3.9.63 \(\int \sec (c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [863]

3.9.63.1 Optimal result
3.9.63.2 Mathematica [A] (verified)
3.9.63.3 Rubi [A] (verified)
3.9.63.4 Maple [A] (verified)
3.9.63.5 Fricas [A] (verification not implemented)
3.9.63.6 Sympy [F]
3.9.63.7 Maxima [A] (verification not implemented)
3.9.63.8 Giac [B] (verification not implemented)
3.9.63.9 Mupad [B] (verification not implemented)

3.9.63.1 Optimal result

Integrand size = 37, antiderivative size = 101 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {(b B+a (2 A+C)) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(3 A b+3 a B+2 b C) \tan (c+d x)}{3 d}+\frac {(b B+a C) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {b C \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

output
1/2*(B*b+a*(2*A+C))*arctanh(sin(d*x+c))/d+1/3*(3*A*b+3*B*a+2*C*b)*tan(d*x+ 
c)/d+1/2*(B*b+C*a)*sec(d*x+c)*tan(d*x+c)/d+1/3*b*C*sec(d*x+c)^2*tan(d*x+c) 
/d
 
3.9.63.2 Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 (b B+a (2 A+C)) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (6 A b+6 a B+6 b C+3 (b B+a C) \sec (c+d x)+2 b C \tan ^2(c+d x)\right )}{6 d} \]

input
Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c 
+ d*x]^2),x]
 
output
(3*(b*B + a*(2*A + C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(6*A*b + 6*a*B 
 + 6*b*C + 3*(b*B + a*C)*Sec[c + d*x] + 2*b*C*Tan[c + d*x]^2))/(6*d)
 
3.9.63.3 Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.270, Rules used = {3042, 4564, 3042, 4535, 3042, 4254, 24, 4534, 3042, 4257}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4564

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4535

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4254

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

\(\Big \downarrow \) 24

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

\(\Big \downarrow \) 4534

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

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

3.9.63.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4534
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

rule 4535
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* 
(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b   Int[(b*Cs 
c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) 
, x] /; FreeQ[{b, e, f, A, B, C, m}, x]
 

rule 4564
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*Cot[e + f*x]*((d*Csc[e + f*x])^ 
n/(f*(n + 2))), x] + Simp[1/(n + 2)   Int[(d*Csc[e + f*x])^n*Simp[A*a*(n + 
2) + (B*a*(n + 2) + b*(C*(n + 1) + A*(n + 2)))*Csc[e + f*x] + (a*C + B*b)*( 
n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] 
&&  !LtQ[n, -1]
 
3.9.63.4 Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.05

method result size
parts \(\frac {\left (A b +a B \right ) \tan \left (d x +c \right )}{d}+\frac {\left (B b +C a \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) \(106\)
derivativedivides \(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) a +C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \tan \left (d x +c \right ) b +B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) \(131\)
default \(\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) a +C a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \tan \left (d x +c \right ) b +B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) \(131\)
parallelrisch \(\frac {-\left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\left (A +\frac {C}{2}\right ) a +\frac {B b}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+\left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right ) \left (\left (A +\frac {C}{2}\right ) a +\frac {B b}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (a B +b \left (A +\frac {2 C}{3}\right )\right ) \sin \left (3 d x +3 c \right )+\left (B b +C a \right ) \sin \left (2 d x +2 c \right )+\left (a B +b \left (A +2 C \right )\right ) \sin \left (d x +c \right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) \(169\)
norman \(\frac {\frac {4 \left (3 A b +3 a B +C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}-\frac {\left (2 A b +2 a B -B b -C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (2 A b +2 a B +B b +C a +2 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (2 a A +B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 a A +B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) \(173\)
risch \(-\frac {i \left (3 B b \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C a \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-6 a B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 C b \,{\mathrm e}^{2 i \left (d x +c \right )}-3 B b \,{\mathrm e}^{i \left (d x +c \right )}-3 C a \,{\mathrm e}^{i \left (d x +c \right )}-6 A b -6 a B -4 C b \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a A}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a A}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) \(270\)

input
int(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_ 
RETURNVERBOSE)
 
output
(A*b+B*a)/d*tan(d*x+c)+(B*b+C*a)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d 
*x+c)+tan(d*x+c)))-C*b/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a*A/d*ln(sec(d 
*x+c)+tan(d*x+c))
 
3.9.63.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.27 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left ({\left (2 \, A + C\right )} a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A + C\right )} a + B b\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (3 \, B a + {\left (3 \, A + 2 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 2 \, C b + 3 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, a 
lgorithm="fricas")
 
output
1/12*(3*((2*A + C)*a + B*b)*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*((2*A 
 + C)*a + B*b)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(3*B*a + (3*A 
+ 2*C)*b)*cos(d*x + c)^2 + 2*C*b + 3*(C*a + B*b)*cos(d*x + c))*sin(d*x + c 
))/(d*cos(d*x + c)^3)
 
3.9.63.6 Sympy [F]

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

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 
output
Integral((a + b*sec(c + d*x))*(A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec 
(c + d*x), x)
 
3.9.63.7 Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.53 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b - 3 \, C a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a \tan \left (d x + c\right ) + 12 \, A b \tan \left (d x + c\right )}{12 \, d} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, a 
lgorithm="maxima")
 
output
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*b - 3*C*a*(2*sin(d*x + c)/(sin 
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3*B*b* 
(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x 
 + c) - 1)) + 12*A*a*log(sec(d*x + c) + tan(d*x + c)) + 12*B*a*tan(d*x + c 
) + 12*A*b*tan(d*x + c))/d
 
3.9.63.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 261 vs. \(2 (93) = 186\).

Time = 0.32 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.58 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (2 \, A a + C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, A a + C a + B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

input
integrate(sec(d*x+c)*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, a 
lgorithm="giac")
 
output
1/6*(3*(2*A*a + C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*A*a + 
 C*a + B*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a*tan(1/2*d*x + 1/ 
2*c)^5 - 3*C*a*tan(1/2*d*x + 1/2*c)^5 + 6*A*b*tan(1/2*d*x + 1/2*c)^5 - 3*B 
*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*a*tan(1/2* 
d*x + 1/2*c)^3 - 12*A*b*tan(1/2*d*x + 1/2*c)^3 - 4*C*b*tan(1/2*d*x + 1/2*c 
)^3 + 6*B*a*tan(1/2*d*x + 1/2*c) + 3*C*a*tan(1/2*d*x + 1/2*c) + 6*A*b*tan( 
1/2*d*x + 1/2*c) + 3*B*b*tan(1/2*d*x + 1/2*c) + 6*C*b*tan(1/2*d*x + 1/2*c) 
)/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
 
3.9.63.9 Mupad [B] (verification not implemented)

Time = 19.83 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.88 \[ \int \sec (c+d x) (a+b \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,a+\frac {B\,b}{2}+\frac {C\,a}{2}\right )}{4\,A\,a+2\,B\,b+2\,C\,a}\right )\,\left (2\,A\,a+B\,b+C\,a\right )}{d}-\frac {\left (2\,A\,b+2\,B\,a-B\,b-C\,a+2\,C\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A\,b-4\,B\,a-\frac {4\,C\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,b+2\,B\,a+B\,b+C\,a+2\,C\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

input
int(((a + b/cos(c + d*x))*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/cos(c + 
 d*x),x)
 
output
(atanh((4*tan(c/2 + (d*x)/2)*(A*a + (B*b)/2 + (C*a)/2))/(4*A*a + 2*B*b + 2 
*C*a))*(2*A*a + B*b + C*a))/d - (tan(c/2 + (d*x)/2)^5*(2*A*b + 2*B*a - B*b 
 - C*a + 2*C*b) + tan(c/2 + (d*x)/2)*(2*A*b + 2*B*a + B*b + C*a + 2*C*b) - 
 tan(c/2 + (d*x)/2)^3*(4*A*b + 4*B*a + (4*C*b)/3))/(d*(3*tan(c/2 + (d*x)/2 
)^2 - 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 - 1))