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

Optimal. Leaf size=112 \[ \frac {2 a^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4 \sqrt {a-b} \sqrt {a+b}}-\frac {a x \left (2 a^2+b^2\right )}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}-\frac {a \sinh (x) \cosh (x)}{2 b^2}+\frac {\sinh (x) \cosh ^2(x)}{3 b} \]

[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)

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Rubi [A]  time = 0.31, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2793, 3049, 3023, 2735, 2659, 208} \[ -\frac {a x \left (2 a^2+b^2\right )}{2 b^4}+\frac {\left (3 a^2+2 b^2\right ) \sinh (x)}{3 b^3}+\frac {2 a^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{b^4 \sqrt {a-b} \sqrt {a+b}}-\frac {a \sinh (x) \cosh (x)}{2 b^2}+\frac {\sinh (x) \cosh ^2(x)}{3 b} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[x]^4/(a + b*Cosh[x]),x]

[Out]

-(a*(2*a^2 + b^2)*x)/(2*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 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]

Rule 3049

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*} \int \frac {\cosh ^4(x)}{a+b \cosh (x)} \, dx &=\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 ) \operatorname {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 \tanh ^{-1}\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*}

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Mathematica [A]  time = 0.20, size = 99, normalized size = 0.88 \[ \frac {-6 a x \left (2 a^2+b^2\right )+3 b \left (4 a^2+3 b^2\right ) \sinh (x)-\frac {24 a^4 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}-3 a b^2 \sinh (2 x)+b^3 \sinh (3 x)}{12 b^4} \]

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 0.80, size = 1625, normalized size = 14.51 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.12, size = 133, normalized size = 1.19 \[ \frac {2 \, a^{4} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} b^{4}} + \frac {b^{2} e^{\left (3 \, x\right )} - 3 \, a b e^{\left (2 \, x\right )} + 12 \, a^{2} e^{x} + 9 \, b^{2} e^{x}}{24 \, b^{3}} - \frac {{\left (2 \, a^{3} + a b^{2}\right )} x}{2 \, b^{4}} + \frac {{\left (3 \, a b^{2} e^{x} - b^{3} - 3 \, {\left (4 \, a^{2} b + 3 \, b^{3}\right )} e^{\left (2 \, x\right )}\right )} e^{\left (-3 \, x\right )}}{24 \, b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [B]  time = 0.07, size = 264, normalized size = 2.36 \[ -\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{3}}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {a^{2}}{b^{3} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )-1\right )}-\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )-1\right )}+\frac {a^{3} \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{b^{4}}+\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )-1\right )}{2 b^{2}}-\frac {1}{3 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{3}}+\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}+\frac {1}{2 b \left (\tanh \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {a^{2}}{b^{3} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a}{2 b^{2} \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {1}{b \left (\tanh \left (\frac {x}{2}\right )+1\right )}-\frac {a^{3} \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{b^{4}}-\frac {a \ln \left (\tanh \left (\frac {x}{2}\right )+1\right )}{2 b^{2}}+\frac {2 a^{4} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(x)^4/(a+b*cosh(x)),x)

[Out]

-1/3/b/(tanh(1/2*x)-1)^3-1/2/b^2/(tanh(1/2*x)-1)^2*a-1/2/b/(tanh(1/2*x)-1)^2-1/b^3/(tanh(1/2*x)-1)*a^2-1/2/b^2
/(tanh(1/2*x)-1)*a-1/b/(tanh(1/2*x)-1)+a^3/b^4*ln(tanh(1/2*x)-1)+1/2*a/b^2*ln(tanh(1/2*x)-1)-1/3/b/(tanh(1/2*x
)+1)^3+1/2/b^2/(tanh(1/2*x)+1)^2*a+1/2/b/(tanh(1/2*x)+1)^2-1/b^3/(tanh(1/2*x)+1)*a^2-1/2/b^2/(tanh(1/2*x)+1)*a
-1/b/(tanh(1/2*x)+1)-a^3/b^4*ln(tanh(1/2*x)+1)-1/2*a/b^2*ln(tanh(1/2*x)+1)+2*a^4/b^4/((a+b)*(a-b))^(1/2)*arcta
nh((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)^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 details)Is 4*a^2-4*b^2 positive or negative?

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mupad [B]  time = 1.26, size = 209, normalized size = 1.87 \[ \frac {{\mathrm {e}}^{3\,x}}{24\,b}-\frac {{\mathrm {e}}^{-3\,x}}{24\,b}-\frac {x\,\left (2\,a^3+a\,b^2\right )}{2\,b^4}+\frac {{\mathrm {e}}^x\,\left (4\,a^2+3\,b^2\right )}{8\,b^3}+\frac {a\,{\mathrm {e}}^{-2\,x}}{8\,b^2}-\frac {a\,{\mathrm {e}}^{2\,x}}{8\,b^2}-\frac {{\mathrm {e}}^{-x}\,\left (4\,a^2+3\,b^2\right )}{8\,b^3}+\frac {a^4\,\ln \left (-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}-\frac {2\,a^4\,\left (b+a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a+b}\,\sqrt {a-b}}\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {a^4\,\ln \left (\frac {2\,a^4\,\left (b+a\,{\mathrm {e}}^x\right )}{b^5\,\sqrt {a+b}\,\sqrt {a-b}}-\frac {2\,a^4\,{\mathrm {e}}^x}{b^5}\right )}{b^4\,\sqrt {a+b}\,\sqrt {a-b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

[Out]

Timed out

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