)^(1/2)*(a - b)^(1/2))