\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [912]

Optimal result

Integrand size = 33, antiderivative size = 138 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {A x}{a^2}-\frac {2 \left (2 a^2 A b-A b^3-a^3 B+a^2 b C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.98 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.17 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (A x (b+a \cos (c+d x))-\frac {2 i \left (A b^3+a^3 B-a^2 b (2 A+C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x)) (\cos (c)-i \sin (c))}{\left (a^2-b^2\right )^{3/2} d \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {\left (A b^2+a (-b B+a C)\right ) (-b \sin (c)+a \sin (d x))}{(a-b) (a+b) d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}\right )}{a^2 (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.15, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4548, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4548

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4407

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 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 4548

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

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.31

method	result	size
derivativedivides	\(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {-\frac {2 \left (A \,b^{2}-B a b +C \,a^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (2 A \,a^{2} b -A \,b^{3}-B \,a^{3}+a^{2} b C \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}}{d}\)	\(181\)
default	\(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {-\frac {2 \left (A \,b^{2}-B a b +C \,a^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (2 A \,a^{2} b -A \,b^{3}-B \,a^{3}+a^{2} b C \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}}{d}\)	\(181\)
risch	\(\frac {A x}{a^{2}}+\frac {2 i \left (A \,b^{2}-B a b +C \,a^{2}\right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{2} \left (a^{2}-b^{2}\right ) d \left ({\mathrm e}^{2 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A \,b^{3}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) b C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A \,b^{3}}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) b C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\)	\(755\)

Input:

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

Output:

1/d*(2*A/a^2*arctan(tan(1/2*d*x+1/2*c))+2/a^2*(-(A*b^2-B*a*b+C*a^2)*a/(a^2 
-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*A*a^2*b-A*b^3-B*a^3+C*a^2*b)/(a+b)/(a-b)/((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. 266 vs. \(2 (127) = 254\).

Time = 0.12 (sec) , antiderivative size = 594, normalized size of antiderivative = 4.30 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {2 \, {\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x - {\left (B a^{3} b - {\left (2 \, A + C\right )} a^{2} b^{2} + A b^{4} + {\left (B a^{4} - {\left (2 \, A + C\right )} a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (C a^{5} - B a^{4} b + {\left (A - C\right )} a^{3} b^{2} + B a^{2} b^{3} - A a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac {{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x + {\left (B a^{3} b - {\left (2 \, A + C\right )} a^{2} b^{2} + A b^{4} + {\left (B a^{4} - {\left (2 \, A + C\right )} a^{3} b + A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{5} - B a^{4} b + {\left (A - C\right )} a^{3} b^{2} + B a^{2} b^{3} - A a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.61 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (B a^{3} - 2 \, A a^{2} b - C a^{2} b + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {{\left (d x + c\right )} A}{a^{2}} - \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 21.49 (sec) , antiderivative size = 4544, normalized size of antiderivative = 32.93 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(2*A*atan(((A*((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 + B^2*a^6 - 2*A 
^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C 
^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2*A*C*a^2*b^4 + 4 
*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (A*((32*(A*a^4*b^5 - B* 
a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 - B*a^6*b^3 + B*a^7*b^2 + C*a^5*b^4 
- C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a 
^3*b^3 - a^4*b^2) - (A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 
 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a 
^3*b^2)))*1i)/a^2))/a^2 + (A*((32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 
+ B^2*a^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3* 
A^2*a^4*b^2 + C^2*a^4*b^2 - 4*A*B*a^5*b - 2*B*C*a^5*b + 2*A*B*a^3*b^3 - 2* 
A*C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (A*((32* 
(A*a^4*b^5 - B*a^9 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 - B*a^6*b^3 + B*a^7*b 
^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + B*a^8*b + C*a^8*b))/( 
a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4* 
b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^ 
5 - a^2*b^3 - a^3*b^2)))*1i)/a^2))/a^2)/((64*(A^3*b^5 + A*B^2*a^5 - A^2*B* 
a^5 - A^3*a*b^4 + 2*A^3*a^4*b - 3*A^3*a^2*b^3 + 2*A^3*a^3*b^2 - 3*A^2*B*a^ 
4*b - A^2*C*a*b^4 + A^2*C*a^4*b + A^2*B*a^2*b^3 + A^2*B*a^3*b^2 + A*C^2*a^ 
3*b^2 - A^2*C*a^2*b^3 + 3*A^2*C*a^3*b^2 - 2*A*B*C*a^4*b))/(a^5*b + a^6 ...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 497, normalized size of antiderivative = 3.60 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a^{3} b -2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a^{2} b c +2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) \cos \left (d x +c \right ) a \,b^{3}-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a^{2} b^{2}-2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) a \,b^{2} c +2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right ) b^{4}+\cos \left (d x +c \right ) a^{5} d x -2 \cos \left (d x +c \right ) a^{3} b^{2} d x +\cos \left (d x +c \right ) a \,b^{4} d x +\sin \left (d x +c \right ) a^{4} c -\sin \left (d x +c \right ) a^{2} b^{2} c +a^{4} b d x -2 a^{2} b^{3} d x +b^{5} d x}{a d \left (\cos \left (d x +c \right ) a^{5}-2 \cos \left (d x +c \right ) a^{3} b^{2}+\cos \left (d x +c \right ) a \,b^{4}+a^{4} b -2 a^{2} b^{3}+b^{5}\right )} \] Input:

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

Output:

( - 2*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 - 2*sqrt( - a**2 + b**2)*atan((t 
an((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x) 
*a**2*b*c + 2*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 - 2*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 - 2*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**2*c + 2*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 + cos(c + d*x) 
*a**5*d*x - 2*cos(c + d*x)*a**3*b**2*d*x + cos(c + d*x)*a*b**4*d*x + sin(c 
 + d*x)*a**4*c - sin(c + d*x)*a**2*b**2*c + a**4*b*d*x - 2*a**2*b**3*d*x + 
 b**5*d*x)/(a*d*(cos(c + d*x)*a**5 - 2*cos(c + d*x)*a**3*b**2 + cos(c + d* 
x)*a*b**4 + a**4*b - 2*a**2*b**3 + b**5))