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

Optimal result

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

1/2*x^2/a^2-b*(2*a^2-b^2)*arctanh((a-b)^(1/2)*tan(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*tan(d*x^2+c)/a/(a^2-b^2)/d/(a+ 
b*sec(d*x^2+c))
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.24 \[ \int \frac {x}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx=\frac {-\frac {2 b \left (-2 a^2+b^2\right ) \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+\frac {a \left (a^2-b^2\right ) \left (c+d x^2\right ) \cos \left (c+d x^2\right )+b \left (\left (a^2-b^2\right ) \left (c+d x^2\right )+a b \sin \left (c+d x^2\right )\right )}{b+a \cos \left (c+d x^2\right )}}{2 a^2 (a-b) (a+b) d} \] Input:

Integrate[x/(a + b*Sec[c + d*x^2])^2,x]
 

Output:

((-2*b*(-2*a^2 + b^2)*ArcTanh[((-a + b)*Tan[(c + d*x^2)/2])/Sqrt[a^2 - b^2 
]])/Sqrt[a^2 - b^2] + (a*(a^2 - b^2)*(c + d*x^2)*Cos[c + d*x^2] + b*((a^2 
- b^2)*(c + d*x^2) + a*b*Sin[c + d*x^2]))/(b + a*Cos[c + d*x^2]))/(2*a^2*( 
a - b)*(a + b)*d)
 

Rubi [A] (verified)

Time = 0.71 (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 = {4692, 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.

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

\(\Big \downarrow \) 4692

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4272

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{2} \left (\frac {\frac {x^2 \left (a^2-b^2\right )}{a}-\frac {2 b \left (2 a^2-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} \left (c+d x^2\right )\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}+\frac {b^2 \tan \left (c+d x^2\right )}{a d \left (a^2-b^2\right ) \left (a+b \sec \left (c+d x^2\right )\right )}\right )\)

Input:

Int[x/(a + b*Sec[c + d*x^2])^2,x]
 

Output:

((((a^2 - b^2)*x^2)/a - (2*b*(2*a^2 - b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d 
*x^2)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/(a*(a^2 - b^2)) + ( 
b^2*Tan[c + d*x^2])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x^2])))/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 4692

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Sec[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]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b \left (-\frac {a b \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2}}}{2 d}\)	\(162\)
default	\(\frac {\frac {2 \arctan \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {2 b \left (-\frac {a b \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (2 a^{2}-b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d \,x^{2}}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2}}}{2 d}\)	\(162\)
risch	\(\frac {x^{2}}{2 a^{2}}+\frac {i b^{2} \left (b \,{\mathrm e}^{i \left (d \,x^{2}+c \right )}+a \right )}{a^{2} \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d \,x^{2}+c \right )}+2 b \,{\mathrm e}^{i \left (d \,x^{2}+c \right )}+a \right )}+\frac {b \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}-\frac {i a^{2}-i b^{2}-b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d \,x^{2}+c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{2 \sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}\)	\(424\)

Input:

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

Output:

1/2/d*(2/a^2*arctan(tan(1/2*d*x^2+1/2*c))+2*b/a^2*(-a*b/(a^2-b^2)*tan(1/2* 
d*x^2+1/2*c)/(tan(1/2*d*x^2+1/2*c)^2*a-tan(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)*arctanh((a-b)*tan(1/2*d*x^2+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. 229 vs. \(2 (110) = 220\).

Time = 0.10 (sec) , antiderivative size = 525, normalized size of antiderivative = 4.27 \[ \int \frac {x}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx=\left [\frac {2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} \cos \left (d x^{2} + c\right ) + 2 \, {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x^{2} + {\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x^{2} + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x^{2} + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x^{2} + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x^{2} + c\right ) + a\right )} \sin \left (d x^{2} + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x^{2} + c\right )^{2} + 2 \, a b \cos \left (d x^{2} + c\right ) + b^{2}}\right ) + 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x^{2} + c\right )}{4 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x^{2} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac {{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d x^{2} \cos \left (d x^{2} + c\right ) + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d x^{2} - {\left (2 \, a^{2} b^{2} - b^{4} + {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x^{2} + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x^{2} + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x^{2} + c\right )}\right ) + {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x^{2} + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x^{2} + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}\right ] \] Input:

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

Output:

[1/4*(2*(a^5 - 2*a^3*b^2 + a*b^4)*d*x^2*cos(d*x^2 + c) + 2*(a^4*b - 2*a^2* 
b^3 + b^5)*d*x^2 + (2*a^2*b^2 - b^4 + (2*a^3*b - a*b^3)*cos(d*x^2 + c))*sq 
rt(a^2 - b^2)*log((2*a*b*cos(d*x^2 + c) - (a^2 - 2*b^2)*cos(d*x^2 + c)^2 - 
 2*sqrt(a^2 - b^2)*(b*cos(d*x^2 + c) + a)*sin(d*x^2 + c) + 2*a^2 - b^2)/(a 
^2*cos(d*x^2 + c)^2 + 2*a*b*cos(d*x^2 + c) + b^2)) + 2*(a^3*b^2 - a*b^4)*s 
in(d*x^2 + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x^2 + c) + (a^6*b - 2* 
a^4*b^3 + a^2*b^5)*d), 1/2*((a^5 - 2*a^3*b^2 + a*b^4)*d*x^2*cos(d*x^2 + c) 
 + (a^4*b - 2*a^2*b^3 + b^5)*d*x^2 - (2*a^2*b^2 - b^4 + (2*a^3*b - a*b^3)* 
cos(d*x^2 + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x^2 + c 
) + a)/((a^2 - b^2)*sin(d*x^2 + c))) + (a^3*b^2 - a*b^4)*sin(d*x^2 + c))/( 
(a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x^2 + c) + (a^6*b - 2*a^4*b^3 + a^2*b^ 
5)*d)]
 

Sympy [F]

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

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

Output:

Integral(x/(a + b*sec(c + d*x**2))**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8871 vs. \(2 (110) = 220\).

Time = 23.06 (sec) , antiderivative size = 8871, normalized size of antiderivative = 72.12 \[ \int \frac {x}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((a^6 - 2*a^4*b^2 + a^2*b^4)*d*x^2*cos(2*d*x^2 + 2*c)^2 + 4*(a^4*b^2 - 
 2*a^2*b^4 + b^6)*d*x^2*cos(d*x^2 + c)^2 + (a^6 - 2*a^4*b^2 + a^2*b^4)*d*x 
^2*sin(2*d*x^2 + 2*c)^2 + 4*(a^4*b^2 - 2*a^2*b^4 + b^6)*d*x^2*sin(d*x^2 + 
c)^2 + 4*(a^5*b - 2*a^3*b^3 + a*b^5)*d*x^2*cos(d*x^2 + c) + (a^6 - 2*a^4*b 
^2 + a^2*b^4)*d*x^2 + (2*a^4*b - a^2*b^3 + (2*a^4*b - a^2*b^3)*cos(2*d*x^2 
 + 2*c)^2 + 4*(2*a^2*b^3 - b^5)*cos(d*x^2 + c)^2 + (2*a^4*b - a^2*b^3)*sin 
(2*d*x^2 + 2*c)^2 + 4*(2*a^3*b^2 - a*b^4)*sin(2*d*x^2 + 2*c)*sin(d*x^2 + c 
) + 4*(2*a^2*b^3 - b^5)*sin(d*x^2 + c)^2 + 2*(2*a^4*b - a^2*b^3 + 2*(2*a^3 
*b^2 - a*b^4)*cos(d*x^2 + c))*cos(2*d*x^2 + 2*c) + 4*(2*a^3*b^2 - a*b^4)*c 
os(d*x^2 + c))*sqrt(-a^2 + b^2)*arctan2(2*(4*(a^6 - a^4*b^2)*cos(d*x^2 + 2 
*c)^4*cos(c)*sin(c) - 4*(a^6 - a^4*b^2)*cos(c)*sin(d*x^2 + 2*c)^4*sin(c) + 
 4*(3*(a^5*b - a^3*b^3)*cos(c)^2*sin(c) + (a^5*b - a^3*b^3)*sin(c)^3)*cos( 
d*x^2 + 2*c)^3 - 4*((a^5*b - a^3*b^3)*cos(c)^3 + 3*(a^5*b - a^3*b^3)*cos(c 
)*sin(c)^2 + ((a^6 - a^4*b^2)*cos(c)^2 - (a^6 - a^4*b^2)*sin(c)^2)*cos(d*x 
^2 + 2*c))*sin(d*x^2 + 2*c)^3 - 4*((a^6 - 5*a^4*b^2 + 4*a^2*b^4)*cos(c)^3* 
sin(c) + (a^6 - 5*a^4*b^2 + 4*a^2*b^4)*cos(c)*sin(c)^3)*cos(d*x^2 + 2*c)^2 
 + 4*((a^6 - 5*a^4*b^2 + 4*a^2*b^4)*cos(c)^3*sin(c) + (a^6 - 5*a^4*b^2 + 4 
*a^2*b^4)*cos(c)*sin(c)^3 - 3*((a^5*b - a^3*b^3)*cos(c)^2*sin(c) - (a^5*b 
- a^3*b^3)*sin(c)^3)*cos(d*x^2 + 2*c))*sin(d*x^2 + 2*c)^2 - 4*((a^5*b - 3* 
a^3*b^3 + 2*a*b^5)*cos(c)^4*sin(c) + 2*(a^5*b - 3*a^3*b^3 + 2*a*b^5)*co...
 

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.59 \[ \int \frac {x}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx=-\frac {b^{2} \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )}{{\left (a^{3} d - a b^{2} d\right )} {\left (a \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )^{2} - a - b\right )}} + \frac {{\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x^{2} + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x^{2} + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} d - a^{2} b^{2} d\right )} \sqrt {-a^{2} + b^{2}}} + \frac {d x^{2} + c}{2 \, a^{2} d} \] Input:

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

Output:

-b^2*tan(1/2*d*x^2 + 1/2*c)/((a^3*d - a*b^2*d)*(a*tan(1/2*d*x^2 + 1/2*c)^2 
 - b*tan(1/2*d*x^2 + 1/2*c)^2 - a - b)) + (2*a^2*b - b^3)*(pi*floor(1/2*(d 
*x^2 + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x^2 + 1/2*c) - b* 
tan(1/2*d*x^2 + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4*d - a^2*b^2*d)*sqrt(-a^2 
+ b^2)) + 1/2*(d*x^2 + c)/(a^2*d)
 

Mupad [B] (verification not implemented)

Time = 19.98 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.76 \[ \int \frac {x}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx=\frac {\frac {b^2}{d\,\left (a\,b^2\,1{}\mathrm {i}-a^3\,1{}\mathrm {i}\right )}+\frac {b^3\,{\mathrm {e}}^{1{}\mathrm {i}\,d\,x^2+c\,1{}\mathrm {i}}}{a\,d\,\left (a\,b^2\,1{}\mathrm {i}-a^3\,1{}\mathrm {i}\right )}}{a+a\,{\mathrm {e}}^{2{}\mathrm {i}\,d\,x^2+c\,2{}\mathrm {i}}+2\,b\,{\mathrm {e}}^{1{}\mathrm {i}\,d\,x^2+c\,1{}\mathrm {i}}}+\frac {x^2}{2\,a^2}+\frac {b\,\ln \left (2\,b\,x\,{\mathrm {e}}^{1{}\mathrm {i}\,d\,x^2+c\,1{}\mathrm {i}}\,\left (2\,a^2-b^2\right )-\frac {b\,x\,\left (a^2-b^2\right )\,\left (2\,a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{1{}\mathrm {i}\,d\,x^2+c\,1{}\mathrm {i}}\right )\,2{}\mathrm {i}}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{2\,a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {b\,\ln \left (2\,b\,x\,{\mathrm {e}}^{1{}\mathrm {i}\,d\,x^2+c\,1{}\mathrm {i}}\,\left (2\,a^2-b^2\right )+\frac {b\,x\,\left (a^2-b^2\right )\,\left (2\,a^2-b^2\right )\,\left (a+b\,{\mathrm {e}}^{1{}\mathrm {i}\,d\,x^2+c\,1{}\mathrm {i}}\right )\,2{}\mathrm {i}}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (2\,a^2-b^2\right )}{2\,a^2\,d\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:

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

Output:

(b^2/(d*(a*b^2*1i - a^3*1i)) + (b^3*exp(c*1i + d*x^2*1i))/(a*d*(a*b^2*1i - 
 a^3*1i)))/(a + a*exp(c*2i + d*x^2*2i) + 2*b*exp(c*1i + d*x^2*1i)) + x^2/( 
2*a^2) + (b*log(2*b*x*exp(c*1i + d*x^2*1i)*(2*a^2 - b^2) - (b*x*(a^2 - b^2 
)*(2*a^2 - b^2)*(a + b*exp(c*1i + d*x^2*1i))*2i)/((a + b)^(3/2)*(a - b)^(3 
/2)))*(2*a^2 - b^2))/(2*a^2*d*(a + b)^(3/2)*(a - b)^(3/2)) - (b*log(2*b*x* 
exp(c*1i + d*x^2*1i)*(2*a^2 - b^2) + (b*x*(a^2 - b^2)*(2*a^2 - b^2)*(a + b 
*exp(c*1i + d*x^2*1i))*2i)/((a + b)^(3/2)*(a - b)^(3/2)))*(2*a^2 - b^2))/( 
2*a^2*d*(a + b)^(3/2)*(a - b)^(3/2))
 

Reduce [F]

\[ \int \frac {x}{\left (a+b \sec \left (c+d x^2\right )\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x**2)/2)*a - tan((c + d*x**2)/ 
2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x**2)*a**2 + 2*sqrt( - a**2 + b**2)* 
atan((tan((c + d*x**2)/2)*a - tan((c + d*x**2)/2)*b)/sqrt( - a**2 + b**2)) 
*cos(c + d*x**2)*a*b - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x**2)/2)*a 
- tan((c + d*x**2)/2)*b)/sqrt( - a**2 + b**2))*a*b + 2*sqrt( - a**2 + b**2 
)*atan((tan((c + d*x**2)/2)*a - tan((c + d*x**2)/2)*b)/sqrt( - a**2 + b**2 
))*b**2 + 8*cos(c + d*x**2)*int(x/(tan((c + d*x**2)/2)**4*a**2 - 2*tan((c 
+ d*x**2)/2)**4*a*b + tan((c + d*x**2)/2)**4*b**2 - 2*tan((c + d*x**2)/2)* 
*2*a**2 + 2*tan((c + d*x**2)/2)**2*b**2 + a**2 + 2*a*b + b**2),x)*a**5*d - 
 16*cos(c + d*x**2)*int(x/(tan((c + d*x**2)/2)**4*a**2 - 2*tan((c + d*x**2 
)/2)**4*a*b + tan((c + d*x**2)/2)**4*b**2 - 2*tan((c + d*x**2)/2)**2*a**2 
+ 2*tan((c + d*x**2)/2)**2*b**2 + a**2 + 2*a*b + b**2),x)*a**3*b**2*d + 8* 
cos(c + d*x**2)*int(x/(tan((c + d*x**2)/2)**4*a**2 - 2*tan((c + d*x**2)/2) 
**4*a*b + tan((c + d*x**2)/2)**4*b**2 - 2*tan((c + d*x**2)/2)**2*a**2 + 2* 
tan((c + d*x**2)/2)**2*b**2 + a**2 + 2*a*b + b**2),x)*a*b**4*d + 8*int(x/( 
tan((c + d*x**2)/2)**4*a**2 - 2*tan((c + d*x**2)/2)**4*a*b + tan((c + d*x* 
*2)/2)**4*b**2 - 2*tan((c + d*x**2)/2)**2*a**2 + 2*tan((c + d*x**2)/2)**2* 
b**2 + a**2 + 2*a*b + b**2),x)*a**4*b*d - 16*int(x/(tan((c + d*x**2)/2)**4 
*a**2 - 2*tan((c + d*x**2)/2)**4*a*b + tan((c + d*x**2)/2)**4*b**2 - 2*tan 
((c + d*x**2)/2)**2*a**2 + 2*tan((c + d*x**2)/2)**2*b**2 + a**2 + 2*a*b...