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

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

[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
)

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Rubi [A]  time = 0.471966, antiderivative size = 114, normalized size of antiderivative = 1., 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 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 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 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 3770

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

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{\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 = 0.381951, size = 101, normalized size = 0.89 \[ \frac{-\frac{12 b^4 \tan ^{-1}\left (\frac{(a-b) \tanh \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-6 b \left (a^2+2 b^2\right ) \tan ^{-1}\left (\tanh \left (\frac{x}{2}\right )\right )+a \tanh (x) \left (2 a^2 \text{sech}^2(x)+4 a^2-3 a b \text{sech}(x)+6 b^2\right )}{6 a^4} \]

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)

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Maple [B]  time = 0.029, size = 239, normalized size = 2.1 \begin{align*} 2\,{\frac{{b}^{4}}{{a}^{4}\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 ) }+2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{5}}{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{b}{{a}^{2}} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}+2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{5}{b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+{\frac{4}{3\,a} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}+4\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}{b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{\tanh \left ( x/2 \right ) }{a \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}+2\,{\frac{\tanh \left ( x/2 \right ){b}^{2}}{{a}^{3} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{3}}}-{\frac{b}{{a}^{2}}\tanh \left ({\frac{x}{2}} \right ) \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-3}}-{\frac{b}{{a}^{2}}\arctan \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) }-2\,{\frac{\arctan \left ( \tanh \left ( x/2 \right ) \right ){b}^{3}}{{a}^{4}}} \end{align*}

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

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.90383, size = 5894, normalized size = 51.7 \begin{align*} \text{result too large to display} \end{align*}

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))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{sech}^{4}{\left (x \right )}}{a + b \cosh{\left (x \right )}}\, dx \end{align*}

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)

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Giac [A]  time = 1.27504, size = 166, normalized size = 1.46 \begin{align*} \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}} \end{align*}

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
)