\(\int \frac {1}{(a+b \text {sech}^2(c+d x))^3} \, dx\) [164]

Optimal result

Integrand size = 14, antiderivative size = 146 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {x}{a^3}-\frac {\sqrt {b} \left (15 a^2+20 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{8 a^3 (a+b)^{5/2} d}-\frac {b \tanh (c+d x)}{4 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )^2}-\frac {b (7 a+4 b) \tanh (c+d x)}{8 a^2 (a+b)^2 d \left (a+b-b \tanh ^2(c+d x)\right )} \] Output:

x/a^3-1/8*b^(1/2)*(15*a^2+20*a*b+8*b^2)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^ 
(1/2))/a^3/(a+b)^(5/2)/d-1/4*b*tanh(d*x+c)/a/(a+b)/d/(a+b-b*tanh(d*x+c)^2) 
^2-1/8*b*(7*a+4*b)*tanh(d*x+c)/a^2/(a+b)^2/d/(a+b-b*tanh(d*x+c)^2)
                                                                                    
                                                                                    
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(301\) vs. \(2(146)=292\).

Time = 5.40 (sec) , antiderivative size = 301, normalized size of antiderivative = 2.06 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^6(c+d x) \left (8 x (a+2 b+a \cosh (2 (c+d x)))^2-\frac {b \left (15 a^2+20 a b+8 b^2\right ) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (a+2 b+a \cosh (2 (c+d x)))^2 (\cosh (2 c)-\sinh (2 c))}{(a+b)^{5/2} d \sqrt {b (\cosh (c)-\sinh (c))^4}}-\frac {4 b^2 \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{(a+b) d}+\frac {b (a+2 b+a \cosh (2 (c+d x))) \text {sech}(2 c) \left (\left (9 a^2+28 a b+16 b^2\right ) \sinh (2 c)-3 a (3 a+2 b) \sinh (2 d x)\right )}{(a+b)^2 d}\right )}{64 a^3 \left (a+b \text {sech}^2(c+d x)\right )^3} \] Input:

Integrate[(a + b*Sech[c + d*x]^2)^(-3),x]
 

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^6*(8*x*(a + 2*b + a*Cosh[2* 
(c + d*x)])^2 - (b*(15*a^2 + 20*a*b + 8*b^2)*ArcTanh[(Sech[d*x]*(Cosh[2*c] 
 - Sinh[2*c])*((a + 2*b)*Sinh[d*x] - a*Sinh[2*c + d*x]))/(2*Sqrt[a + b]*Sq 
rt[b*(Cosh[c] - Sinh[c])^4])]*(a + 2*b + a*Cosh[2*(c + d*x)])^2*(Cosh[2*c] 
 - Sinh[2*c]))/((a + b)^(5/2)*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) - (4*b^2*Se 
ch[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/((a + b)*d) + (b*(a + 2*b + 
 a*Cosh[2*(c + d*x)])*Sech[2*c]*((9*a^2 + 28*a*b + 16*b^2)*Sinh[2*c] - 3*a 
*(3*a + 2*b)*Sinh[2*d*x]))/((a + b)^2*d)))/(64*a^3*(a + b*Sech[c + d*x]^2) 
^3)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.643, Rules used = {3042, 4616, 316, 25, 402, 25, 397, 219, 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.

\(\displaystyle \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

\(\displaystyle \frac {\int \frac {1}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^3}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 316

\(\displaystyle \frac {-\frac {\int -\frac {3 b \tanh ^2(c+d x)+4 a+b}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^2}d\tanh (c+d x)}{4 a (a+b)}-\frac {b \tanh (c+d x)}{4 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {3 b \tanh ^2(c+d x)+4 a+b}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )^2}d\tanh (c+d x)}{4 a (a+b)}-\frac {b \tanh (c+d x)}{4 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {-\frac {\int -\frac {8 a^2+9 b a+4 b^2+b (7 a+4 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{2 a (a+b)}-\frac {b (7 a+4 b) \tanh (c+d x)}{2 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{4 a (a+b)}-\frac {b \tanh (c+d x)}{4 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {8 a^2+9 b a+4 b^2+b (7 a+4 b) \tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right ) \left (-b \tanh ^2(c+d x)+a+b\right )}d\tanh (c+d x)}{2 a (a+b)}-\frac {b (7 a+4 b) \tanh (c+d x)}{2 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{4 a (a+b)}-\frac {b \tanh (c+d x)}{4 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{a}-\frac {b \left (15 a^2+20 a b+8 b^2\right ) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{2 a (a+b)}-\frac {b (7 a+4 b) \tanh (c+d x)}{2 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{4 a (a+b)}-\frac {b \tanh (c+d x)}{4 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \text {arctanh}(\tanh (c+d x))}{a}-\frac {b \left (15 a^2+20 a b+8 b^2\right ) \int \frac {1}{-b \tanh ^2(c+d x)+a+b}d\tanh (c+d x)}{a}}{2 a (a+b)}-\frac {b (7 a+4 b) \tanh (c+d x)}{2 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{4 a (a+b)}-\frac {b \tanh (c+d x)}{4 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \text {arctanh}(\tanh (c+d x))}{a}-\frac {\sqrt {b} \left (15 a^2+20 a b+8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{2 a (a+b)}-\frac {b (7 a+4 b) \tanh (c+d x)}{2 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )}}{4 a (a+b)}-\frac {b \tanh (c+d x)}{4 a (a+b) \left (a-b \tanh ^2(c+d x)+b\right )^2}}{d}\)

Input:

Int[(a + b*Sech[c + d*x]^2)^(-3),x]
 

Output:

(-1/4*(b*Tanh[c + d*x])/(a*(a + b)*(a + b - b*Tanh[c + d*x]^2)^2) + (((8*( 
a + b)^2*ArcTanh[Tanh[c + d*x]])/a - (Sqrt[b]*(15*a^2 + 20*a*b + 8*b^2)*Ar 
cTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a*(a + b)) 
 - (b*(7*a + 4*b)*Tanh[c + d*x])/(2*a*(a + b)*(a + b - b*Tanh[c + d*x]^2)) 
)/(4*a*(a + b)))/d
 

Defintions of rubi rules used

rule 25

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

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

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 316

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) 
), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d))   Int[(a + b*x^2)^(p + 1)*(c + d*x 
^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x 
], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  ! 
( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, 
 p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

rule 3042

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

rule 4616

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = 
FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Subst[Int[(a + b + b*ff^2*x^2)^p 
/(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] 
&& NeQ[a + b, 0] && NeQ[p, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(351\) vs. \(2(132)=264\).

Time = 1.25 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.41

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {2 b \left (\frac {-\frac {a \left (9 a +4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right )}-\frac {a \left (27 a^{2}+11 a b -4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2}}-\frac {a \left (27 a^{2}+11 a b -4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2}}-\frac {a \left (9 a +4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (15 a^{2}+20 a b +8 b^{2}\right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{8 a^{2}+16 a b +8 b^{2}}\right )}{a^{3}}}{d}\)	\(352\)
default	\(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}+\frac {2 b \left (\frac {-\frac {a \left (9 a +4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{8 \left (a +b \right )}-\frac {a \left (27 a^{2}+11 a b -4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{8 \left (a +b \right )^{2}}-\frac {a \left (27 a^{2}+11 a b -4 b^{2}\right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 \left (a +b \right )^{2}}-\frac {a \left (9 a +4 b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 \left (a +b \right )}}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\left (15 a^{2}+20 a b +8 b^{2}\right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{8 a^{2}+16 a b +8 b^{2}}\right )}{a^{3}}}{d}\)	\(352\)
risch	\(\frac {x}{a^{3}}+\frac {b \left (9 a^{3} {\mathrm e}^{6 d x +6 c}+28 a^{2} b \,{\mathrm e}^{6 d x +6 c}+16 a \,b^{2} {\mathrm e}^{6 d x +6 c}+27 a^{3} {\mathrm e}^{4 d x +4 c}+90 a^{2} b \,{\mathrm e}^{4 d x +4 c}+120 a \,b^{2} {\mathrm e}^{4 d x +4 c}+48 b^{3} {\mathrm e}^{4 d x +4 c}+27 a^{3} {\mathrm e}^{2 d x +2 c}+68 a^{2} b \,{\mathrm e}^{2 d x +2 c}+32 a \,b^{2} {\mathrm e}^{2 d x +2 c}+9 a^{3}+6 a^{2} b \right )}{4 a^{3} \left (a +b \right )^{2} d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}+\frac {15 \sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right )}{16 \left (a +b \right )^{3} d a}+\frac {5 \sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right ) b}{4 \left (a +b \right )^{3} d \,a^{2}}+\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{3} d \,a^{3}}-\frac {15 \sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right )}{16 \left (a +b \right )^{3} d a}-\frac {5 \sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right ) b}{4 \left (a +b \right )^{3} d \,a^{2}}-\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{3} d \,a^{3}}\)	\(529\)

Input:

int(1/(a+b*sech(d*x+c)^2)^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/a^3 
*b*((-1/8*a*(9*a+4*b)/(a+b)*tanh(1/2*d*x+1/2*c)^7-1/8*a*(27*a^2+11*a*b-4*b 
^2)/(a+b)^2*tanh(1/2*d*x+1/2*c)^5-1/8*a*(27*a^2+11*a*b-4*b^2)/(a+b)^2*tanh 
(1/2*d*x+1/2*c)^3-1/8*a*(9*a+4*b)/(a+b)*tanh(1/2*d*x+1/2*c))/(tanh(1/2*d*x 
+1/2*c)^4*a+tanh(1/2*d*x+1/2*c)^4*b+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d 
*x+1/2*c)^2*b+a+b)^2+1/8*(15*a^2+20*a*b+8*b^2)/(a^2+2*a*b+b^2)*(-1/4/b^(1/ 
2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*tanh(1/2*d*x+1/2*c)* 
b^(1/2)+(a+b)^(1/2))+1/4/b^(1/2)/(a+b)^(1/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1 
/2*c)^2-2*tanh(1/2*d*x+1/2*c)*b^(1/2)+(a+b)^(1/2)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3131 vs. \(2 (138) = 276\).

Time = 0.40 (sec) , antiderivative size = 6538, normalized size of antiderivative = 44.78 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {1}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(1/(a+b*sech(d*x+c)**2)**3,x)
 

Output:

Integral((a + b*sech(c + d*x)**2)**(-3), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (138) = 276\).

Time = 0.15 (sec) , antiderivative size = 402, normalized size of antiderivative = 2.75 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{16 \, {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {9 \, a^{3} b + 6 \, a^{2} b^{2} + {\left (27 \, a^{3} b + 68 \, a^{2} b^{2} + 32 \, a b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, {\left (9 \, a^{3} b + 30 \, a^{2} b^{2} + 40 \, a b^{3} + 16 \, b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (9 \, a^{3} b + 28 \, a^{2} b^{2} + 16 \, a b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{4 \, {\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2} + 4 \, {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} + 2 \, a^{4} b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{7} + 14 \, a^{6} b + 27 \, a^{5} b^{2} + 24 \, a^{4} b^{3} + 8 \, a^{3} b^{4}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{7} + 4 \, a^{6} b + 5 \, a^{5} b^{2} + 2 \, a^{4} b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )} + {\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac {d x + c}{a^{3} d} \] Input:

integrate(1/(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

1/16*(15*a^2*b + 20*a*b^2 + 8*b^3)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*s 
qrt((a + b)*b))/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^5 
+ 2*a^4*b + a^3*b^2)*sqrt((a + b)*b)*d) - 1/4*(9*a^3*b + 6*a^2*b^2 + (27*a 
^3*b + 68*a^2*b^2 + 32*a*b^3)*e^(-2*d*x - 2*c) + 3*(9*a^3*b + 30*a^2*b^2 + 
 40*a*b^3 + 16*b^4)*e^(-4*d*x - 4*c) + (9*a^3*b + 28*a^2*b^2 + 16*a*b^3)*e 
^(-6*d*x - 6*c))/((a^7 + 2*a^6*b + a^5*b^2 + 4*(a^7 + 4*a^6*b + 5*a^5*b^2 
+ 2*a^4*b^3)*e^(-2*d*x - 2*c) + 2*(3*a^7 + 14*a^6*b + 27*a^5*b^2 + 24*a^4* 
b^3 + 8*a^3*b^4)*e^(-4*d*x - 4*c) + 4*(a^7 + 4*a^6*b + 5*a^5*b^2 + 2*a^4*b 
^3)*e^(-6*d*x - 6*c) + (a^7 + 2*a^6*b + a^5*b^2)*e^(-8*d*x - 8*c))*d) + (d 
*x + c)/(a^3*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 327 vs. \(2 (138) = 276\).

Time = 0.23 (sec) , antiderivative size = 327, normalized size of antiderivative = 2.24 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {-a b - b^{2}}} - \frac {2 \, {\left (9 \, a^{3} b e^{\left (6 \, d x + 6 \, c\right )} + 28 \, a^{2} b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 16 \, a b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 27 \, a^{3} b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a^{2} b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 120 \, a b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 48 \, b^{4} e^{\left (4 \, d x + 4 \, c\right )} + 27 \, a^{3} b e^{\left (2 \, d x + 2 \, c\right )} + 68 \, a^{2} b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 32 \, a b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 9 \, a^{3} b + 6 \, a^{2} b^{2}\right )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}^{2}} - \frac {8 \, {\left (d x + c\right )}}{a^{3}}}{8 \, d} \] Input:

integrate(1/(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

-1/8*((15*a^2*b + 20*a*b^2 + 8*b^3)*arctan(1/2*(a*e^(2*d*x + 2*c) + a + 2* 
b)/sqrt(-a*b - b^2))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt(-a*b - b^2)) - 2*(9*a 
^3*b*e^(6*d*x + 6*c) + 28*a^2*b^2*e^(6*d*x + 6*c) + 16*a*b^3*e^(6*d*x + 6* 
c) + 27*a^3*b*e^(4*d*x + 4*c) + 90*a^2*b^2*e^(4*d*x + 4*c) + 120*a*b^3*e^( 
4*d*x + 4*c) + 48*b^4*e^(4*d*x + 4*c) + 27*a^3*b*e^(2*d*x + 2*c) + 68*a^2* 
b^2*e^(2*d*x + 2*c) + 32*a*b^3*e^(2*d*x + 2*c) + 9*a^3*b + 6*a^2*b^2)/((a^ 
5 + 2*a^4*b + a^3*b^2)*(a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + 4*b*e^(2 
*d*x + 2*c) + a)^2) - 8*(d*x + c)/a^3)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {1}{{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^3} \,d x \] Input:

int(1/(a + b/cosh(c + d*x)^2)^3,x)
 

Output:

int(1/(a + b/cosh(c + d*x)^2)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 4752, normalized size of antiderivative = 32.55 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*sech(d*x+c)^2)^3,x)
 

Output:

( - 15*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + 
 b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**5 - 50*e**(8*c + 8*d*x)*sqrt(b)* 
sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sq 
rt(a))*a**4*b - 48*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt 
(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**3*b**2 - 16*e**(8*c 
+ 8*d*x)*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) 
+ e**(c + d*x)*sqrt(a))*a**2*b**3 - 15*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b 
)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**5 - 
 50*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - 
a - 2*b) + e**(c + d*x)*sqrt(a))*a**4*b - 48*e**(8*c + 8*d*x)*sqrt(b)*sqrt 
(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))* 
a**3*b**2 - 16*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqr 
t(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**2*b**3 + 15*e**(8*c + 8*d*x 
)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 
 2*b)*a**5 + 50*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a 
+ b) + e**(2*c + 2*d*x)*a + a + 2*b)*a**4*b + 48*e**(8*c + 8*d*x)*sqrt(b)* 
sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a**3 
*b**2 + 16*e**(8*c + 8*d*x)*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) 
+ e**(2*c + 2*d*x)*a + a + 2*b)*a**2*b**3 - 60*e**(6*c + 6*d*x)*sqrt(b)*sq 
rt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*s...