\(\int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx\) [93]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 12, antiderivative size = 173 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx=\frac {x}{a^3}-\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 (a-b)^{5/2} (a+b)^{5/2} d}+\frac {b^2 \tanh (c+d x)}{2 a \left (a^2-b^2\right ) d (a+b \text {sech}(c+d x))^2}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{2 a^2 \left (a^2-b^2\right )^2 d (a+b \text {sech}(c+d x))} \] Output:

x/a^3-b*(6*a^4-5*a^2*b^2+2*b^4)*arctan((a-b)^(1/2)*tanh(1/2*d*x+1/2*c)/(a+ 
b)^(1/2))/a^3/(a-b)^(5/2)/(a+b)^(5/2)/d+1/2*b^2*tanh(d*x+c)/a/(a^2-b^2)/d/ 
(a+b*sech(d*x+c))^2+1/2*b^2*(5*a^2-2*b^2)*tanh(d*x+c)/a^2/(a^2-b^2)^2/d/(a 
+b*sech(d*x+c))
 

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.18 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx=\frac {(b+a \cosh (c+d x)) \text {sech}^3(c+d x) \left (2 (c+d x) (b+a \cosh (c+d x))^2+\frac {2 b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {(-a+b) \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right ) (b+a \cosh (c+d x))^2}{\left (a^2-b^2\right )^{5/2}}+\frac {a b^3 \sinh (c+d x)}{(-a+b) (a+b)}+\frac {3 a b^2 \left (2 a^2-b^2\right ) (b+a \cosh (c+d x)) \sinh (c+d x)}{(a-b)^2 (a+b)^2}\right )}{2 a^3 d (a+b \text {sech}(c+d x))^3} \] Input:

Integrate[(a + b*Sech[c + d*x])^(-3),x]
 

Output:

((b + a*Cosh[c + d*x])*Sech[c + d*x]^3*(2*(c + d*x)*(b + a*Cosh[c + d*x])^ 
2 + (2*b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[((-a + b)*Tanh[(c + d*x)/2])/S 
qrt[a^2 - b^2]]*(b + a*Cosh[c + d*x])^2)/(a^2 - b^2)^(5/2) + (a*b^3*Sinh[c 
 + d*x])/((-a + b)*(a + b)) + (3*a*b^2*(2*a^2 - b^2)*(b + a*Cosh[c + d*x]) 
*Sinh[c + d*x])/((a - b)^2*(a + b)^2)))/(2*a^3*d*(a + b*Sech[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.08 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.25, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.083, Rules used = {3042, 4272, 25, 3042, 4548, 25, 3042, 4407, 3042, 4318, 3042, 3138, 221}

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.

\(\displaystyle \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4272

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

\(\displaystyle \frac {\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}-\frac {\int -\frac {2 \left (a^2-b^2\right )^2-a b \left (4 a^2-b^2\right ) \text {sech}(c+d x)}{a+b \text {sech}(c+d x)}dx}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {2 \left (a^2-b^2\right )^2-a b \left (4 a^2-b^2\right ) \text {sech}(c+d x)}{a+b \text {sech}(c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}+\frac {\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {\int \frac {2 \left (a^2-b^2\right )^2-a b \left (4 a^2-b^2\right ) \csc \left (i c+i d x+\frac {\pi }{2}\right )}{a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}dx}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {\frac {\frac {2 x \left (a^2-b^2\right )^2}{a}-\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \int \frac {\text {sech}(c+d x)}{a+b \text {sech}(c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}+\frac {\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {\frac {2 x \left (a^2-b^2\right )^2}{a}-\frac {b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \int \frac {\csc \left (i c+i d x+\frac {\pi }{2}\right )}{a+b \csc \left (i c+i d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 4318

\(\displaystyle \frac {\frac {\frac {2 x \left (a^2-b^2\right )^2}{a}-\frac {\left (6 a^4-5 a^2 b^2+2 b^4\right ) \int \frac {1}{\frac {a \cosh (c+d x)}{b}+1}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}}{2 a \left (a^2-b^2\right )}+\frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}+\frac {\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {\frac {2 x \left (a^2-b^2\right )^2}{a}-\frac {\left (6 a^4-5 a^2 b^2+2 b^4\right ) \int \frac {1}{\frac {a \sin \left (i c+i d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

\(\displaystyle \frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}+\frac {\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {\frac {2 x \left (a^2-b^2\right )^2}{a}+\frac {2 i \left (6 a^4-5 a^2 b^2+2 b^4\right ) \int \frac {1}{\frac {a+b}{b}-\left (1-\frac {a}{b}\right ) \tanh ^2\left (\frac {1}{2} (c+d x)\right )}d\left (i \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{a d}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {b^2 \tanh (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))^2}+\frac {\frac {b^2 \left (5 a^2-2 b^2\right ) \tanh (c+d x)}{a d \left (a^2-b^2\right ) (a+b \text {sech}(c+d x))}+\frac {\frac {2 x \left (a^2-b^2\right )^2}{a}-\frac {2 b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}}{2 a \left (a^2-b^2\right )}\)

Input:

Int[(a + b*Sech[c + d*x])^(-3),x]
 

Output:

(b^2*Tanh[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Sech[c + d*x])^2) + (((2*(a^ 
2 - b^2)^2*x)/a - (2*b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(Sqrt[a - b]*Tan 
h[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/(a*(a^2 - b^2 
)) + (b^2*(5*a^2 - 2*b^2)*Tanh[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sech[c + 
d*x])))/(2*a*(a^2 - b^2))
 

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 4272
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[b^2*Cot[ 
c + d*x]*((a + b*Csc[c + d*x])^(n + 1)/(a*d*(n + 1)*(a^2 - b^2))), x] + Sim 
p[1/(a*(n + 1)*(a^2 - b^2))   Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^2 - 
b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x 
], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integ 
erQ[2*n]
 

rule 4318
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4407
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
 (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a   Int[Csc[e + f* 
x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c 
- a*d, 0]
 

rule 4548
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 - 
a*b*B + a^2*C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 
 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2))   Int[(a + b*Csc[e + f*x])^( 
m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x 
] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, 
 b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
 
Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.45

method result size
derivativedivides \(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {2 b \left (\frac {-\frac {\left (6 a^{2}+a b -2 b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-a b -2 b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (6 a^{4}-5 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}}{d}\) \(251\)
default \(\frac {-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {2 b \left (\frac {-\frac {\left (6 a^{2}+a b -2 b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}-\frac {\left (6 a^{2}-a b -2 b^{2}\right ) a b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (6 a^{4}-5 a^{2} b^{2}+2 b^{4}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{2 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3}}}{d}\) \(251\)
risch \(\frac {x}{a^{3}}-\frac {b^{2} \left (7 a^{3} b \,{\mathrm e}^{3 d x +3 c}-4 a \,b^{3} {\mathrm e}^{3 d x +3 c}+6 a^{4} {\mathrm e}^{2 d x +2 c}+9 a^{2} b^{2} {\mathrm e}^{2 d x +2 c}-6 b^{4} {\mathrm e}^{2 d x +2 c}+17 a^{3} b \,{\mathrm e}^{d x +c}-8 b^{3} {\mathrm e}^{d x +c} a +6 a^{4}-3 a^{2} b^{2}\right )}{a^{3} \left (a^{2}-b^{2}\right )^{2} d \left (a \,{\mathrm e}^{2 d x +2 c}+2 b \,{\mathrm e}^{d x +c}+a \right )^{2}}-\frac {3 b a \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}+\frac {5 b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}-\frac {b^{5} \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}+\frac {3 b a \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d}-\frac {5 b^{3} \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d a}+\frac {b^{5} \ln \left ({\mathrm e}^{d x +c}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right )^{2} \left (a -b \right )^{2} d \,a^{3}}\) \(631\)

Input:

int(1/(a+b*sech(d*x+c))^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/a^3*ln(tanh(1/2*d*x+1/2*c)-1)+1/a^3*ln(tanh(1/2*d*x+1/2*c)+1)-2*b/ 
a^3*((-1/2*(6*a^2+a*b-2*b^2)*a*b/(a-b)/(a^2+2*a*b+b^2)*tanh(1/2*d*x+1/2*c) 
^3-1/2*(6*a^2-a*b-2*b^2)*a*b/(a+b)/(a^2-2*a*b+b^2)*tanh(1/2*d*x+1/2*c))/(t 
anh(1/2*d*x+1/2*c)^2*a-tanh(1/2*d*x+1/2*c)^2*b+a+b)^2+1/2*(6*a^4-5*a^2*b^2 
+2*b^4)/(a^4-2*a^2*b^2+b^4)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tanh(1/2*d*x+ 
1/2*c)/((a-b)*(a+b))^(1/2))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2000 vs. \(2 (160) = 320\).

Time = 0.17 (sec) , antiderivative size = 4125, normalized size of antiderivative = 23.84 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sech(d*x+c))^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx=\int \frac {1}{\left (a + b \operatorname {sech}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(1/(a+b*sech(d*x+c))**3,x)
 

Output:

Integral((a + b*sech(c + d*x))**(-3), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*sech(d*x+c))^3,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*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.51 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx=-\frac {\frac {{\left (6 \, a^{4} b - 5 \, a^{2} b^{3} + 2 \, b^{5}\right )} \arctan \left (\frac {a e^{\left (d x + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {7 \, a^{3} b^{3} e^{\left (3 \, d x + 3 \, c\right )} - 4 \, a b^{5} e^{\left (3 \, d x + 3 \, c\right )} + 6 \, a^{4} b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a^{2} b^{4} e^{\left (2 \, d x + 2 \, c\right )} - 6 \, b^{6} e^{\left (2 \, d x + 2 \, c\right )} + 17 \, a^{3} b^{3} e^{\left (d x + c\right )} - 8 \, a b^{5} e^{\left (d x + c\right )} + 6 \, a^{4} b^{2} - 3 \, a^{2} b^{4}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} {\left (a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (d x + c\right )} + a\right )}^{2}} - \frac {d x + c}{a^{3}}}{d} \] Input:

integrate(1/(a+b*sech(d*x+c))^3,x, algorithm="giac")
 

Output:

-((6*a^4*b - 5*a^2*b^3 + 2*b^5)*arctan((a*e^(d*x + c) + b)/sqrt(a^2 - b^2) 
)/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 - b^2)) + (7*a^3*b^3*e^(3*d*x + 3* 
c) - 4*a*b^5*e^(3*d*x + 3*c) + 6*a^4*b^2*e^(2*d*x + 2*c) + 9*a^2*b^4*e^(2* 
d*x + 2*c) - 6*b^6*e^(2*d*x + 2*c) + 17*a^3*b^3*e^(d*x + c) - 8*a*b^5*e^(d 
*x + c) + 6*a^4*b^2 - 3*a^2*b^4)/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a*e^(2*d*x 
+ 2*c) + 2*b*e^(d*x + c) + a)^2) - (d*x + c)/a^3)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx=\int \frac {1}{{\left (a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 1649, normalized size of antiderivative = 9.53 \[ \int \frac {1}{(a+b \text {sech}(c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*sech(d*x+c))^3,x)
 

Output:

( - 24*e**(4*c + 4*d*x)*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/sqrt(a 
**2 - b**2))*a**6*b + 20*e**(4*c + 4*d*x)*sqrt(a**2 - b**2)*atan((e**(c + 
d*x)*a + b)/sqrt(a**2 - b**2))*a**4*b**3 - 8*e**(4*c + 4*d*x)*sqrt(a**2 - 
b**2)*atan((e**(c + d*x)*a + b)/sqrt(a**2 - b**2))*a**2*b**5 - 96*e**(3*c 
+ 3*d*x)*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/sqrt(a**2 - b**2))*a* 
*5*b**2 + 80*e**(3*c + 3*d*x)*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/ 
sqrt(a**2 - b**2))*a**3*b**4 - 32*e**(3*c + 3*d*x)*sqrt(a**2 - b**2)*atan( 
(e**(c + d*x)*a + b)/sqrt(a**2 - b**2))*a*b**6 - 48*e**(2*c + 2*d*x)*sqrt( 
a**2 - b**2)*atan((e**(c + d*x)*a + b)/sqrt(a**2 - b**2))*a**6*b - 56*e**( 
2*c + 2*d*x)*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/sqrt(a**2 - b**2) 
)*a**4*b**3 + 64*e**(2*c + 2*d*x)*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + 
 b)/sqrt(a**2 - b**2))*a**2*b**5 - 32*e**(2*c + 2*d*x)*sqrt(a**2 - b**2)*a 
tan((e**(c + d*x)*a + b)/sqrt(a**2 - b**2))*b**7 - 96*e**(c + d*x)*sqrt(a* 
*2 - b**2)*atan((e**(c + d*x)*a + b)/sqrt(a**2 - b**2))*a**5*b**2 + 80*e** 
(c + d*x)*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/sqrt(a**2 - b**2))*a 
**3*b**4 - 32*e**(c + d*x)*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/sqr 
t(a**2 - b**2))*a*b**6 - 24*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/sq 
rt(a**2 - b**2))*a**6*b + 20*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b)/s 
qrt(a**2 - b**2))*a**4*b**3 - 8*sqrt(a**2 - b**2)*atan((e**(c + d*x)*a + b 
)/sqrt(a**2 - b**2))*a**2*b**5 + 4*e**(4*c + 4*d*x)*a**8*d*x - 12*e**(4...