\(\int \frac {1}{(a+b \text {sech}^2(c+d x))^2} \, dx\) [153]

Optimal result

Integrand size = 14, antiderivative size = 93 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {x}{a^2}-\frac {\sqrt {b} (3 a+2 b) \text {arctanh}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 (a+b)^{3/2} d}-\frac {b \tanh (c+d x)}{2 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )} \] Output:

x/a^2-1/2*b^(1/2)*(3*a+2*b)*arctanh(b^(1/2)*tanh(d*x+c)/(a+b)^(1/2))/a^2/( 
a+b)^(3/2)/d-1/2*b*tanh(d*x+c)/a/(a+b)/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. \(221\) vs. \(2(93)=186\).

Time = 2.57 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.38 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x))) \text {sech}^4(c+d x) \left (2 x (a+2 b+a \cosh (2 (c+d x)))-\frac {b (3 a+2 b) \text {arctanh}\left (\frac {\text {sech}(d x) (\cosh (2 c)-\sinh (2 c)) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right ) (a+2 b+a \cosh (2 (c+d x))) (\cosh (2 c)-\sinh (2 c))}{(a+b)^{3/2} d \sqrt {b (\cosh (c)-\sinh (c))^4}}+\frac {b \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{(a+b) d}\right )}{8 a^2 \left (a+b \text {sech}^2(c+d x)\right )^2} \] Input:

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

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])*Sech[c + d*x]^4*(2*x*(a + 2*b + a*Cosh[2* 
(c + d*x)]) - (b*(3*a + 2*b)*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]*Sqrt[b*(Cosh[c] - 
Sinh[c])^4])]*(a + 2*b + a*Cosh[2*(c + d*x)])*(Cosh[2*c] - Sinh[2*c]))/((a 
 + b)^(3/2)*d*Sqrt[b*(Cosh[c] - Sinh[c])^4]) + (b*Sech[2*c]*((a + 2*b)*Sin 
h[2*c] - a*Sinh[2*d*x]))/((a + b)*d)))/(8*a^2*(a + b*Sech[c + d*x]^2)^2)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.20, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4616, 316, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4616

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

\(\Big \downarrow \) 316

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(((2*(a + b)*ArcTanh[Tanh[c + d*x]])/a - (Sqrt[b]*(3*a + 2*b)*ArcTanh[(Sqr 
t[b]*Tanh[c + d*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a*(a + b)) - (b*Tanh 
[c + d*x])/(2*a*(a + b)*(a + b - b*Tanh[c + d*x]^2)))/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 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. \(253\) vs. \(2(81)=162\).

Time = 0.54 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.73

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {2 b \left (\frac {-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a +b \right )}-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right )}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (3 a +2 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{2 a +2 b}\right )}{a^{2}}}{d}\)	\(254\)
default	\(\frac {\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}+\frac {2 b \left (\frac {-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a +b \right )}-\frac {a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right )}}{\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}+\frac {\left (3 a +2 b \right ) \left (-\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}+\frac {\ln \left (\sqrt {a +b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b}+\sqrt {a +b}\right )}{4 \sqrt {b}\, \sqrt {a +b}}\right )}{2 a +2 b}\right )}{a^{2}}}{d}\)	\(254\)
risch	\(\frac {x}{a^{2}}+\frac {b \left (a \,{\mathrm e}^{2 d x +2 c}+2 b \,{\mathrm e}^{2 d x +2 c}+a \right )}{a^{2} \left (a +b \right ) d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {3 \sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right )}{4 \left (a +b \right )^{2} d a}+\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {a +2 \sqrt {b \left (a +b \right )}+2 b}{a}\right ) b}{2 \left (a +b \right )^{2} d \,a^{2}}-\frac {3 \sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right )}{4 \left (a +b \right )^{2} d a}-\frac {\sqrt {b \left (a +b \right )}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {-a +2 \sqrt {b \left (a +b \right )}-2 b}{a}\right ) b}{2 \left (a +b \right )^{2} d \,a^{2}}\)	\(288\)

Input:

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

Output:

1/d*(1/a^2*ln(tanh(1/2*d*x+1/2*c)+1)-1/a^2*ln(tanh(1/2*d*x+1/2*c)-1)+2*b/a 
^2*((-1/2*a/(a+b)*tanh(1/2*d*x+1/2*c)^3-1/2*a/(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)+1/2*(3*a+2*b)/(a+b)*(-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*ta 
nh(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. 707 vs. \(2 (84) = 168\).

Time = 0.45 (sec) , antiderivative size = 1690, normalized size of antiderivative = 18.17 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(4*(a^2 + a*b)*d*x*cosh(d*x + c)^4 + 16*(a^2 + a*b)*d*x*cosh(d*x + c) 
*sinh(d*x + c)^3 + 4*(a^2 + a*b)*d*x*sinh(d*x + c)^4 + 4*(a^2 + a*b)*d*x + 
 4*(2*(a^2 + 3*a*b + 2*b^2)*d*x + a*b + 2*b^2)*cosh(d*x + c)^2 + 4*(6*(a^2 
 + a*b)*d*x*cosh(d*x + c)^2 + 2*(a^2 + 3*a*b + 2*b^2)*d*x + a*b + 2*b^2)*s 
inh(d*x + c)^2 + ((3*a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(3*a^2 + 2*a*b)*cosh 
(d*x + c)*sinh(d*x + c)^3 + (3*a^2 + 2*a*b)*sinh(d*x + c)^4 + 2*(3*a^2 + 8 
*a*b + 4*b^2)*cosh(d*x + c)^2 + 2*(3*(3*a^2 + 2*a*b)*cosh(d*x + c)^2 + 3*a 
^2 + 8*a*b + 4*b^2)*sinh(d*x + c)^2 + 3*a^2 + 2*a*b + 4*((3*a^2 + 2*a*b)*c 
osh(d*x + c)^3 + (3*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqr 
t(b/(a + b))*log((a^2*cosh(d*x + c)^4 + 4*a^2*cosh(d*x + c)*sinh(d*x + c)^ 
3 + a^2*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*a^2*cosh( 
d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^2 + a^2 + 8*a*b + 8*b^2 + 4*(a^2*c 
osh(d*x + c)^3 + (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c) + 4*((a^2 + a* 
b)*cosh(d*x + c)^2 + 2*(a^2 + a*b)*cosh(d*x + c)*sinh(d*x + c) + (a^2 + a* 
b)*sinh(d*x + c)^2 + a^2 + 3*a*b + 2*b^2)*sqrt(b/(a + b)))/(a*cosh(d*x + c 
)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)* 
cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a 
*cosh(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + a)) + 4*a*b + 
8*(2*(a^2 + a*b)*d*x*cosh(d*x + c)^3 + (2*(a^2 + 3*a*b + 2*b^2)*d*x + a*b 
+ 2*b^2)*cosh(d*x + c))*sinh(d*x + c))/((a^4 + a^3*b)*d*cosh(d*x + c)^4...
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (84) = 168\).

Time = 0.13 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.01 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {{\left (3 \, a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {a b + {\left (a b + 2 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + a^{3} b + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + a^{3} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{a^{2} d} \] Input:

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

Output:

1/4*(3*a*b + 2*b^2)*log((a*e^(-2*d*x - 2*c) + a + 2*b - 2*sqrt((a + b)*b)) 
/(a*e^(-2*d*x - 2*c) + a + 2*b + 2*sqrt((a + b)*b)))/((a^3 + a^2*b)*sqrt(( 
a + b)*b)*d) - (a*b + (a*b + 2*b^2)*e^(-2*d*x - 2*c))/((a^4 + a^3*b + 2*(a 
^4 + 3*a^3*b + 2*a^2*b^2)*e^(-2*d*x - 2*c) + (a^4 + a^3*b)*e^(-4*d*x - 4*c 
))*d) + (d*x + c)/(a^2*d)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.75 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {-a b - b^{2}}} - \frac {2 \, {\left (a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a b\right )}}{{\left (a^{3} + a^{2} b\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}} - \frac {2 \, {\left (d x + c\right )}}{a^{2}}}{2 \, d} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1283, normalized size of antiderivative = 13.80 \[ \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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