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

Optimal. Leaf size=114 \[ \frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b}}-\frac {b \tanh (x) \text {sech}(x)}{2 a^2}-\frac {b \left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}+\frac {\left (2 a^2+3 b^2\right ) \tanh (x)}{3 a^3}+\frac {\tanh (x) \text {sech}^2(x)}{3 a} \]

[Out]

-1/2*b*(a^2+2*b^2)*arctan(sinh(x))/a^4+2*b^4*arctanh((a-b)^(1/2)*tanh(1/2*x)/(a+b)^(1/2))/a^4/(a-b)^(1/2)/(a+b
)^(1/2)+1/3*(2*a^2+3*b^2)*tanh(x)/a^3-1/2*b*sech(x)*tanh(x)/a^2+1/3*sech(x)^2*tanh(x)/a

________________________________________________________________________________________

Rubi [A]  time = 0.47, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2802, 3055, 3001, 3770, 2659, 208} \[ \frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2+3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}-\frac {b \tanh (x) \text {sech}(x)}{2 a^2}+\frac {\tanh (x) \text {sech}^2(x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sech[x]^4/(a + b*Cosh[x]),x]

[Out]

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

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 2802

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 + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*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] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\text {sech}^4(x)}{a+b \cosh (x)} \, dx &=\frac {\text {sech}^2(x) \tanh (x)}{3 a}+\frac {\int \frac {\left (-3 b+2 a \cosh (x)+2 b \cosh ^2(x)\right ) \text {sech}^3(x)}{a+b \cosh (x)} \, dx}{3 a}\\ &=-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}+\frac {\int \frac {\left (2 \left (2 a^2+3 b^2\right )+a b \cosh (x)-3 b^2 \cosh ^2(x)\right ) \text {sech}^2(x)}{a+b \cosh (x)} \, dx}{6 a^2}\\ &=\frac {\left (2 a^2+3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}+\frac {\int \frac {\left (-3 b \left (a^2+2 b^2\right )-3 a b^2 \cosh (x)\right ) \text {sech}(x)}{a+b \cosh (x)} \, dx}{6 a^3}\\ &=\frac {\left (2 a^2+3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}+\frac {b^4 \int \frac {1}{a+b \cosh (x)} \, dx}{a^4}-\frac {\left (b \left (a^2+2 b^2\right )\right ) \int \text {sech}(x) \, dx}{2 a^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}+\frac {\left (2 a^2+3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}+\frac {\left (2 b^4\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-(a-b) x^2} \, dx,x,\tanh \left (\frac {x}{2}\right )\right )}{a^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) \tan ^{-1}(\sinh (x))}{2 a^4}+\frac {2 b^4 \tanh ^{-1}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^4 \sqrt {a-b} \sqrt {a+b}}+\frac {\left (2 a^2+3 b^2\right ) \tanh (x)}{3 a^3}-\frac {b \text {sech}(x) \tanh (x)}{2 a^2}+\frac {\text {sech}^2(x) \tanh (x)}{3 a}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.43, size = 101, normalized size = 0.89 \[ \frac {-6 b \left (a^2+2 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac {x}{2}\right )\right )+a \tanh (x) \left (2 a^2 \text {sech}^2(x)+4 a^2-3 a b \text {sech}(x)+6 b^2\right )-\frac {12 b^4 \tan ^{-1}\left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}}{6 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sech[x]^4/(a + b*Cosh[x]),x]

[Out]

(-6*b*(a^2 + 2*b^2)*ArcTan[Tanh[x/2]] - (12*b^4*ArcTan[((a - b)*Tanh[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2]
+ a*(4*a^2 + 6*b^2 - 3*a*b*Sech[x] + 2*a^2*Sech[x]^2)*Tanh[x])/(6*a^4)

________________________________________________________________________________________

fricas [B]  time = 1.11, size = 2483, normalized size = 21.78 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*cosh(x)),x, algorithm="fricas")

[Out]

[-1/3*(3*(a^4*b - a^2*b^3)*cosh(x)^5 + 3*(a^4*b - a^2*b^3)*sinh(x)^5 + 4*a^5 + 2*a^3*b^2 - 6*a*b^4 + 6*(a^3*b^
2 - a*b^4)*cosh(x)^4 + 3*(2*a^3*b^2 - 2*a*b^4 + 5*(a^4*b - a^2*b^3)*cosh(x))*sinh(x)^4 + 6*(5*(a^4*b - a^2*b^3
)*cosh(x)^2 + 4*(a^3*b^2 - a*b^4)*cosh(x))*sinh(x)^3 + 12*(a^5 - a*b^4)*cosh(x)^2 + 6*(2*a^5 - 2*a*b^4 + 5*(a^
4*b - a^2*b^3)*cosh(x)^3 + 6*(a^3*b^2 - a*b^4)*cosh(x)^2)*sinh(x)^2 - 3*(b^4*cosh(x)^6 + 6*b^4*cosh(x)*sinh(x)
^5 + b^4*sinh(x)^6 + 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)^2 + 3*(5*b^4*cosh(x)^2 + b^4)*sinh(x)^4 + b^4 + 4*(5*b^4*
cosh(x)^3 + 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh(x)^4 + 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2 + 6*(b^4*cosh(x)^
5 + 2*b^4*cosh(x)^3 + b^4*cosh(x))*sinh(x))*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^4*b + a^2*b^3 - 2*b^5)*cosh(x)^6 + 6*(a^4
*b + a^2*b^3 - 2*b^5)*cosh(x)*sinh(x)^5 + (a^4*b + a^2*b^3 - 2*b^5)*sinh(x)^6 + a^4*b + a^2*b^3 - 2*b^5 + 3*(a
^4*b + a^2*b^3 - 2*b^5)*cosh(x)^4 + 3*(a^4*b + a^2*b^3 - 2*b^5 + 5*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^2)*sinh(x
)^4 + 4*(5*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^3 + 3*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x))*sinh(x)^3 + 3*(a^4*b + a
^2*b^3 - 2*b^5)*cosh(x)^2 + 3*(a^4*b + a^2*b^3 - 2*b^5 + 5*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^4 + 6*(a^4*b + a^
2*b^3 - 2*b^5)*cosh(x)^2)*sinh(x)^2 + 6*((a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^5 + 2*(a^4*b + a^2*b^3 - 2*b^5)*cos
h(x)^3 + (a^4*b + a^2*b^3 - 2*b^5)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - 3*(a^4*b - a^2*b^3)*cosh(x) -
3*(a^4*b - a^2*b^3 - 5*(a^4*b - a^2*b^3)*cosh(x)^4 - 8*(a^3*b^2 - a*b^4)*cosh(x)^3 - 8*(a^5 - a*b^4)*cosh(x))
*sinh(x))/((a^6 - a^4*b^2)*cosh(x)^6 + 6*(a^6 - a^4*b^2)*cosh(x)*sinh(x)^5 + (a^6 - a^4*b^2)*sinh(x)^6 + a^6 -
a^4*b^2 + 3*(a^6 - a^4*b^2)*cosh(x)^4 + 3*(a^6 - a^4*b^2 + 5*(a^6 - a^4*b^2)*cosh(x)^2)*sinh(x)^4 + 4*(5*(a^6
- a^4*b^2)*cosh(x)^3 + 3*(a^6 - a^4*b^2)*cosh(x))*sinh(x)^3 + 3*(a^6 - a^4*b^2)*cosh(x)^2 + 3*(a^6 - a^4*b^2
+ 5*(a^6 - a^4*b^2)*cosh(x)^4 + 6*(a^6 - a^4*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 - a^4*b^2)*cosh(x)^5 + 2*(a^6
- a^4*b^2)*cosh(x)^3 + (a^6 - a^4*b^2)*cosh(x))*sinh(x)), -1/3*(3*(a^4*b - a^2*b^3)*cosh(x)^5 + 3*(a^4*b - a^
2*b^3)*sinh(x)^5 + 4*a^5 + 2*a^3*b^2 - 6*a*b^4 + 6*(a^3*b^2 - a*b^4)*cosh(x)^4 + 3*(2*a^3*b^2 - 2*a*b^4 + 5*(a
^4*b - a^2*b^3)*cosh(x))*sinh(x)^4 + 6*(5*(a^4*b - a^2*b^3)*cosh(x)^2 + 4*(a^3*b^2 - a*b^4)*cosh(x))*sinh(x)^3
+ 12*(a^5 - a*b^4)*cosh(x)^2 + 6*(2*a^5 - 2*a*b^4 + 5*(a^4*b - a^2*b^3)*cosh(x)^3 + 6*(a^3*b^2 - a*b^4)*cosh(
x)^2)*sinh(x)^2 + 6*(b^4*cosh(x)^6 + 6*b^4*cosh(x)*sinh(x)^5 + b^4*sinh(x)^6 + 3*b^4*cosh(x)^4 + 3*b^4*cosh(x)
^2 + 3*(5*b^4*cosh(x)^2 + b^4)*sinh(x)^4 + b^4 + 4*(5*b^4*cosh(x)^3 + 3*b^4*cosh(x))*sinh(x)^3 + 3*(5*b^4*cosh
(x)^4 + 6*b^4*cosh(x)^2 + b^4)*sinh(x)^2 + 6*(b^4*cosh(x)^5 + 2*b^4*cosh(x)^3 + b^4*cosh(x))*sinh(x))*sqrt(-a^
2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cosh(x) + b*sinh(x) + a)/(a^2 - b^2)) + 3*((a^4*b + a^2*b^3 - 2*b^5)*cosh
(x)^6 + 6*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)*sinh(x)^5 + (a^4*b + a^2*b^3 - 2*b^5)*sinh(x)^6 + a^4*b + a^2*b^3
- 2*b^5 + 3*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^4 + 3*(a^4*b + a^2*b^3 - 2*b^5 + 5*(a^4*b + a^2*b^3 - 2*b^5)*cos
h(x)^2)*sinh(x)^4 + 4*(5*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^3 + 3*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x))*sinh(x)^3
+ 3*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^2 + 3*(a^4*b + a^2*b^3 - 2*b^5 + 5*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^4 +
6*(a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^2)*sinh(x)^2 + 6*((a^4*b + a^2*b^3 - 2*b^5)*cosh(x)^5 + 2*(a^4*b + a^2*b^
3 - 2*b^5)*cosh(x)^3 + (a^4*b + a^2*b^3 - 2*b^5)*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - 3*(a^4*b - a^2*
b^3)*cosh(x) - 3*(a^4*b - a^2*b^3 - 5*(a^4*b - a^2*b^3)*cosh(x)^4 - 8*(a^3*b^2 - a*b^4)*cosh(x)^3 - 8*(a^5 - a
*b^4)*cosh(x))*sinh(x))/((a^6 - a^4*b^2)*cosh(x)^6 + 6*(a^6 - a^4*b^2)*cosh(x)*sinh(x)^5 + (a^6 - a^4*b^2)*sin
h(x)^6 + a^6 - a^4*b^2 + 3*(a^6 - a^4*b^2)*cosh(x)^4 + 3*(a^6 - a^4*b^2 + 5*(a^6 - a^4*b^2)*cosh(x)^2)*sinh(x)
^4 + 4*(5*(a^6 - a^4*b^2)*cosh(x)^3 + 3*(a^6 - a^4*b^2)*cosh(x))*sinh(x)^3 + 3*(a^6 - a^4*b^2)*cosh(x)^2 + 3*(
a^6 - a^4*b^2 + 5*(a^6 - a^4*b^2)*cosh(x)^4 + 6*(a^6 - a^4*b^2)*cosh(x)^2)*sinh(x)^2 + 6*((a^6 - a^4*b^2)*cosh
(x)^5 + 2*(a^6 - a^4*b^2)*cosh(x)^3 + (a^6 - a^4*b^2)*cosh(x))*sinh(x))]

________________________________________________________________________________________

giac [A]  time = 0.12, size = 123, normalized size = 1.08 \[ \frac {2 \, b^{4} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{4}} - \frac {{\left (a^{2} b + 2 \, b^{3}\right )} \arctan \left (e^{x}\right )}{a^{4}} - \frac {3 \, a b e^{\left (5 \, x\right )} + 6 \, b^{2} e^{\left (4 \, x\right )} + 12 \, a^{2} e^{\left (2 \, x\right )} + 12 \, b^{2} e^{\left (2 \, x\right )} - 3 \, a b e^{x} + 4 \, a^{2} + 6 \, b^{2}}{3 \, a^{3} {\left (e^{\left (2 \, x\right )} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^4/(a+b*cosh(x)),x, algorithm="giac")

[Out]

2*b^4*arctan((b*e^x + a)/sqrt(-a^2 + b^2))/(sqrt(-a^2 + b^2)*a^4) - (a^2*b + 2*b^3)*arctan(e^x)/a^4 - 1/3*(3*a
*b*e^(5*x) + 6*b^2*e^(4*x) + 12*a^2*e^(2*x) + 12*b^2*e^(2*x) - 3*a*b*e^x + 4*a^2 + 6*b^2)/(a^3*(e^(2*x) + 1)^3
)

________________________________________________________________________________________

maple [B]  time = 0.11, size = 239, normalized size = 2.10 \[ \frac {2 b^{4} \arctanh \left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{4} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {2 \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {\left (\tanh ^{5}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2 \left (\tanh ^{5}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{3 a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {4 \left (\tanh ^{3}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2 \tanh \left (\frac {x}{2}\right )}{a \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {2 \tanh \left (\frac {x}{2}\right ) b^{2}}{a^{3} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {\tanh \left (\frac {x}{2}\right ) b}{a^{2} \left (\tanh ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {b \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {2 \arctan \left (\tanh \left (\frac {x}{2}\right )\right ) b^{3}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^4/a^4/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tanh(1/2*x)/((a+b)*(a-b))^(1/2))+2/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*
x)^5+1/a^2/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5*b+2/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^5*b^2+4/3/a/(tanh(1/2*x)^
2+1)^3*tanh(1/2*x)^3+4/a^3/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)^3*b^2+2/a/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)+2/a^3/(ta
nh(1/2*x)^2+1)^3*tanh(1/2*x)*b^2-1/a^2/(tanh(1/2*x)^2+1)^3*tanh(1/2*x)*b-1/a^2*b*arctan(tanh(1/2*x))-2/a^4*arc
tan(tanh(1/2*x))*b^3

________________________________________________________________________________________

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(sech(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?

________________________________________________________________________________________

mupad [B]  time = 4.53, size = 547, normalized size = 4.80 \[ \frac {8}{3\,\left (a+3\,a\,{\mathrm {e}}^{2\,x}+3\,a\,{\mathrm {e}}^{4\,x}+a\,{\mathrm {e}}^{6\,x}\right )}-\frac {4}{a+2\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{4\,x}}-\frac {2\,b^2}{a^3\,{\mathrm {e}}^{2\,x}+a^3}+\frac {b^3\,\left (\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{a^4}-\frac {b\,{\mathrm {e}}^x}{a^2\,{\mathrm {e}}^{2\,x}+a^2}+\frac {2\,b\,{\mathrm {e}}^x}{2\,a^2\,{\mathrm {e}}^{2\,x}+a^2\,{\mathrm {e}}^{4\,x}+a^2}+\frac {b\,\left (\ln \left ({\mathrm {e}}^x-\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{2\,a^2}+\frac {b^4\,\ln \left (32\,a^3\,b^4-24\,b^6\,\sqrt {a^2-b^2}-48\,a\,b^6+16\,a^5\,b^2+24\,b^7\,{\mathrm {e}}^x+32\,a^6\,b\,{\mathrm {e}}^x+40\,a^2\,b^4\,\sqrt {a^2-b^2}+16\,a^4\,b^2\,\sqrt {a^2-b^2}-112\,a^2\,b^5\,{\mathrm {e}}^x+56\,a^4\,b^3\,{\mathrm {e}}^x+72\,a^3\,b^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-72\,a\,b^5\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}+32\,a^5\,b\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^4\,\sqrt {a^2-b^2}}-\frac {b^4\,\ln \left (24\,b^6\,\sqrt {a^2-b^2}-48\,a\,b^6+32\,a^3\,b^4+16\,a^5\,b^2+24\,b^7\,{\mathrm {e}}^x+32\,a^6\,b\,{\mathrm {e}}^x-40\,a^2\,b^4\,\sqrt {a^2-b^2}-16\,a^4\,b^2\,\sqrt {a^2-b^2}-112\,a^2\,b^5\,{\mathrm {e}}^x+56\,a^4\,b^3\,{\mathrm {e}}^x-72\,a^3\,b^3\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}+72\,a\,b^5\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-32\,a^5\,b\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^4\,\sqrt {a^2-b^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cosh(x)^4*(a + b*cosh(x))),x)

[Out]

8/(3*(a + 3*a*exp(2*x) + 3*a*exp(4*x) + a*exp(6*x))) - 4/(a + 2*a*exp(2*x) + a*exp(4*x)) - (2*b^2)/(a^3*exp(2*
x) + a^3) + (b^3*(log(exp(x) - 1i)*1i - log(exp(x) + 1i)*1i))/a^4 - (b*exp(x))/(a^2*exp(2*x) + a^2) + (2*b*exp
(x))/(2*a^2*exp(2*x) + a^2*exp(4*x) + a^2) + (b*(log(exp(x) - 1i)*1i - log(exp(x) + 1i)*1i))/(2*a^2) + (b^4*lo
g(32*a^3*b^4 - 24*b^6*(a^2 - b^2)^(1/2) - 48*a*b^6 + 16*a^5*b^2 + 24*b^7*exp(x) + 32*a^6*b*exp(x) + 40*a^2*b^4
*(a^2 - b^2)^(1/2) + 16*a^4*b^2*(a^2 - b^2)^(1/2) - 112*a^2*b^5*exp(x) + 56*a^4*b^3*exp(x) + 72*a^3*b^3*exp(x)
*(a^2 - b^2)^(1/2) - 72*a*b^5*exp(x)*(a^2 - b^2)^(1/2) + 32*a^5*b*exp(x)*(a^2 - b^2)^(1/2)))/(a^4*(a^2 - b^2)^
(1/2)) - (b^4*log(24*b^6*(a^2 - b^2)^(1/2) - 48*a*b^6 + 32*a^3*b^4 + 16*a^5*b^2 + 24*b^7*exp(x) + 32*a^6*b*exp
(x) - 40*a^2*b^4*(a^2 - b^2)^(1/2) - 16*a^4*b^2*(a^2 - b^2)^(1/2) - 112*a^2*b^5*exp(x) + 56*a^4*b^3*exp(x) - 7
2*a^3*b^3*exp(x)*(a^2 - b^2)^(1/2) + 72*a*b^5*exp(x)*(a^2 - b^2)^(1/2) - 32*a^5*b*exp(x)*(a^2 - b^2)^(1/2)))/(
a^4*(a^2 - b^2)^(1/2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {sech}^{4}{\relax (x )}}{a + b \cosh {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________