3.5.30 \(\int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [430]

3.5.30.1 Optimal result
3.5.30.2 Mathematica [A] (verified)
3.5.30.3 Rubi [A] (verified)
3.5.30.4 Maple [A] (verified)
3.5.30.5 Fricas [A] (verification not implemented)
3.5.30.6 Sympy [F]
3.5.30.7 Maxima [B] (verification not implemented)
3.5.30.8 Giac [A] (verification not implemented)
3.5.30.9 Mupad [B] (verification not implemented)

3.5.30.1 Optimal result

Integrand size = 33, antiderivative size = 162 \[ \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^3 A x+\frac {a^3 (28 A+20 B+15 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^3 (4 A+4 B+3 C) \tan (c+d x)}{8 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac {(4 B+3 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 a d}+\frac {(12 A+20 B+15 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{24 d} \]

output
a^3*A*x+1/8*a^3*(28*A+20*B+15*C)*arctanh(sin(d*x+c))/d+5/8*a^3*(4*A+4*B+3* 
C)*tan(d*x+c)/d+1/4*C*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+1/12*(4*B+3*C)*(a^2+ 
a^2*sec(d*x+c))^2*tan(d*x+c)/a/d+1/24*(12*A+20*B+15*C)*(a^3+a^3*sec(d*x+c) 
)*tan(d*x+c)/d
 
3.5.30.2 Mathematica [A] (verified)

Time = 5.50 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.58 \[ \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (24 A d x+(84 A+60 B+45 C) \text {arctanh}(\sin (c+d x))+3 \left (24 A+32 (B+C)+(4 A+12 B+15 C) \sec (c+d x)+2 C \sec ^3(c+d x)\right ) \tan (c+d x)+8 (B+3 C) \tan ^3(c+d x)\right )}{24 d} \]

input
Integrate[(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x 
]
 
output
(a^3*(24*A*d*x + (84*A + 60*B + 45*C)*ArcTanh[Sin[c + d*x]] + 3*(24*A + 32 
*(B + C) + (4*A + 12*B + 15*C)*Sec[c + d*x] + 2*C*Sec[c + d*x]^3)*Tan[c + 
d*x] + 8*(B + 3*C)*Tan[c + d*x]^3))/(24*d)
 
3.5.30.3 Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4542, 3042, 4405, 3042, 4405, 27, 3042, 4402, 3042, 4254, 24, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4542

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4405

\(\displaystyle \frac {\frac {1}{3} \int (\sec (c+d x) a+a)^2 \left (12 A a^2+(12 A+20 B+15 C) \sec (c+d x) a^2\right )dx+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (12 A a^2+(12 A+20 B+15 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 4405

\(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int 3 (\sec (c+d x) a+a) \left (8 A a^3+5 (4 A+4 B+3 C) \sec (c+d x) a^3\right )dx+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int (\sec (c+d x) a+a) \left (8 A a^3+5 (4 A+4 B+3 C) \sec (c+d x) a^3\right )dx+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (8 A a^3+5 (4 A+4 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 4402

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (5 a^4 (4 A+4 B+3 C) \int \sec ^2(c+d x)dx+a^4 (28 A+20 B+15 C) \int \sec (c+d x)dx+8 a^4 A x\right )+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (a^4 (28 A+20 B+15 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+5 a^4 (4 A+4 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+8 a^4 A x\right )+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (-\frac {5 a^4 (4 A+4 B+3 C) \int 1d(-\tan (c+d x))}{d}+a^4 (28 A+20 B+15 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 a^4 A x\right )+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (a^4 (28 A+20 B+15 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^4 (4 A+4 B+3 C) \tan (c+d x)}{d}+8 a^4 A x\right )+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\frac {a^4 (28 A+20 B+15 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^4 (4 A+4 B+3 C) \tan (c+d x)}{d}+8 a^4 A x\right )+\frac {(12 A+20 B+15 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )+\frac {(4 B+3 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}\)

input
Int[(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 
output
(C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d) + (((4*B + 3*C)*(a^2 + a^2*S 
ec[c + d*x])^2*Tan[c + d*x])/(3*d) + (((12*A + 20*B + 15*C)*(a^4 + a^4*Sec 
[c + d*x])*Tan[c + d*x])/(2*d) + (3*(8*a^4*A*x + (a^4*(28*A + 20*B + 15*C) 
*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(4*A + 4*B + 3*C)*Tan[c + d*x])/d))/2)/ 
3)/(4*a)
 

3.5.30.3.1 Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 4402
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + 
 (c_)), x_Symbol] :> Simp[a*c*x, x] + (Simp[b*d   Int[Csc[e + f*x]^2, x], x 
] + Simp[(b*c + a*d)   Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f 
}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
 

rule 4405
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m 
 - 1)/(f*m)), x] + Simp[1/m   Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + 
 (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, 
f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 
*m]
 

rule 4542
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/(b*(m + 1))   I 
nt[(a + b*Csc[e + f*x])^m*Simp[A*b*(m + 1) + (a*C*m + b*B*(m + 1))*Csc[e + 
f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] 
 &&  !LtQ[m, -2^(-1)]
 
3.5.30.4 Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.27

method result size
parts \(a^{3} A x +\frac {\left (3 a^{3} A +B \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {\left (B \,a^{3}+3 a^{3} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (a^{3} A +3 B \,a^{3}+3 a^{3} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 a^{3} A +3 B \,a^{3}+a^{3} C \right ) \tan \left (d x +c \right )}{d}+\frac {a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(206\)
parallelrisch \(\frac {a^{3} \left (-14 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {5 B}{7}+\frac {15 C}{28}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+14 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {5 B}{7}+\frac {15 C}{28}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 d x A \cos \left (2 d x +2 c \right )+d x A \cos \left (4 d x +4 c \right )+\left (6 A +\frac {26 B}{3}+10 C \right ) \sin \left (2 d x +2 c \right )+\left (A +3 B +\frac {15 C}{4}\right ) \sin \left (3 d x +3 c \right )+\left (3 C +3 A +\frac {11 B}{3}\right ) \sin \left (4 d x +4 c \right )+\left (3 B +\frac {23 C}{4}+A \right ) \sin \left (d x +c \right )+3 d x A \right )}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) \(226\)
norman \(\frac {a^{3} A x +a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-4 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-4 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-\frac {5 a^{3} \left (4 A +4 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {a^{3} \left (28 A +49 C +44 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (204 A +220 B +165 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {a^{3} \left (228 A +292 B +219 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a^{3} \left (28 A +20 B +15 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{3} \left (28 A +20 B +15 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) \(270\)
derivativedivides \(\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \tan \left (d x +c \right )}{d}\) \(297\)
default \(\frac {a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+3 a^{3} A \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \tan \left (d x +c \right )}{d}\) \(297\)
risch \(a^{3} A x -\frac {i a^{3} \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+36 B \,{\mathrm e}^{7 i \left (d x +c \right )}+45 C \,{\mathrm e}^{7 i \left (d x +c \right )}-72 A \,{\mathrm e}^{6 i \left (d x +c \right )}-72 B \,{\mathrm e}^{6 i \left (d x +c \right )}-24 C \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+36 B \,{\mathrm e}^{5 i \left (d x +c \right )}+69 C \,{\mathrm e}^{5 i \left (d x +c \right )}-216 A \,{\mathrm e}^{4 i \left (d x +c \right )}-264 B \,{\mathrm e}^{4 i \left (d x +c \right )}-216 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-36 B \,{\mathrm e}^{3 i \left (d x +c \right )}-69 C \,{\mathrm e}^{3 i \left (d x +c \right )}-216 A \,{\mathrm e}^{2 i \left (d x +c \right )}-280 B \,{\mathrm e}^{2 i \left (d x +c \right )}-264 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,{\mathrm e}^{i \left (d x +c \right )}-36 B \,{\mathrm e}^{i \left (d x +c \right )}-45 C \,{\mathrm e}^{i \left (d x +c \right )}-72 A -88 B -72 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) \(424\)

input
int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVER 
BOSE)
 
output
a^3*A*x+(3*A*a^3+B*a^3)/d*ln(sec(d*x+c)+tan(d*x+c))-(B*a^3+3*C*a^3)/d*(-2/ 
3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(A*a^3+3*B*a^3+3*C*a^3)/d*(1/2*sec(d*x+c)*t 
an(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(3*A*a^3+3*B*a^3+C*a^3)/d*tan(d*x 
+c)+a^3*C/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)))
 
3.5.30.5 Fricas [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.07 \[ \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {48 \, A a^{3} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (28 \, A + 20 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (28 \, A + 20 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (9 \, A + 11 \, B + 9 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 12 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, C a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

input
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm= 
"fricas")
 
output
1/48*(48*A*a^3*d*x*cos(d*x + c)^4 + 3*(28*A + 20*B + 15*C)*a^3*cos(d*x + c 
)^4*log(sin(d*x + c) + 1) - 3*(28*A + 20*B + 15*C)*a^3*cos(d*x + c)^4*log( 
-sin(d*x + c) + 1) + 2*(8*(9*A + 11*B + 9*C)*a^3*cos(d*x + c)^3 + 3*(4*A + 
 12*B + 15*C)*a^3*cos(d*x + c)^2 + 8*(B + 3*C)*a^3*cos(d*x + c) + 6*C*a^3) 
*sin(d*x + c))/(d*cos(d*x + c)^4)
 
3.5.30.6 Sympy [F]

\[ \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{3} \left (\int A\, dx + \int 3 A \sec {\left (c + d x \right )}\, dx + \int 3 A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec {\left (c + d x \right )}\, dx + \int 3 B \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

input
integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 
output
a**3*(Integral(A, x) + Integral(3*A*sec(c + d*x), x) + Integral(3*A*sec(c 
+ d*x)**2, x) + Integral(A*sec(c + d*x)**3, x) + Integral(B*sec(c + d*x), 
x) + Integral(3*B*sec(c + d*x)**2, x) + Integral(3*B*sec(c + d*x)**3, x) + 
 Integral(B*sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**2, x) + Integra 
l(3*C*sec(c + d*x)**3, x) + Integral(3*C*sec(c + d*x)**4, x) + Integral(C* 
sec(c + d*x)**5, x))
 
3.5.30.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 352 vs. \(2 (152) = 304\).

Time = 0.24 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.17 \[ \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {48 \, {\left (d x + c\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 48 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 144 \, A a^{3} \tan \left (d x + c\right ) + 144 \, B a^{3} \tan \left (d x + c\right ) + 48 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \]

input
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm= 
"maxima")
 
output
1/48*(48*(d*x + c)*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 48 
*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 3*C*a^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*A*a^3*(2*sin(d*x + c)/(sin(d*x + c 
)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 36*B*a^3*(2*si 
n(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) 
 - 1)) - 36*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) 
+ 1) + log(sin(d*x + c) - 1)) + 144*A*a^3*log(sec(d*x + c) + tan(d*x + c)) 
 + 48*B*a^3*log(sec(d*x + c) + tan(d*x + c)) + 144*A*a^3*tan(d*x + c) + 14 
4*B*a^3*tan(d*x + c) + 48*C*a^3*tan(d*x + c))/d
 
3.5.30.8 Giac [A] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 301, normalized size of antiderivative = 1.86 \[ \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (d x + c\right )} A a^{3} + 3 \, {\left (28 \, A a^{3} + 20 \, B a^{3} + 15 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (28 \, A a^{3} + 20 \, B a^{3} + 15 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 45 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 204 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 228 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 219 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 84 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 147 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

input
integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm= 
"giac")
 
output
1/24*(24*(d*x + c)*A*a^3 + 3*(28*A*a^3 + 20*B*a^3 + 15*C*a^3)*log(abs(tan( 
1/2*d*x + 1/2*c) + 1)) - 3*(28*A*a^3 + 20*B*a^3 + 15*C*a^3)*log(abs(tan(1/ 
2*d*x + 1/2*c) - 1)) - 2*(60*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*B*a^3*tan(1 
/2*d*x + 1/2*c)^7 + 45*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 204*A*a^3*tan(1/2*d* 
x + 1/2*c)^5 - 220*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 165*C*a^3*tan(1/2*d*x + 
1/2*c)^5 + 228*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 292*B*a^3*tan(1/2*d*x + 1/2* 
c)^3 + 219*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 84*A*a^3*tan(1/2*d*x + 1/2*c) - 
132*B*a^3*tan(1/2*d*x + 1/2*c) - 147*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2* 
d*x + 1/2*c)^2 - 1)^4)/d
 
3.5.30.9 Mupad [B] (verification not implemented)

Time = 18.63 (sec) , antiderivative size = 611, normalized size of antiderivative = 3.77 \[ \int (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {9\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {63\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {45\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+\frac {135\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{8}+9\,A\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {3\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {9\,A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{2}+13\,B\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {9\,B\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {11\,B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{2}+15\,C\,a^3\,\sin \left (2\,c+2\,d\,x\right )+\frac {45\,C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{8}+\frac {9\,C\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{2}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {9\,B\,a^3\,\sin \left (c+d\,x\right )}{2}+\frac {69\,C\,a^3\,\sin \left (c+d\,x\right )}{8}+12\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )+3\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (4\,c+4\,d\,x\right )+42\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )+\frac {21\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (4\,c+4\,d\,x\right )}{2}+30\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )+\frac {15\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (4\,c+4\,d\,x\right )}{2}+\frac {45\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {45\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\cos \left (4\,c+4\,d\,x\right )}{8}}{12\,d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {\cos \left (4\,c+4\,d\,x\right )}{8}+\frac {3}{8}\right )} \]

input
int((a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
 
output
(9*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (63*A*a^3*atanh(sin 
(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + (45*B*a^3*atanh(sin(c/2 + (d*x)/2 
)/cos(c/2 + (d*x)/2)))/2 + (135*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + ( 
d*x)/2)))/8 + 9*A*a^3*sin(2*c + 2*d*x) + (3*A*a^3*sin(3*c + 3*d*x))/2 + (9 
*A*a^3*sin(4*c + 4*d*x))/2 + 13*B*a^3*sin(2*c + 2*d*x) + (9*B*a^3*sin(3*c 
+ 3*d*x))/2 + (11*B*a^3*sin(4*c + 4*d*x))/2 + 15*C*a^3*sin(2*c + 2*d*x) + 
(45*C*a^3*sin(3*c + 3*d*x))/8 + (9*C*a^3*sin(4*c + 4*d*x))/2 + (3*A*a^3*si 
n(c + d*x))/2 + (9*B*a^3*sin(c + d*x))/2 + (69*C*a^3*sin(c + d*x))/8 + 12* 
A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + 3*A*a 
^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x) + 42*A*a^3 
*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + (21*A*a^3 
*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/2 + 30*B*a 
^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + (15*B*a 
^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/2 + (45* 
C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2 + ( 
45*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/8) 
/(12*d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))