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

Optimal. Leaf size=93 \[ -\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 d (a+b)^{3/2}}+\frac {x}{a^2}-\frac {b \tanh (c+d x)}{2 a d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4128, 414, 522, 206, 208} \[ -\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tanh (c+d x)}{\sqrt {a+b}}\right )}{2 a^2 d (a+b)^{3/2}}+\frac {x}{a^2}-\frac {b \tanh (c+d x)}{2 a d (a+b) \left (a-b \tanh ^2(c+d x)+b\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sech[c + d*x]^2)^(-2),x]

[Out]

x/a^2 - (Sqrt[b]*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Tanh[c + d*x])/Sqrt[a + b]])/(2*a^2*(a + b)^(3/2)*d) - (b*Tanh[c
+ d*x])/(2*a*(a + b)*d*(a + b - b*Tanh[c + d*x]^2))

Rule 206

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

Rule 414

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

Rule 522

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

Rule 4128

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[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]

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \left (a+b-b x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=-\frac {b \tanh (c+d x)}{2 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}-\frac {\operatorname {Subst}\left (\int \frac {-2 a-b-b x^2}{\left (1-x^2\right ) \left (a+b-b x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 a (a+b) d}\\ &=-\frac {b \tanh (c+d x)}{2 a (a+b) d \left (a+b-b \tanh ^2(c+d x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{a^2 d}-\frac {(b (3 a+2 b)) \operatorname {Subst}\left (\int \frac {1}{a+b-b x^2} \, dx,x,\tanh (c+d x)\right )}{2 a^2 (a+b) d}\\ &=\frac {x}{a^2}-\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\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 )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 2.04, size = 221, normalized size = 2.38 \[ \frac {\text {sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (2 x (a \cosh (2 (c+d x))+a+2 b)+\frac {b \text {sech}(2 c) ((a+2 b) \sinh (2 c)-a \sinh (2 d x))}{d (a+b)}-\frac {b (3 a+2 b) (\cosh (2 c)-\sinh (2 c)) (a \cosh (2 (c+d x))+a+2 b) \tanh ^{-1}\left (\frac {(\cosh (2 c)-\sinh (2 c)) \text {sech}(d x) ((a+2 b) \sinh (d x)-a \sinh (2 c+d x))}{2 \sqrt {a+b} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )}{d (a+b)^{3/2} \sqrt {b (\cosh (c)-\sinh (c))^4}}\right )}{8 a^2 \left (a+b \text {sech}^2(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((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] - Si
nh[c])^4]) + (b*Sech[2*c]*((a + 2*b)*Sinh[2*c] - a*Sinh[2*d*x]))/((a + b)*d)))/(8*a^2*(a + b*Sech[c + d*x]^2)^
2)

________________________________________________________________________________________

fricas [B]  time = 0.46, size = 1690, normalized size = 18.17 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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)*sinh(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)*cosh(d*x + c)^3 + (3*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c
))*sqrt(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*cosh(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*co
sh(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*c
osh(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 + 4*(a^4 + a^3*b)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 + a^3
*b)*d*sinh(d*x + c)^4 + 2*(a^4 + 3*a^3*b + 2*a^2*b^2)*d*cosh(d*x + c)^2 + 2*(3*(a^4 + a^3*b)*d*cosh(d*x + c)^2
+ (a^4 + 3*a^3*b + 2*a^2*b^2)*d)*sinh(d*x + c)^2 + (a^4 + a^3*b)*d + 4*((a^4 + a^3*b)*d*cosh(d*x + c)^3 + (a^
4 + 3*a^3*b + 2*a^2*b^2)*d*cosh(d*x + c))*sinh(d*x + c)), 1/2*(2*(a^2 + a*b)*d*x*cosh(d*x + c)^4 + 8*(a^2 + a*
b)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*(a^2 + a*b)*d*x*sinh(d*x + c)^4 + 2*(a^2 + a*b)*d*x + 2*(2*(a^2 + 3*a
*b + 2*b^2)*d*x + a*b + 2*b^2)*cosh(d*x + c)^2 + 2*(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)*sinh(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)*cosh(d*x + c)
^3 + (3*a^2 + 8*a*b + 4*b^2)*cosh(d*x + c))*sinh(d*x + c))*sqrt(-b/(a + b))*arctan(1/2*(a*cosh(d*x + c)^2 + 2*
a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x + c)^2 + a + 2*b)*sqrt(-b/(a + b))/b) + 2*a*b + 4*(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 + 4*(a^4 + a^3*b)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^4 + a^3*b)*d*sinh(d*x + c)^4 + 2*(a^4
+ 3*a^3*b + 2*a^2*b^2)*d*cosh(d*x + c)^2 + 2*(3*(a^4 + a^3*b)*d*cosh(d*x + c)^2 + (a^4 + 3*a^3*b + 2*a^2*b^2)*
d)*sinh(d*x + c)^2 + (a^4 + a^3*b)*d + 4*((a^4 + a^3*b)*d*cosh(d*x + c)^3 + (a^4 + 3*a^3*b + 2*a^2*b^2)*d*cosh
(d*x + c))*sinh(d*x + c))]

________________________________________________________________________________________

giac [A]  time = 0.44, size = 163, normalized size = 1.75 \[ -\frac {\frac {{\left (3 \, a b + 2 \, b^{2}\right )} \arctan \left (\frac {a e^{\left (2 \, d x + 2 \, c\right )} + a + 2 \, b}{2 \, \sqrt {-a b - b^{2}}}\right )}{{\left (a^{3} + a^{2} b\right )} \sqrt {-a b - b^{2}}} - \frac {2 \, {\left (a b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + a b\right )}}{{\left (a^{3} + a^{2} b\right )} {\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}} - \frac {2 \, {\left (d x + c\right )}}{a^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

________________________________________________________________________________________

maple [B]  time = 0.38, size = 423, normalized size = 4.55 \[ -\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \,a^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {b \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right ) a \left (a +b \right )}-\frac {b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{d a \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right ) \left (a +b \right )}+\frac {3 \sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d a \left (a +b \right )^{\frac {3}{2}}}-\frac {3 \sqrt {b}\, \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{4 d a \left (a +b \right )^{\frac {3}{2}}}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{2} \left (a +b \right )^{\frac {3}{2}}}-\frac {b^{\frac {3}{2}} \ln \left (\sqrt {a +b}\, \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \sqrt {b}\, \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a +b}\right )}{2 d \,a^{2} \left (a +b \right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)+1/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)-1/d/a*b/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1
/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)/(a+b)*tanh(1/2*d*x+1/2*c)^3-1/d/a*b/(
tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)/(a+b)
*tanh(1/2*d*x+1/2*c)+3/4/d/a*b^(1/2)/(a+b)^(3/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*b^(1/2)*tanh(1/2*d*x+1
/2*c)+(a+b)^(1/2))-3/4/d/a*b^(1/2)/(a+b)^(3/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2
*c)+(a+b)^(1/2))+1/2/d/a^2*b^(3/2)/(a+b)^(3/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2-2*b^(1/2)*tanh(1/2*d*x+1/2
*c)+(a+b)^(1/2))-1/2/d/a^2*b^(3/2)/(a+b)^(3/2)*ln((a+b)^(1/2)*tanh(1/2*d*x+1/2*c)^2+2*b^(1/2)*tanh(1/2*d*x+1/2
*c)+(a+b)^(1/2))

________________________________________________________________________________________

maxima [B]  time = 0.69, size = 187, normalized size = 2.01 \[ \frac {{\left (3 \, a b + 2 \, b^{2}\right )} \log \left (\frac {a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b - 2 \, \sqrt {{\left (a + b\right )} b}}{a e^{\left (-2 \, d x - 2 \, c\right )} + a + 2 \, b + 2 \, \sqrt {{\left (a + b\right )} b}}\right )}{4 \, {\left (a^{3} + a^{2} b\right )} \sqrt {{\left (a + b\right )} b} d} - \frac {a b + {\left (a b + 2 \, b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (a^{4} + a^{3} b + 2 \, {\left (a^{4} + 3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{4} + a^{3} b\right )} e^{\left (-4 \, d x - 4 \, c\right )}\right )} d} + \frac {d x + c}{a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*s
qrt((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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (a+\frac {b}{{\mathrm {cosh}\left (c+d\,x\right )}^2}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/cosh(c + d*x)^2)^2,x)

[Out]

int(1/(a + b/cosh(c + d*x)^2)^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \operatorname {sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________