\(\int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx\) [60]

Optimal result

Integrand size = 13, antiderivative size = 87 \[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=\frac {\left (a^2+2 b^2\right ) \arctan (\sinh (x))}{2 a^3}-\frac {2 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b}}-\frac {b \tanh (x)}{a^2}+\frac {\text {sech}(x) \tanh (x)}{2 a} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.94 \[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=\frac {2 \left (a^2+2 b^2\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )+\frac {4 b^3 \arctan \left (\frac {(a-b) \tanh \left (\frac {x}{2}\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}+a (-2 b+a \text {sech}(x)) \tanh (x)}{2 a^3} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.13, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.923, Rules used = {3042, 3281, 25, 3042, 3534, 25, 3042, 3480, 3042, 3138, 221, 4257}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin \left (\frac {\pi }{2}+i x\right )^3 \left (a+b \sin \left (\frac {\pi }{2}+i x\right )\right )}dx\)

\(\Big \downarrow \) 3281

\(\displaystyle \frac {\int -\frac {\left (-b \cosh ^2(x)-a \cosh (x)+2 b\right ) \text {sech}^2(x)}{a+b \cosh (x)}dx}{2 a}+\frac {\tanh (x) \text {sech}(x)}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\int \frac {\left (-b \cosh ^2(x)-a \cosh (x)+2 b\right ) \text {sech}^2(x)}{a+b \cosh (x)}dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\int \frac {-b \sin \left (i x+\frac {\pi }{2}\right )^2-a \sin \left (i x+\frac {\pi }{2}\right )+2 b}{\sin \left (i x+\frac {\pi }{2}\right )^2 \left (a+b \sin \left (i x+\frac {\pi }{2}\right )\right )}dx}{2 a}\)

\(\Big \downarrow \) 3534

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {\int -\frac {\left (a^2+b \cosh (x) a+2 b^2\right ) \text {sech}(x)}{a+b \cosh (x)}dx}{a}+\frac {2 b \tanh (x)}{a}}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {2 b \tanh (x)}{a}-\frac {\int \frac {\left (a^2+b \cosh (x) a+2 b^2\right ) \text {sech}(x)}{a+b \cosh (x)}dx}{a}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {2 b \tanh (x)}{a}-\frac {\int \frac {a^2+b \sin \left (i x+\frac {\pi }{2}\right ) a+2 b^2}{\sin \left (i x+\frac {\pi }{2}\right ) \left (a+b \sin \left (i x+\frac {\pi }{2}\right )\right )}dx}{a}}{2 a}\)

\(\Big \downarrow \) 3480

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {2 b \tanh (x)}{a}-\frac {\frac {\left (a^2+2 b^2\right ) \int \text {sech}(x)dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \cosh (x)}dx}{a}}{a}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {2 b \tanh (x)}{a}-\frac {\frac {\left (a^2+2 b^2\right ) \int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b^3 \int \frac {1}{a+b \sin \left (i x+\frac {\pi }{2}\right )}dx}{a}}{a}}{2 a}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {2 b \tanh (x)}{a}-\frac {-\frac {4 b^3 \int \frac {1}{-\left ((a-b) \tanh ^2\left (\frac {x}{2}\right )\right )+a+b}d\tanh \left (\frac {x}{2}\right )}{a}+\frac {\left (a^2+2 b^2\right ) \int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}}{a}}{2 a}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {2 b \tanh (x)}{a}-\frac {-\frac {4 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}+\frac {\left (a^2+2 b^2\right ) \int \csc \left (i x+\frac {\pi }{2}\right )dx}{a}}{a}}{2 a}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\tanh (x) \text {sech}(x)}{2 a}-\frac {\frac {2 b \tanh (x)}{a}-\frac {\frac {\left (a^2+2 b^2\right ) \arctan (\sinh (x))}{a}-\frac {4 b^3 \text {arctanh}\left (\frac {\sqrt {a-b} \tanh \left (\frac {x}{2}\right )}{\sqrt {a+b}}\right )}{a \sqrt {a-b} \sqrt {a+b}}}{a}}{2 a}\)

Input:

Int[Sech[x]^3/(a + b*Cosh[x]),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 221

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-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] + Simp[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)*Si 
n[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 3480

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b 
- a*B)/(b*c - a*d)   Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(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 3534

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[(-(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] + Simp[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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ 
[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) | 
| EqQ[a, 0])))
 

rule 4257

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

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.25

method	result	size
default	\(-\frac {2 b^{3} \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tanh \left (\frac {x}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{3} \sqrt {\left (a +b \right ) \left (a -b \right )}}+\frac {\frac {2 \left (\left (-\frac {1}{2} a^{2}-a b \right ) \tanh \left (\frac {x}{2}\right )^{3}+\left (\frac {1}{2} a^{2}-a b \right ) \tanh \left (\frac {x}{2}\right )\right )}{\left (\tanh \left (\frac {x}{2}\right )^{2}+1\right )^{2}}+\left (a^{2}+2 b^{2}\right ) \arctan \left (\tanh \left (\frac {x}{2}\right )\right )}{a^{3}}\)	\(109\)
risch	\(\frac {a \,{\mathrm e}^{3 x}+2 \,{\mathrm e}^{2 x} b -a \,{\mathrm e}^{x}+2 b}{\left ({\mathrm e}^{2 x}+1\right )^{2} a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}+a^{2}-b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{x}+\frac {a \sqrt {a^{2}-b^{2}}-a^{2}+b^{2}}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, a^{3}}+\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{2 a}+\frac {i \ln \left ({\mathrm e}^{x}+i\right ) b^{2}}{a^{3}}-\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{2 a}-\frac {i \ln \left ({\mathrm e}^{x}-i\right ) b^{2}}{a^{3}}\)	\(209\)

Input:

int(sech(x)^3/(a+b*cosh(x)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 651 vs. \(2 (73) = 146\).

Time = 0.17 (sec) , antiderivative size = 1370, normalized size of antiderivative = 15.75 \[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=\text {Too large to display} \] Input:

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

Output:

[(2*a^3*b - 2*a*b^3 + (a^4 - a^2*b^2)*cosh(x)^3 + (a^4 - a^2*b^2)*sinh(x)^ 
3 + 2*(a^3*b - a*b^3)*cosh(x)^2 + (2*a^3*b - 2*a*b^3 + 3*(a^4 - a^2*b^2)*c 
osh(x))*sinh(x)^2 + (b^3*cosh(x)^4 + 4*b^3*cosh(x)*sinh(x)^3 + b^3*sinh(x) 
^4 + 2*b^3*cosh(x)^2 + b^3 + 2*(3*b^3*cosh(x)^2 + b^3)*sinh(x)^2 + 4*(b^3* 
cosh(x)^3 + b^3*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)) + ((a^4 + a^2*b^2 - 2*b^4 
)*cosh(x)^4 + 4*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)*sinh(x)^3 + (a^4 + a^2*b^2 
 - 2*b^4)*sinh(x)^4 + a^4 + a^2*b^2 - 2*b^4 + 2*(a^4 + a^2*b^2 - 2*b^4)*co 
sh(x)^2 + 2*(a^4 + a^2*b^2 - 2*b^4 + 3*(a^4 + a^2*b^2 - 2*b^4)*cosh(x)^2)* 
sinh(x)^2 + 4*((a^4 + a^2*b^2 - 2*b^4)*cosh(x)^3 + (a^4 + a^2*b^2 - 2*b^4) 
*cosh(x))*sinh(x))*arctan(cosh(x) + sinh(x)) - (a^4 - a^2*b^2)*cosh(x) - ( 
a^4 - a^2*b^2 - 3*(a^4 - a^2*b^2)*cosh(x)^2 - 4*(a^3*b - a*b^3)*cosh(x))*s 
inh(x))/(a^5 - a^3*b^2 + (a^5 - a^3*b^2)*cosh(x)^4 + 4*(a^5 - a^3*b^2)*cos 
h(x)*sinh(x)^3 + (a^5 - a^3*b^2)*sinh(x)^4 + 2*(a^5 - a^3*b^2)*cosh(x)^2 + 
 2*(a^5 - a^3*b^2 + 3*(a^5 - a^3*b^2)*cosh(x)^2)*sinh(x)^2 + 4*((a^5 - a^3 
*b^2)*cosh(x)^3 + (a^5 - a^3*b^2)*cosh(x))*sinh(x)), (2*a^3*b - 2*a*b^3 + 
(a^4 - a^2*b^2)*cosh(x)^3 + (a^4 - a^2*b^2)*sinh(x)^3 + 2*(a^3*b - a*b^3)* 
cosh(x)^2 + (2*a^3*b - 2*a*b^3 + 3*(a^4 - a^2*b^2)*cosh(x))*sinh(x)^2 +...
 

Sympy [F]

\[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=\int \frac {\operatorname {sech}^{3}{\left (x \right )}}{a + b \cosh {\left (x \right )}}\, dx \] Input:

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=-\frac {2 \, b^{3} \arctan \left (\frac {b e^{x} + a}{\sqrt {-a^{2} + b^{2}}}\right )}{\sqrt {-a^{2} + b^{2}} a^{3}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} \arctan \left (e^{x}\right )}{a^{3}} + \frac {a e^{\left (3 \, x\right )} + 2 \, b e^{\left (2 \, x\right )} - a e^{x} + 2 \, b}{a^{2} {\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.41 (sec) , antiderivative size = 476, normalized size of antiderivative = 5.47 \[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=\frac {{\mathrm {e}}^x}{a+a\,{\mathrm {e}}^{2\,x}}-\frac {2\,{\mathrm {e}}^x}{a+2\,a\,{\mathrm {e}}^{2\,x}+a\,{\mathrm {e}}^{4\,x}}+\frac {2\,b}{a^2\,{\mathrm {e}}^{2\,x}+a^2}-\frac {\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a}-\frac {b^2\,\left (\ln \left (1+{\mathrm {e}}^x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left ({\mathrm {e}}^x+1{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{a^3}-\frac {b^3\,\ln \left (16\,a^5\,b-48\,a\,b^5-24\,b^5\,\sqrt {a^2-b^2}+32\,a^3\,b^3+32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x+16\,a^4\,b\,\sqrt {a^2-b^2}+40\,a^2\,b^3\,\sqrt {a^2-b^2}+32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x+72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^3\,\sqrt {a^2-b^2}}+\frac {b^3\,\ln \left (16\,a^5\,b-48\,a\,b^5+24\,b^5\,\sqrt {a^2-b^2}+32\,a^3\,b^3+32\,a^6\,{\mathrm {e}}^x+24\,b^6\,{\mathrm {e}}^x-16\,a^4\,b\,\sqrt {a^2-b^2}-40\,a^2\,b^3\,\sqrt {a^2-b^2}-32\,a^5\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}-112\,a^2\,b^4\,{\mathrm {e}}^x+56\,a^4\,b^2\,{\mathrm {e}}^x-72\,a^3\,b^2\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}+72\,a\,b^4\,{\mathrm {e}}^x\,\sqrt {a^2-b^2}\right )}{a^3\,\sqrt {a^2-b^2}} \] Input:

int(1/(cosh(x)^3*(a + b*cosh(x))),x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 361, normalized size of antiderivative = 4.15 \[ \int \frac {\text {sech}^3(x)}{a+b \cosh (x)} \, dx=\frac {e^{4 x} \mathit {atan} \left (e^{x}\right ) a^{4}+e^{4 x} \mathit {atan} \left (e^{x}\right ) a^{2} b^{2}-2 e^{4 x} \mathit {atan} \left (e^{x}\right ) b^{4}+2 e^{2 x} \mathit {atan} \left (e^{x}\right ) a^{4}+2 e^{2 x} \mathit {atan} \left (e^{x}\right ) a^{2} b^{2}-4 e^{2 x} \mathit {atan} \left (e^{x}\right ) b^{4}+\mathit {atan} \left (e^{x}\right ) a^{4}+\mathit {atan} \left (e^{x}\right ) a^{2} b^{2}-2 \mathit {atan} \left (e^{x}\right ) b^{4}+2 e^{4 x} \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b^{3}+4 e^{2 x} \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b^{3}+2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{x} b +a}{\sqrt {-a^{2}+b^{2}}}\right ) b^{3}-e^{4 x} a^{3} b +e^{4 x} a \,b^{3}+e^{3 x} a^{4}-e^{3 x} a^{2} b^{2}-e^{x} a^{4}+e^{x} a^{2} b^{2}+a^{3} b -a \,b^{3}}{a^{3} \left (e^{4 x} a^{2}-e^{4 x} b^{2}+2 e^{2 x} a^{2}-2 e^{2 x} b^{2}+a^{2}-b^{2}\right )} \] Input:

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

Output:

(e**(4*x)*atan(e**x)*a**4 + e**(4*x)*atan(e**x)*a**2*b**2 - 2*e**(4*x)*ata 
n(e**x)*b**4 + 2*e**(2*x)*atan(e**x)*a**4 + 2*e**(2*x)*atan(e**x)*a**2*b** 
2 - 4*e**(2*x)*atan(e**x)*b**4 + atan(e**x)*a**4 + atan(e**x)*a**2*b**2 - 
2*atan(e**x)*b**4 + 2*e**(4*x)*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqrt 
( - a**2 + b**2))*b**3 + 4*e**(2*x)*sqrt( - a**2 + b**2)*atan((e**x*b + a) 
/sqrt( - a**2 + b**2))*b**3 + 2*sqrt( - a**2 + b**2)*atan((e**x*b + a)/sqr 
t( - a**2 + b**2))*b**3 - e**(4*x)*a**3*b + e**(4*x)*a*b**3 + e**(3*x)*a** 
4 - e**(3*x)*a**2*b**2 - e**x*a**4 + e**x*a**2*b**2 + a**3*b - a*b**3)/(a* 
*3*(e**(4*x)*a**2 - e**(4*x)*b**2 + 2*e**(2*x)*a**2 - 2*e**(2*x)*b**2 + a* 
*2 - b**2))