3.7.56 \(\int (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [656]

3.7.56.1 Optimal result
3.7.56.2 Mathematica [A] (verified)
3.7.56.3 Rubi [A] (verified)
3.7.56.4 Maple [A] (verified)
3.7.56.5 Fricas [A] (verification not implemented)
3.7.56.6 Sympy [F]
3.7.56.7 Maxima [A] (verification not implemented)
3.7.56.8 Giac [B] (verification not implemented)
3.7.56.9 Mupad [B] (verification not implemented)

3.7.56.1 Optimal result

Integrand size = 25, antiderivative size = 167 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \tan (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C (a+b \sec (c+d x))^2 \tan (c+d x)}{4 d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d} \]

output
a^3*A*x+1/8*b*(12*a^2*(2*A+C)+b^2*(4*A+3*C))*arctanh(sin(d*x+c))/d+1/2*a*( 
6*A*b^2+(a^2+4*b^2)*C)*tan(d*x+c)/d+1/8*b*(2*C*a^2+b^2*(4*A+3*C))*sec(d*x+ 
c)*tan(d*x+c)/d+1/4*a*C*(a+b*sec(d*x+c))^2*tan(d*x+c)/d+1/4*C*(a+b*sec(d*x 
+c))^3*tan(d*x+c)/d
 
3.7.56.2 Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.79 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 a^3 A d x+b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))+\left (8 a \left (3 A b^2+\left (a^2+3 b^2\right ) C\right )+b \left (4 A b^2+3 \left (4 a^2+b^2\right ) C\right ) \sec (c+d x)+2 b^3 C \sec ^3(c+d x)\right ) \tan (c+d x)+8 a b^2 C \tan ^3(c+d x)}{8 d} \]

input
Integrate[(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]
 
output
(8*a^3*A*d*x + b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*ArcTanh[Sin[c + d*x] 
] + (8*a*(3*A*b^2 + (a^2 + 3*b^2)*C) + b*(4*A*b^2 + 3*(4*a^2 + b^2)*C)*Sec 
[c + d*x] + 2*b^3*C*Sec[c + d*x]^3)*Tan[c + d*x] + 8*a*b^2*C*Tan[c + d*x]^ 
3)/(8*d)
 
3.7.56.3 Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4545, 3042, 4544, 27, 3042, 4536, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4545

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int 3 (a+b \sec (c+d x)) \left (4 A a^2+b (8 A+5 C) \sec (c+d x) a+\left (2 C a^2+b^2 (4 A+3 C)\right ) \sec ^2(c+d x)\right )dx+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\int (a+b \sec (c+d x)) \left (4 A a^2+b (8 A+5 C) \sec (c+d x) a+\left (2 C a^2+b^2 (4 A+3 C)\right ) \sec ^2(c+d x)\right )dx+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (4 A a^2+b (8 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (2 C a^2+b^2 (4 A+3 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 4536

\(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \left (8 A a^3+4 \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sec ^2(c+d x) a+b \left (12 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \sec (c+d x)\right )dx+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{4} \left (\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (8 a^3 A x+\frac {b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {4 a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \tan (c+d x)}{d}\right )+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}\)

input
Int[(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]
 
output
(C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + ((b*(2*a^2*C + b^2*(4*A + 
3*C))*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a*C*(a + b*Sec[c + d*x])^2*Tan[c 
 + d*x])/d + (8*a^3*A*x + (b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*ArcTanh[ 
Sin[c + d*x]])/d + (4*a*(6*A*b^2 + (a^2 + 4*b^2)*C)*Tan[c + d*x])/d)/2)/4
 

3.7.56.3.1 Defintions of rubi rules used

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 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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

rule 4536
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + 
 f*x]*(Cot[e + f*x]/(2*f)), x] + Simp[1/2   Int[Simp[2*A*a + (2*B*a + b*(2* 
A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a 
, b, e, f, A, B, C}, x]
 

rule 4544
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot 
[e + f*x]*((a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[( 
a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m 
)*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
 

rule 4545
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. 
) + (a_))^(m_.), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^ 
m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[(a + b*Csc[e + f*x])^(m - 1)*Simp 
[a*A*(m + 1) + (A*b*(m + 1) + b*C*m)*Csc[e + f*x] + a*C*m*Csc[e + f*x]^2, x 
], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 
 0]
 
3.7.56.4 Maple [A] (verified)

Time = 1.05 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.10

method result size
parts \(a^{3} A x +\frac {\left (A \,b^{3}+3 a^{2} b C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 a A \,b^{2}+a^{3} C \right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {3 C a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) \(183\)
derivativedivides \(\frac {a^{3} A \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \tan \left (d x +c \right ) a \,b^{2}-3 C a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(209\)
default \(\frac {a^{3} A \left (d x +c \right )+a^{3} C \tan \left (d x +c \right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \tan \left (d x +c \right ) a \,b^{2}-3 C a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(209\)
parallelrisch \(\frac {-96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (\left (\frac {A}{6}+\frac {C}{8}\right ) b^{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (\left (\frac {A}{6}+\frac {C}{8}\right ) b^{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 a^{3} A x d \cos \left (2 d x +2 c \right )+8 a^{3} A x d \cos \left (4 d x +4 c \right )+48 a \left (b^{2} \left (A +\frac {4 C}{3}\right )+\frac {C \,a^{2}}{3}\right ) \sin \left (2 d x +2 c \right )+8 \left (\left (A +\frac {3 C}{4}\right ) b^{2}+3 C \,a^{2}\right ) b \sin \left (3 d x +3 c \right )+24 a \left (b^{2} \left (A +\frac {2 C}{3}\right )+\frac {C \,a^{2}}{3}\right ) \sin \left (4 d x +4 c \right )+8 b \left (\left (A +\frac {11 C}{4}\right ) b^{2}+3 C \,a^{2}\right ) \sin \left (d x +c \right )+24 a^{3} A x d}{8 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) \(298\)
norman \(\frac {a^{3} A x +a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {\left (24 a A \,b^{2}-4 A \,b^{3}+8 a^{3} C -12 a^{2} b C +24 C a \,b^{2}-5 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (24 a A \,b^{2}+4 A \,b^{3}+8 a^{3} C +12 a^{2} b C +24 C a \,b^{2}+5 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (72 a A \,b^{2}-4 A \,b^{3}+24 a^{3} C -12 a^{2} b C +40 C a \,b^{2}+3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}-\frac {\left (72 a A \,b^{2}+4 A \,b^{3}+24 a^{3} C +12 a^{2} b C +40 C a \,b^{2}-3 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}-4 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {b \left (24 a^{2} A +4 A \,b^{2}+12 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {b \left (24 a^{2} A +4 A \,b^{2}+12 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) \(404\)
risch \(a^{3} A x -\frac {i \left (4 A \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+12 C \,a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+3 C \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-24 A a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-8 C \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}+4 A \,b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+12 C \,a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}+11 C \,b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-72 A a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-24 C \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-48 C a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-4 A \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-12 C \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}-11 C \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-72 A a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-24 C \,a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-64 C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-4 A \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-12 C \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-24 a A \,b^{2}-8 a^{3} C -16 C a \,b^{2}\right )}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} A}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,a^{2}}{2 d}-\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} A}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,a^{2}}{2 d}+\frac {3 b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) \(537\)

input
int((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
 
output
a^3*A*x+(A*b^3+3*C*a^2*b)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+t 
an(d*x+c)))+(3*A*a*b^2+C*a^3)/d*tan(d*x+c)+C*b^3/d*(-(-1/4*sec(d*x+c)^3-3/ 
8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+3*A*a^2*b/d*ln(sec 
(d*x+c)+tan(d*x+c))-3*C*a*b^2/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)
 
3.7.56.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.19 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, A a^{3} d x \cos \left (d x + c\right )^{4} + {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, C a b^{2} \cos \left (d x + c\right ) + 2 \, C b^{3} + 8 \, {\left (C a^{3} + {\left (3 \, A + 2 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (12 \, C a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \]

input
integrate((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
 
output
1/16*(16*A*a^3*d*x*cos(d*x + c)^4 + (12*(2*A + C)*a^2*b + (4*A + 3*C)*b^3) 
*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (12*(2*A + C)*a^2*b + (4*A + 3*C)* 
b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*C*a*b^2*cos(d*x + c) + 2 
*C*b^3 + 8*(C*a^3 + (3*A + 2*C)*a*b^2)*cos(d*x + c)^3 + (12*C*a^2*b + (4*A 
 + 3*C)*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
 
3.7.56.6 Sympy [F]

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

input
integrate((a+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)
 
output
Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**3, x)
 
3.7.56.7 Maxima [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.52 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (d x + c\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{2} - C b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 4 \, A b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{2} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 16 \, C a^{3} \tan \left (d x + c\right ) + 48 \, A a b^{2} \tan \left (d x + c\right )}{16 \, d} \]

input
integrate((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
 
output
1/16*(16*(d*x + c)*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a*b^2 - 
C*b^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + 
 c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*C*a^2 
*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin( 
d*x + c) - 1)) - 4*A*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d* 
x + c) + 1) + log(sin(d*x + c) - 1)) + 48*A*a^2*b*log(sec(d*x + c) + tan(d 
*x + c)) + 16*C*a^3*tan(d*x + c) + 48*A*a*b^2*tan(d*x + c))/d
 
3.7.56.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 526 vs. \(2 (157) = 314\).

Time = 0.36 (sec) , antiderivative size = 526, normalized size of antiderivative = 3.15 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, {\left (d x + c\right )} A a^{3} + {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, C b^{3} \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 )}^{4}}}{8 \, d} \]

input
integrate((a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")
 
output
1/8*(8*(d*x + c)*A*a^3 + (24*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 3*C*b^3)*log 
(abs(tan(1/2*d*x + 1/2*c) + 1)) - (24*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 3*C 
*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(8*C*a^3*tan(1/2*d*x + 1/2*c) 
^7 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 
 + 24*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 4*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 5* 
C*b^3*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2* 
b*tan(1/2*d*x + 1/2*c)^5 - 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 40*C*a*b^2* 
tan(1/2*d*x + 1/2*c)^5 + 4*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*C*b^3*tan(1/2* 
d*x + 1/2*c)^5 + 24*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*b*tan(1/2*d*x 
+ 1/2*c)^3 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 40*C*a*b^2*tan(1/2*d*x + 
1/2*c)^3 + 4*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 
 - 8*C*a^3*tan(1/2*d*x + 1/2*c) - 12*C*a^2*b*tan(1/2*d*x + 1/2*c) - 24*A*a 
*b^2*tan(1/2*d*x + 1/2*c) - 24*C*a*b^2*tan(1/2*d*x + 1/2*c) - 4*A*b^3*tan( 
1/2*d*x + 1/2*c) - 5*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 
 1)^4)/d
 
3.7.56.9 Mupad [B] (verification not implemented)

Time = 17.89 (sec) , antiderivative size = 1547, normalized size of antiderivative = 9.26 \[ \int (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

input
int((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^3,x)
 
output
((3*A*a^3*atan((64*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^6*sin(c/2 + (d*x) 
/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) + 192*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 
576*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 72*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 1 
44*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2 + (d*x)/2) + 240*A* 
C*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/ 
2 + (d*x)/2)*(64*A^2*a^6 + 16*A^2*b^6 + 9*C^2*b^6 + 192*A^2*a^2*b^4 + 576* 
A^2*a^4*b^2 + 72*C^2*a^2*b^4 + 144*C^2*a^4*b^2 + 24*A*C*b^6 + 240*A*C*a^2* 
b^4 + 576*A*C*a^4*b^2))))/4 + (3*A*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + 
(d*x)/2)))/8 + (9*C*b^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/32 + 
 (A*b^3*sin(3*c + 3*d*x))/8 + (C*a^3*sin(2*c + 2*d*x))/4 + (C*a^3*sin(4*c 
+ 4*d*x))/8 + (3*C*b^3*sin(3*c + 3*d*x))/32 + (A*b^3*sin(c + d*x))/8 + (11 
*C*b^3*sin(c + d*x))/32 + (3*C*a^2*b*sin(c + d*x))/8 + (9*A*a^2*b*atanh(si 
n(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/4 + (9*C*a^2*b*atanh(sin(c/2 + (d*x) 
/2)/cos(c/2 + (d*x)/2)))/8 + (3*A*a*b^2*sin(2*c + 2*d*x))/4 + (3*A*a*b^2*s 
in(4*c + 4*d*x))/8 + C*a*b^2*sin(2*c + 2*d*x) + (3*C*a^2*b*sin(3*c + 3*d*x 
))/8 + (C*a*b^2*sin(4*c + 4*d*x))/4 + A*a^3*atan((64*A^2*a^6*sin(c/2 + (d* 
x)/2) + 16*A^2*b^6*sin(c/2 + (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) + 192 
*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 72* 
C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 144*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A 
*C*b^6*sin(c/2 + (d*x)/2) + 240*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 576*A*...