3.1.27 \(\int \frac {x}{(a+b \text {sech}(c+d x^2))^2} \, dx\) [27]

3.1.27.1 Optimal result
3.1.27.2 Mathematica [A] (verified)
3.1.27.3 Rubi [A] (verified)
3.1.27.4 Maple [A] (verified)
3.1.27.5 Fricas [B] (verification not implemented)
3.1.27.6 Sympy [F]
3.1.27.7 Maxima [F(-2)]
3.1.27.8 Giac [A] (verification not implemented)
3.1.27.9 Mupad [B] (verification not implemented)

3.1.27.1 Optimal result

Integrand size = 16, antiderivative size = 123 \[ \int \frac {x}{\left (a+b \text {sech}\left (c+d x^2\right )\right )^2} \, dx=\frac {x^2}{2 a^2}-\frac {b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b^2 \tanh \left (c+d x^2\right )}{2 a \left (a^2-b^2\right ) d \left (a+b \text {sech}\left (c+d x^2\right )\right )} \]

output
1/2*x^2/a^2-b*(2*a^2-b^2)*arctan((a-b)^(1/2)*tanh(1/2*d*x^2+1/2*c)/(a+b)^( 
1/2))/a^2/(a-b)^(3/2)/(a+b)^(3/2)/d+1/2*b^2*tanh(d*x^2+c)/a/(a^2-b^2)/d/(a 
+b*sech(d*x^2+c))
 
3.1.27.2 Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.79 \[ \int \frac {x}{\left (a+b \text {sech}\left (c+d x^2\right )\right )^2} \, dx=\frac {a \left (\left (a^2-b^2\right )^{3/2} \left (c+d x^2\right )+\left (4 a^2 b-2 b^3\right ) \arctan \left (\frac {(-a+b) \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )\right ) \cosh \left (c+d x^2\right )+b \left (\left (a^2-b^2\right )^{3/2} \left (c+d x^2\right )+\left (4 a^2 b-2 b^3\right ) \arctan \left (\frac {(-a+b) \tanh \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )+a b \sqrt {a^2-b^2} \sinh \left (c+d x^2\right )\right )}{2 a^2 (a-b) (a+b) \sqrt {a^2-b^2} d \left (b+a \cosh \left (c+d x^2\right )\right )} \]

input
Integrate[x/(a + b*Sech[c + d*x^2])^2,x]
 
output
(a*((a^2 - b^2)^(3/2)*(c + d*x^2) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tan 
h[(c + d*x^2)/2])/Sqrt[a^2 - b^2]])*Cosh[c + d*x^2] + b*((a^2 - b^2)^(3/2) 
*(c + d*x^2) + (4*a^2*b - 2*b^3)*ArcTan[((-a + b)*Tanh[(c + d*x^2)/2])/Sqr 
t[a^2 - b^2]] + a*b*Sqrt[a^2 - b^2]*Sinh[c + d*x^2]))/(2*a^2*(a - b)*(a + 
b)*Sqrt[a^2 - b^2]*d*(b + a*Cosh[c + d*x^2]))
 
3.1.27.3 Rubi [A] (verified)

Time = 0.74 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.19, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.688, Rules used = {5959, 3042, 4272, 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.

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

\(\Big \downarrow \) 5959

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4272

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

input
Int[x/(a + b*Sech[c + d*x^2])^2,x]
 
output
((((a^2 - b^2)*x^2)/a - (2*b*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tanh[(c + d 
*x^2)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/(a*(a^2 - b^2)) + ( 
b^2*Tanh[c + d*x^2])/(a*(a^2 - b^2)*d*(a + b*Sech[c + d*x^2])))/2
 

3.1.27.3.1 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 5959
Int[(x_)^(m_.)*((a_.) + (b_.)*Sech[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbo 
l] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sech[c + d*x] 
)^p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m 
 + 1)/n], 0] && IntegerQ[p]
 
3.1.27.4 Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.44

method result size
derivativedivides \(\frac {\frac {\ln \left (1+\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {2 b \left (-\frac {a b \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} a -\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}}{2 d}\) \(177\)
default \(\frac {\frac {\ln \left (1+\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{a^{2}}-\frac {2 b \left (-\frac {a b \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} a -\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a^{2}}-\frac {\ln \left (\tanh \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}}{2 d}\) \(177\)
risch \(\frac {x^{2}}{2 a^{2}}-\frac {b^{2} \left ({\mathrm e}^{d \,x^{2}+c} b +a \right )}{a^{2} \left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 d \,x^{2}+2 c} a +2 \,{\mathrm e}^{d \,x^{2}+c} b +a \right )}-\frac {b \ln \left ({\mathrm e}^{d \,x^{2}+c}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{d \,x^{2}+c}+\frac {b \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{d \,x^{2}+c}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{d \,x^{2}+c}+\frac {b \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, a}\right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}\) \(385\)

input
int(x/(a+b*sech(d*x^2+c))^2,x,method=_RETURNVERBOSE)
 
output
1/2/d*(1/a^2*ln(1+tanh(1/2*d*x^2+1/2*c))-2/a^2*b*(-a*b/(a^2-b^2)*tanh(1/2* 
d*x^2+1/2*c)/(tanh(1/2*d*x^2+1/2*c)^2*a-tanh(1/2*d*x^2+1/2*c)^2*b+a+b)+(2* 
a^2-b^2)/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctan((a-b)*tanh(1/2*d*x^2+1/2*c 
)/((a+b)*(a-b))^(1/2)))-1/a^2*ln(tanh(1/2*d*x^2+1/2*c)-1))
 
3.1.27.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 587 vs. \(2 (110) = 220\).

Time = 0.30 (sec) , antiderivative size = 1314, normalized size of antiderivative = 10.68 \[ \int \frac {x}{\left (a+b \text {sech}\left (c+d x^2\right )\right )^2} \, dx=\text {Too large to display} \]

input
integrate(x/(a+b*sech(d*x^2+c))^2,x, algorithm="fricas")
 
output
[1/2*((a^5 - 2*a^3*b^2 + a*b^4)*d*x^2*cosh(d*x^2 + c)^2 + (a^5 - 2*a^3*b^2 
 + a*b^4)*d*x^2*sinh(d*x^2 + c)^2 - 2*a^3*b^2 + 2*a*b^4 + (a^5 - 2*a^3*b^2 
 + a*b^4)*d*x^2 - (2*a^3*b - a*b^3 + (2*a^3*b - a*b^3)*cosh(d*x^2 + c)^2 + 
 (2*a^3*b - a*b^3)*sinh(d*x^2 + c)^2 + 2*(2*a^2*b^2 - b^4)*cosh(d*x^2 + c) 
 + 2*(2*a^2*b^2 - b^4 + (2*a^3*b - a*b^3)*cosh(d*x^2 + c))*sinh(d*x^2 + c) 
)*sqrt(-a^2 + b^2)*log((a^2*cosh(d*x^2 + c)^2 + a^2*sinh(d*x^2 + c)^2 + 2* 
a*b*cosh(d*x^2 + c) - a^2 + 2*b^2 + 2*(a^2*cosh(d*x^2 + c) + a*b)*sinh(d*x 
^2 + c) + 2*sqrt(-a^2 + b^2)*(a*cosh(d*x^2 + c) + a*sinh(d*x^2 + c) + b))/ 
(a*cosh(d*x^2 + c)^2 + a*sinh(d*x^2 + c)^2 + 2*b*cosh(d*x^2 + c) + 2*(a*co 
sh(d*x^2 + c) + b)*sinh(d*x^2 + c) + a)) - 2*(a^2*b^3 - b^5 - (a^4*b - 2*a 
^2*b^3 + b^5)*d*x^2)*cosh(d*x^2 + c) - 2*(a^2*b^3 - b^5 - (a^5 - 2*a^3*b^2 
 + a*b^4)*d*x^2*cosh(d*x^2 + c) - (a^4*b - 2*a^2*b^3 + b^5)*d*x^2)*sinh(d* 
x^2 + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cosh(d*x^2 + c)^2 + (a^7 - 2*a^5* 
b^2 + a^3*b^4)*d*sinh(d*x^2 + c)^2 + 2*(a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos 
h(d*x^2 + c) + (a^7 - 2*a^5*b^2 + a^3*b^4)*d + 2*((a^7 - 2*a^5*b^2 + a^3*b 
^4)*d*cosh(d*x^2 + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)*sinh(d*x^2 + c)), 
 1/2*((a^5 - 2*a^3*b^2 + a*b^4)*d*x^2*cosh(d*x^2 + c)^2 + (a^5 - 2*a^3*b^2 
 + a*b^4)*d*x^2*sinh(d*x^2 + c)^2 - 2*a^3*b^2 + 2*a*b^4 + (a^5 - 2*a^3*b^2 
 + a*b^4)*d*x^2 + 2*(2*a^3*b - a*b^3 + (2*a^3*b - a*b^3)*cosh(d*x^2 + c)^2 
 + (2*a^3*b - a*b^3)*sinh(d*x^2 + c)^2 + 2*(2*a^2*b^2 - b^4)*cosh(d*x^2...
 
3.1.27.6 Sympy [F]

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

input
integrate(x/(a+b*sech(d*x**2+c))**2,x)
 
output
Integral(x/(a + b*sech(c + d*x**2))**2, x)
 
3.1.27.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {x}{\left (a+b \text {sech}\left (c+d x^2\right )\right )^2} \, dx=\text {Exception raised: ValueError} \]

input
integrate(x/(a+b*sech(d*x^2+c))^2,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
 
3.1.27.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.20 \[ \int \frac {x}{\left (a+b \text {sech}\left (c+d x^2\right )\right )^2} \, dx=-\frac {{\left (2 \, a^{2} b - b^{3}\right )} \arctan \left (\frac {a e^{\left (d x^{2} + c\right )} + b}{\sqrt {a^{2} - b^{2}}}\right )}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt {a^{2} - b^{2}}} - \frac {b^{3} e^{\left (d x^{2} + c\right )} + a b^{2}}{{\left (a^{4} d - a^{2} b^{2} d\right )} {\left (a e^{\left (2 \, d x^{2} + 2 \, c\right )} + 2 \, b e^{\left (d x^{2} + c\right )} + a\right )}} + \frac {d x^{2} + c}{2 \, a^{2} d} \]

input
integrate(x/(a+b*sech(d*x^2+c))^2,x, algorithm="giac")
 
output
-(2*a^2*b - b^3)*arctan((a*e^(d*x^2 + c) + b)/sqrt(a^2 - b^2))/((a^4*d - a 
^2*b^2*d)*sqrt(a^2 - b^2)) - (b^3*e^(d*x^2 + c) + a*b^2)/((a^4*d - a^2*b^2 
*d)*(a*e^(2*d*x^2 + 2*c) + 2*b*e^(d*x^2 + c) + a)) + 1/2*(d*x^2 + c)/(a^2* 
d)
 
3.1.27.9 Mupad [B] (verification not implemented)

Time = 2.58 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.57 \[ \int \frac {x}{\left (a+b \text {sech}\left (c+d x^2\right )\right )^2} \, dx=\frac {\frac {b^2}{d\,\left (a\,b^2-a^3\right )}+\frac {b^3\,{\mathrm {e}}^{d\,x^2+c}}{a\,d\,\left (a\,b^2-a^3\right )}}{a+2\,b\,{\mathrm {e}}^{d\,x^2+c}+a\,{\mathrm {e}}^{2\,d\,x^2+2\,c}}+\frac {x^2}{2\,a^2}+\frac {b\,\ln \left (\frac {2\,b\,x\,{\mathrm {e}}^{d\,x^2+c}\,\left (2\,a^2-b^2\right )}{a^3\,\left (a^2-b^2\right )}-\frac {2\,b\,x\,\left (a+b\,{\mathrm {e}}^{d\,x^2+c}\right )\,\left (2\,a^2-b^2\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{2\,a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}-\frac {b\,\ln \left (\frac {2\,b\,x\,{\mathrm {e}}^{d\,x^2+c}\,\left (2\,a^2-b^2\right )}{a^3\,\left (a^2-b^2\right )}+\frac {2\,b\,x\,\left (a+b\,{\mathrm {e}}^{d\,x^2+c}\right )\,\left (2\,a^2-b^2\right )}{a^3\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{2\,a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (b-a\right )}^{3/2}} \]

input
int(x/(a + b/cosh(c + d*x^2))^2,x)
 
output
(b^2/(d*(a*b^2 - a^3)) + (b^3*exp(c + d*x^2))/(a*d*(a*b^2 - a^3)))/(a + 2* 
b*exp(c + d*x^2) + a*exp(2*c + 2*d*x^2)) + x^2/(2*a^2) + (b*log((2*b*x*exp 
(c + d*x^2)*(2*a^2 - b^2))/(a^3*(a^2 - b^2)) - (2*b*x*(a + b*exp(c + d*x^2 
))*(2*a^2 - b^2))/(a^3*(a + b)^(3/2)*(b - a)^(3/2)))*(2*a^2 - b^2))/(2*a^2 
*d*(a + b)^(3/2)*(b - a)^(3/2)) - (b*log((2*b*x*exp(c + d*x^2)*(2*a^2 - b^ 
2))/(a^3*(a^2 - b^2)) + (2*b*x*(a + b*exp(c + d*x^2))*(2*a^2 - b^2))/(a^3* 
(a + b)^(3/2)*(b - a)^(3/2)))*(2*a^2 - b^2))/(2*a^2*d*(a + b)^(3/2)*(b - a 
)^(3/2))