\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [919]

Optimal result

Integrand size = 39, antiderivative size = 202 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=-\frac {\left (3 a b B-a^2 (2 A+C)-b^2 (A+2 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \tan (c+d x)}{2 b \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:

-(3*B*a*b-a^2*(2*A+C)-b^2*(A+2*C))*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/ 
(a+b)^(1/2))/(a-b)^(5/2)/(a+b)^(5/2)/d-1/2*(A*b^2-a*(B*b-C*a))*tan(d*x+c)/ 
b/(a^2-b^2)/d/(a+b*sec(d*x+c))^2+1/2*(B*a^2*b+2*B*b^3+a^3*C-a*b^2*(3*A+4*C 
))*tan(d*x+c)/b/(a^2-b^2)^2/d/(a+b*sec(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.37 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.03 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {4 i \left (-3 a b B+a^2 (2 A+C)+b^2 (A+2 C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x))^2 (\cos (c)-i \sin (c))}{\left (a^2-b^2\right )^{5/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {a \sec (c) \left (\left (2 A b^4+5 a^3 b B+4 a b^3 B+a^4 C-a^2 b^2 (11 A+10 C)\right ) \sin (d x)+\left (-2 A b^4-3 a^3 b B+a^4 C+a^2 b^2 (5 A+2 C)\right ) \sin (2 c+d x)+a \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \sin (c+2 d x)\right )-\left (a^2+2 b^2\right ) \left (A b^3+2 a^3 B+a b^2 B-a^2 b (4 A+3 C)\right ) \tan (c)}{\left (a^3-a b^2\right )^2}\right )}{2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^3} \] Input:

Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Se 
c[c + d*x])^3,x]
 

Output:

((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2) 
*(((-4*I)*(-3*a*b*B + a^2*(2*A + C) + b^2*(A + 2*C))*ArcTan[((I*Cos[c] + S 
in[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 - b^2]*Sqrt[(C 
os[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])^2*(Cos[c] - I*Sin[c]))/((a^2 - 
 b^2)^(5/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (a*Sec[c]*((2*A*b^4 + 5*a^3*b*B 
 + 4*a*b^3*B + a^4*C - a^2*b^2*(11*A + 10*C))*Sin[d*x] + (-2*A*b^4 - 3*a^3 
*b*B + a^4*C + a^2*b^2*(5*A + 2*C))*Sin[2*c + d*x] + a*(A*b^3 + 2*a^3*B + 
a*b^2*B - a^2*b*(4*A + 3*C))*Sin[c + 2*d*x]) - (a^2 + 2*b^2)*(A*b^3 + 2*a^ 
3*B + a*b^2*B - a^2*b*(4*A + 3*C))*Tan[c])/(a^3 - a*b^2)^2))/(2*d*(A + 2*C 
 + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^3)
 

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {3042, 4568, 3042, 4491, 25, 27, 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.

\(\displaystyle \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3}dx\)

\(\Big \downarrow \) 4568

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4491

\(\displaystyle -\frac {-\frac {\int -\frac {b \left (-\left ((2 A+C) a^2\right )+3 b B a-b^2 (A+2 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\frac {\int \frac {b \left (-\left ((2 A+C) a^2\right )+3 b B a-b^2 (A+2 C)\right ) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {b \left (-\left (a^2 (2 A+C)\right )+3 a b B-b^2 (A+2 C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {b \left (-\left (a^2 (2 A+C)\right )+3 a b B-b^2 (A+2 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 4318

\(\displaystyle -\frac {\frac {\left (-\left (a^2 (2 A+C)\right )+3 a b B-b^2 (A+2 C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\left (-\left (a^2 (2 A+C)\right )+3 a b B-b^2 (A+2 C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a^2-b^2}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 3138

\(\displaystyle -\frac {\frac {2 \left (-\left (a^2 (2 A+C)\right )+3 a b B-b^2 (A+2 C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{d \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {\tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {\frac {2 b \left (-\left (a^2 (2 A+C)\right )+3 a b B-b^2 (A+2 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (a^3 C+a^2 b B-a b^2 (3 A+4 C)+2 b^3 B\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b \left (a^2-b^2\right )}\)

Input:

Int[(Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + 
d*x])^3,x]
 

Output:

-1/2*((A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Sec[c 
+ d*x])^2) - ((2*b*(3*a*b*B - a^2*(2*A + C) - b^2*(A + 2*C))*ArcTanh[(Sqrt 
[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^ 
2)*d) - ((a^2*b*B + 2*b^3*B + a^3*C - a*b^2*(3*A + 4*C))*Tan[c + d*x])/((a 
^2 - b^2)*d*(a + b*Sec[c + d*x])))/(2*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 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 4491

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e 
 + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1 
/((m + 1)*(a^2 - b^2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp 
[(a*A - b*B)*(m + 1) - (A*b - a*B)*(m + 2)*Csc[e + f*x], x], x], x] /; Free 
Q[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m 
, -1]
 

rule 4568

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cot[e + f*x]*((a + b*Csc[e + f*x] 
)^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) 
 Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m 
+ 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Csc[e + f*x], 
x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^ 
2, 0]
 

Maple [A] (verified)

Time = 0.54 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.33

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {\left (4 a A b +A \,b^{2}-2 B \,a^{2}-B a b -2 B \,b^{2}+C \,a^{2}+4 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a A b -A \,b^{2}-2 B \,a^{2}+B a b -2 B \,b^{2}-C \,a^{2}+4 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}+\frac {\left (2 a^{2} A +A \,b^{2}-3 B a b +C \,a^{2}+2 C \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(268\)
default	\(\frac {-\frac {2 \left (-\frac {\left (4 a A b +A \,b^{2}-2 B \,a^{2}-B a b -2 B \,b^{2}+C \,a^{2}+4 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a A b -A \,b^{2}-2 B \,a^{2}+B a b -2 B \,b^{2}-C \,a^{2}+4 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a^{2}-2 a b +b^{2}\right )}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}+\frac {\left (2 a^{2} A +A \,b^{2}-3 B a b +C \,a^{2}+2 C \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\)	\(268\)
risch	\(\text {Expression too large to display}\)	\(1247\)

Input:

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x,method 
=_RETURNVERBOSE)
 

Output:

1/d*(-2*(-1/2*(4*A*a*b+A*b^2-2*B*a^2-B*a*b-2*B*b^2+C*a^2+4*C*a*b)/(a-b)/(a 
^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*(4*A*a*b-A*b^2-2*B*a^2+B*a*b-2*B*b^ 
2-C*a^2+4*C*a*b)/(a+b)/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/ 
2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)^2+(2*A*a^2+A*b^2-3*B*a*b+C*a^2+2*C*b^ 
2)/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+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. 395 vs. \(2 (185) = 370\).

Time = 0.14 (sec) , antiderivative size = 848, normalized size of antiderivative = 4.20 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, 
 algorithm="fricas")
 

Output:

[1/4*(((2*A + C)*a^2*b^2 - 3*B*a*b^3 + (A + 2*C)*b^4 + ((2*A + C)*a^4 - 3* 
B*a^3*b + (A + 2*C)*a^2*b^2)*cos(d*x + c)^2 + 2*((2*A + C)*a^3*b - 3*B*a^2 
*b^2 + (A + 2*C)*a*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + 
 c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a 
)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b 
^2)) + 2*(C*a^5 + B*a^4*b - (3*A + 5*C)*a^3*b^2 + B*a^2*b^3 + (3*A + 4*C)* 
a*b^4 - 2*B*b^5 + (2*B*a^5 - (4*A + 3*C)*a^4*b - B*a^3*b^2 + (5*A + 3*C)*a 
^2*b^3 - B*a*b^4 - A*b^5)*cos(d*x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 
3*a^4*b^4 - a^2*b^6)*d*cos(d*x + c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - 
 a*b^7)*d*cos(d*x + c) + (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d), 1/2*( 
((2*A + C)*a^2*b^2 - 3*B*a*b^3 + (A + 2*C)*b^4 + ((2*A + C)*a^4 - 3*B*a^3* 
b + (A + 2*C)*a^2*b^2)*cos(d*x + c)^2 + 2*((2*A + C)*a^3*b - 3*B*a^2*b^2 + 
 (A + 2*C)*a*b^3)*cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)* 
(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) + (C*a^5 + B*a^4*b - (3*A 
 + 5*C)*a^3*b^2 + B*a^2*b^3 + (3*A + 4*C)*a*b^4 - 2*B*b^5 + (2*B*a^5 - (4* 
A + 3*C)*a^4*b - B*a^3*b^2 + (5*A + 3*C)*a^2*b^3 - B*a*b^4 - A*b^5)*cos(d* 
x + c))*sin(d*x + c))/((a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*cos(d*x + 
 c)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*d*cos(d*x + c) + (a^6*b^ 
2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*d)]
 

Sympy [F]

\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{3}}\, dx \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**3, 
x)
 

Output:

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)/(a + b*sec( 
c + d*x))**3, x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,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*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 510 vs. \(2 (185) = 370\).

Time = 0.30 (sec) , antiderivative size = 510, normalized size of antiderivative = 2.52 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x, 
 algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 16.84 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.39 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A\,b^2-2\,B\,a^2-2\,B\,b^2+C\,a^2+4\,A\,a\,b-B\,a\,b+4\,C\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,b^2+2\,B\,a^2+2\,B\,b^2+C\,a^2-4\,A\,a\,b-B\,a\,b-4\,C\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )}+\frac {\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,A\,a^2+A\,b^2+C\,a^2+2\,C\,b^2-3\,B\,a\,b\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \] Input:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)*(a + b/cos(c + d 
*x))^3),x)
 

Output:

((tan(c/2 + (d*x)/2)^3*(A*b^2 - 2*B*a^2 - 2*B*b^2 + C*a^2 + 4*A*a*b - B*a* 
b + 4*C*a*b))/((a + b)^2*(a - b)) + (tan(c/2 + (d*x)/2)*(A*b^2 + 2*B*a^2 + 
 2*B*b^2 + C*a^2 - 4*A*a*b - B*a*b - 4*C*a*b))/((a + b)*(a^2 - 2*a*b + b^2 
)))/(d*(2*a*b - tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^ 
4*(a^2 - 2*a*b + b^2) + a^2 + b^2)) + (atanh((tan(c/2 + (d*x)/2)*(2*a - 2* 
b)*(a^2 - 2*a*b + b^2))/(2*(a + b)^(1/2)*(a - b)^(5/2)))*(2*A*a^2 + A*b^2 
+ C*a^2 + 2*C*b^2 - 3*B*a*b))/(d*(a + b)^(5/2)*(a - b)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1087, normalized size of antiderivative = 5.38 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

int(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x)
 

Output:

(8*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t( - a**2 + b**2))*cos(c + d*x)*a**4*b + 4*sqrt( - a**2 + b**2)*atan((tan( 
(c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a* 
*3*b*c - 8*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2 
)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**2*b**3 + 8*sqrt( - a**2 + b**2) 
*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos( 
c + d*x)*a*b**3*c - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan( 
(c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**5 - 2*sqrt( - a** 
2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b* 
*2))*sin(c + d*x)**2*a**4*c + 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2 
)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**3*b**2 
- 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sq 
rt( - a**2 + b**2))*sin(c + d*x)**2*a**2*b**2*c + 4*sqrt( - a**2 + b**2)*a 
tan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**5 + 
 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqr 
t( - a**2 + b**2))*a**4*c + 6*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)* 
a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**2*b**2*c - 4*sqrt( - a**2 
 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b** 
2))*a*b**4 + 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d* 
x)/2)*b)/sqrt( - a**2 + b**2))*b**4*c - 2*cos(c + d*x)*sin(c + d*x)*a**...