\(\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [102]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 33, antiderivative size = 197 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 (30 A+23 C) \text {arctanh}(\sin (c+d x))}{16 d}+\frac {a^3 (30 A+23 C) \tan (c+d x)}{10 d}+\frac {3 a^3 (30 A+23 C) \sec (c+d x) \tan (c+d x)}{80 d}+\frac {(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac {C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac {a^3 (30 A+23 C) \tan ^3(c+d x)}{120 d} \] Output:

1/16*a^3*(30*A+23*C)*arctanh(sin(d*x+c))/d+1/10*a^3*(30*A+23*C)*tan(d*x+c) 
/d+3/80*a^3*(30*A+23*C)*sec(d*x+c)*tan(d*x+c)/d+1/120*(30*A+7*C)*(a+a*sec( 
d*x+c))^3*tan(d*x+c)/d+1/6*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^3*tan(d*x+c)/d+ 
1/10*C*(a+a*sec(d*x+c))^4*tan(d*x+c)/a/d+1/120*a^3*(30*A+23*C)*tan(d*x+c)^ 
3/d
 

Mathematica [A] (verified)

Time = 3.98 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.57 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (15 (30 A+23 C) \text {arctanh}(\sin (c+d x))+\left (720 A+544 C+15 (30 A+23 C) \sec (c+d x)+16 (15 A+17 C) \sec ^2(c+d x)+10 (6 A+23 C) \sec ^3(c+d x)+144 C \sec ^4(c+d x)+40 C \sec ^5(c+d x)\right ) \tan (c+d x)\right )}{240 d} \] Input:

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

Output:

(a^3*(15*(30*A + 23*C)*ArcTanh[Sin[c + d*x]] + (720*A + 544*C + 15*(30*A + 
 23*C)*Sec[c + d*x] + 16*(15*A + 17*C)*Sec[c + d*x]^2 + 10*(6*A + 23*C)*Se 
c[c + d*x]^3 + 144*C*Sec[c + d*x]^4 + 40*C*Sec[c + d*x]^5)*Tan[c + d*x]))/ 
(240*d)
 

Rubi [A] (verified)

Time = 0.93 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4577, 3042, 4498, 3042, 4489, 3042, 4278, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4577

\(\displaystyle \frac {\int \sec ^2(c+d x) (\sec (c+d x) a+a)^3 (2 a (3 A+C)+3 a C \sec (c+d x))dx}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (2 a (3 A+C)+3 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

\(\Big \downarrow \) 4498

\(\displaystyle \frac {\frac {\int \sec (c+d x) (\sec (c+d x) a+a)^3 \left (12 C a^2+(30 A+7 C) \sec (c+d x) a^2\right )dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (12 C a^2+(30 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {\frac {3}{4} a^2 (30 A+23 C) \int \sec (c+d x) (\sec (c+d x) a+a)^3dx+\frac {a^2 (30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3}{4} a^2 (30 A+23 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3dx+\frac {a^2 (30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

\(\Big \downarrow \) 4278

\(\displaystyle \frac {\frac {\frac {3}{4} a^2 (30 A+23 C) \int \left (a^3 \sec ^4(c+d x)+3 a^3 \sec ^3(c+d x)+3 a^3 \sec ^2(c+d x)+a^3 \sec (c+d x)\right )dx+\frac {a^2 (30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {\frac {a^2 (30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}+\frac {3}{4} a^2 (30 A+23 C) \left (\frac {5 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^3 \tan ^3(c+d x)}{3 d}+\frac {4 a^3 \tan (c+d x)}{d}+\frac {3 a^3 \tan (c+d x) \sec (c+d x)}{2 d}\right )}{5 a}+\frac {3 C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 d}}{6 a}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}\)

Input:

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

Output:

(C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(6*d) + ((3*C*(a + 
a*Sec[c + d*x])^4*Tan[c + d*x])/(5*d) + ((a^2*(30*A + 7*C)*(a + a*Sec[c + 
d*x])^3*Tan[c + d*x])/(4*d) + (3*a^2*(30*A + 23*C)*((5*a^3*ArcTanh[Sin[c + 
 d*x]])/(2*d) + (4*a^3*Tan[c + d*x])/d + (3*a^3*Sec[c + d*x]*Tan[c + d*x]) 
/(2*d) + (a^3*Tan[c + d*x]^3)/(3*d)))/4)/(5*a))/(6*a)
 

Defintions of rubi rules used

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 4278
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^(m_), x_Symbol] :> Int[ExpandTrig[(a + b*csc[e + f*x])^m*(d*csc[e + f 
*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && I 
GtQ[m, 0] && RationalQ[n]
 

rule 4489
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 
 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B 
, e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b 
*(m + 1), 0] &&  !LtQ[m, -2^(-1)]
 

rule 4498
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( 
csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]* 
((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   Int 
[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B) 
*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a 
*B, 0] &&  !LtQ[m, -1]
 

rule 4577
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C) 
*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*(m + n + 1))), 
x] + Simp[1/(b*(m + n + 1))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n 
*Simp[A*b*(m + n + 1) + b*C*n + a*C*m*Csc[e + f*x], x], x], x] /; FreeQ[{a, 
 b, d, e, f, A, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && 
!LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
 
Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.15

method result size
norman \(\frac {\frac {7 a^{3} \left (14 A +15 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d}-\frac {33 a^{3} \left (30 A +23 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 d}+\frac {17 a^{3} \left (30 A +23 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{24 d}-\frac {a^{3} \left (30 A +23 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{8 d}-\frac {a^{3} \left (342 A +211 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{8 d}+\frac {a^{3} \left (1250 A +969 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{6}}-\frac {a^{3} \left (30 A +23 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16 d}+\frac {a^{3} \left (30 A +23 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16 d}\) \(227\)
parallelrisch \(\frac {19 a^{3} \left (-\frac {225 \left (\frac {\cos \left (6 d x +6 c \right )}{15}+\frac {2 \cos \left (4 d x +4 c \right )}{5}+\cos \left (2 d x +2 c \right )+\frac {2}{3}\right ) \left (A +\frac {23 C}{30}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{76}+\frac {225 \left (\frac {\cos \left (6 d x +6 c \right )}{15}+\frac {2 \cos \left (4 d x +4 c \right )}{5}+\cos \left (2 d x +2 c \right )+\frac {2}{3}\right ) \left (A +\frac {23 C}{30}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{76}+\left (\frac {46 A}{19}+\frac {60 C}{19}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {391 C}{228}+\frac {53 A}{38}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {136 C}{95}+\frac {32 A}{19}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {23 C}{76}+\frac {15 A}{38}\right ) \sin \left (5 d x +5 c \right )+\left (\frac {6 A}{19}+\frac {68 C}{285}\right ) \sin \left (6 d x +6 c \right )+\sin \left (d x +c \right ) \left (A +\frac {75 C}{38}\right )\right )}{2 d \left (\cos \left (6 d x +6 c \right )+6 \cos \left (4 d x +4 c \right )+15 \cos \left (2 d x +2 c \right )+10\right )}\) \(246\)
parts \(\frac {\left (a^{3} A +3 a^{3} C \right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (3 a^{3} A +a^{3} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{3} A \tan \left (d x +c \right )}{d}+\frac {a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}+\frac {3 a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {3 a^{3} C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\) \(250\)
derivativedivides \(\frac {a^{3} A \tan \left (d x +c \right )-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-3 a^{3} C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) \(294\)
default \(\frac {a^{3} A \tan \left (d x +c \right )-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 a^{3} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-3 a^{3} C \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+a^{3} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{3} C \left (-\left (-\frac {\sec \left (d x +c \right )^{5}}{6}-\frac {5 \sec \left (d x +c \right )^{3}}{24}-\frac {5 \sec \left (d x +c \right )}{16}\right ) \tan \left (d x +c \right )+\frac {5 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16}\right )}{d}\) \(294\)
risch \(-\frac {i a^{3} \left (450 A \,{\mathrm e}^{11 i \left (d x +c \right )}+345 C \,{\mathrm e}^{11 i \left (d x +c \right )}-240 A \,{\mathrm e}^{10 i \left (d x +c \right )}+1590 A \,{\mathrm e}^{9 i \left (d x +c \right )}+1955 C \,{\mathrm e}^{9 i \left (d x +c \right )}-2640 A \,{\mathrm e}^{8 i \left (d x +c \right )}-480 C \,{\mathrm e}^{8 i \left (d x +c \right )}+1140 A \,{\mathrm e}^{7 i \left (d x +c \right )}+2250 C \,{\mathrm e}^{7 i \left (d x +c \right )}-7200 A \,{\mathrm e}^{6 i \left (d x +c \right )}-5440 C \,{\mathrm e}^{6 i \left (d x +c \right )}-1140 A \,{\mathrm e}^{5 i \left (d x +c \right )}-2250 C \,{\mathrm e}^{5 i \left (d x +c \right )}-8160 A \,{\mathrm e}^{4 i \left (d x +c \right )}-7680 C \,{\mathrm e}^{4 i \left (d x +c \right )}-1590 A \,{\mathrm e}^{3 i \left (d x +c \right )}-1955 C \,{\mathrm e}^{3 i \left (d x +c \right )}-4080 A \,{\mathrm e}^{2 i \left (d x +c \right )}-3264 C \,{\mathrm e}^{2 i \left (d x +c \right )}-450 A \,{\mathrm e}^{i \left (d x +c \right )}-345 C \,{\mathrm e}^{i \left (d x +c \right )}-720 A -544 C \right )}{120 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}+\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{8 d}+\frac {23 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{16 d}-\frac {15 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{8 d}-\frac {23 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{16 d}\) \(371\)

Input:

int(sec(d*x+c)^2*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVER 
BOSE)
 

Output:

(7/8*a^3*(14*A+15*C)/d*tan(1/2*d*x+1/2*c)-33/20*a^3*(30*A+23*C)/d*tan(1/2* 
d*x+1/2*c)^7+17/24*a^3*(30*A+23*C)/d*tan(1/2*d*x+1/2*c)^9-1/8*a^3*(30*A+23 
*C)/d*tan(1/2*d*x+1/2*c)^11-1/8*a^3*(342*A+211*C)/d*tan(1/2*d*x+1/2*c)^3+1 
/20*a^3*(1250*A+969*C)/d*tan(1/2*d*x+1/2*c)^5)/(tan(1/2*d*x+1/2*c)^2-1)^6- 
1/16*a^3*(30*A+23*C)/d*ln(tan(1/2*d*x+1/2*c)-1)+1/16*a^3*(30*A+23*C)/d*ln( 
tan(1/2*d*x+1/2*c)+1)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.92 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (45 \, A + 34 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} + 15 \, {\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 16 \, {\left (15 \, A + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 10 \, {\left (6 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 144 \, C a^{3} \cos \left (d x + c\right ) + 40 \, C a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \] Input:

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

Output:

1/480*(15*(30*A + 23*C)*a^3*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(30* 
A + 23*C)*a^3*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(16*(45*A + 34*C)* 
a^3*cos(d*x + c)^5 + 15*(30*A + 23*C)*a^3*cos(d*x + c)^4 + 16*(15*A + 17*C 
)*a^3*cos(d*x + c)^3 + 10*(6*A + 23*C)*a^3*cos(d*x + c)^2 + 144*C*a^3*cos( 
d*x + c) + 40*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^6)
 

Sympy [F]

\[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 A \sec ^{4}{\left (c + d x \right )}\, dx + \int A \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{5}{\left (c + d x \right )}\, dx + \int 3 C \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{7}{\left (c + d x \right )}\, dx\right ) \] Input:

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

Output:

a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*sec(c + d*x)**3, x) + 
Integral(3*A*sec(c + d*x)**4, x) + Integral(A*sec(c + d*x)**5, x) + Integr 
al(C*sec(c + d*x)**4, x) + Integral(3*C*sec(c + d*x)**5, x) + Integral(3*C 
*sec(c + d*x)**6, x) + Integral(C*sec(c + d*x)**7, x))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 382 vs. \(2 (183) = 366\).

Time = 0.04 (sec) , antiderivative size = 382, normalized size of antiderivative = 1.94 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 96 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{3} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 5 \, C a^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, C a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 480 \, A a^{3} \tan \left (d x + c\right )}{480 \, d} \] Input:

integrate(sec(d*x+c)^2*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm= 
"maxima")
                                                                                    
                                                                                    
 

Output:

1/480*(480*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 96*(3*tan(d*x + c)^5 
+ 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^3 + 160*(tan(d*x + c)^3 + 3*tan 
(d*x + c))*C*a^3 - 5*C*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33* 
sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 
 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 30*A*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)) - 90*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(s 
in(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 360*A*a^3*(2*sin(d*x + c)/(s 
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 480* 
A*a^3*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.42 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (30 \, A a^{3} + 23 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (30 \, A a^{3} + 23 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (450 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 345 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 2550 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 1955 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 5940 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 4554 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7500 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5814 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5130 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3165 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1470 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{6}}}{240 \, d} \] Input:

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

Output:

1/240*(15*(30*A*a^3 + 23*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3 
0*A*a^3 + 23*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(450*A*a^3*tan( 
1/2*d*x + 1/2*c)^11 + 345*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 2550*A*a^3*tan(1 
/2*d*x + 1/2*c)^9 - 1955*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 5940*A*a^3*tan(1/2 
*d*x + 1/2*c)^7 + 4554*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 7500*A*a^3*tan(1/2*d 
*x + 1/2*c)^5 - 5814*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 5130*A*a^3*tan(1/2*d*x 
 + 1/2*c)^3 + 3165*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1470*A*a^3*tan(1/2*d*x + 
 1/2*c) - 1575*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6) 
/d
 

Mupad [B] (verification not implemented)

Time = 14.06 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.33 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (-\frac {15\,A\,a^3}{4}-\frac {23\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {85\,A\,a^3}{4}+\frac {391\,C\,a^3}{24}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {99\,A\,a^3}{2}-\frac {759\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {125\,A\,a^3}{2}+\frac {969\,C\,a^3}{20}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {171\,A\,a^3}{4}-\frac {211\,C\,a^3}{8}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+\frac {105\,C\,a^3}{8}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (30\,A+23\,C\right )}{8\,d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*((49*A*a^3)/4 + (105*C*a^3)/8) - tan(c/2 + (d*x)/2)^11 
*((15*A*a^3)/4 + (23*C*a^3)/8) - tan(c/2 + (d*x)/2)^3*((171*A*a^3)/4 + (21 
1*C*a^3)/8) + tan(c/2 + (d*x)/2)^9*((85*A*a^3)/4 + (391*C*a^3)/24) - tan(c 
/2 + (d*x)/2)^7*((99*A*a^3)/2 + (759*C*a^3)/20) + tan(c/2 + (d*x)/2)^5*((1 
25*A*a^3)/2 + (969*C*a^3)/20))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + ( 
d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 - 6*tan(c/2 
+ (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (a^3*atanh(tan(c/2 + (d*x)/2 
))*(30*A + 23*C))/(8*d)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 537, normalized size of antiderivative = 2.73 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

(a**3*( - 720*cos(c + d*x)*sin(c + d*x)**5*a - 544*cos(c + d*x)*sin(c + d* 
x)**5*c + 1680*cos(c + d*x)*sin(c + d*x)**3*a + 1360*cos(c + d*x)*sin(c + 
d*x)**3*c - 960*cos(c + d*x)*sin(c + d*x)*a - 960*cos(c + d*x)*sin(c + d*x 
)*c - 450*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**6*a - 345*log(tan((c + d 
*x)/2) - 1)*sin(c + d*x)**6*c + 1350*log(tan((c + d*x)/2) - 1)*sin(c + d*x 
)**4*a + 1035*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*c - 1350*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2*a - 1035*log(tan((c + d*x)/2) - 1)*sin(c 
+ d*x)**2*c + 450*log(tan((c + d*x)/2) - 1)*a + 345*log(tan((c + d*x)/2) - 
 1)*c + 450*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**6*a + 345*log(tan((c + 
 d*x)/2) + 1)*sin(c + d*x)**6*c - 1350*log(tan((c + d*x)/2) + 1)*sin(c + d 
*x)**4*a - 1035*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c + 1350*log(tan 
((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 1035*log(tan((c + d*x)/2) + 1)*sin( 
c + d*x)**2*c - 450*log(tan((c + d*x)/2) + 1)*a - 345*log(tan((c + d*x)/2) 
 + 1)*c - 450*sin(c + d*x)**5*a - 345*sin(c + d*x)**5*c + 960*sin(c + d*x) 
**3*a + 920*sin(c + d*x)**3*c - 510*sin(c + d*x)*a - 615*sin(c + d*x)*c))/ 
(240*d*(sin(c + d*x)**6 - 3*sin(c + d*x)**4 + 3*sin(c + d*x)**2 - 1))