\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx\) [479]

Optimal result

Integrand size = 33, antiderivative size = 148 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {A x}{a^4}-\frac {(55 A-6 B-8 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {2 (80 A-3 B-4 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A-B+C) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(10 A-3 B-4 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \] Output:

A*x/a^4-1/105*(55*A-6*B-8*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2-2/105*(80*A 
-3*B-4*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B+C)*tan(d*x+c)/d/(a+a*se 
c(d*x+c))^4-1/35*(10*A-3*B-4*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(405\) vs. \(2(148)=296\).

Time = 4.10 (sec) , antiderivative size = 405, normalized size of antiderivative = 2.74 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (3675 A d x \cos \left (\frac {d x}{2}\right )+3675 A d x \cos \left (c+\frac {d x}{2}\right )+2205 A d x \cos \left (c+\frac {3 d x}{2}\right )+2205 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+735 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+735 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+105 A d x \cos \left (3 c+\frac {7 d x}{2}\right )+105 A d x \cos \left (4 c+\frac {7 d x}{2}\right )-9940 A \sin \left (\frac {d x}{2}\right )+1260 B \sin \left (\frac {d x}{2}\right )+560 C \sin \left (\frac {d x}{2}\right )+8260 A \sin \left (c+\frac {d x}{2}\right )-1260 B \sin \left (c+\frac {d x}{2}\right )-350 C \sin \left (c+\frac {d x}{2}\right )-7140 A \sin \left (c+\frac {3 d x}{2}\right )+882 B \sin \left (c+\frac {3 d x}{2}\right )+336 C \sin \left (c+\frac {3 d x}{2}\right )+3780 A \sin \left (2 c+\frac {3 d x}{2}\right )-630 B \sin \left (2 c+\frac {3 d x}{2}\right )-210 C \sin \left (2 c+\frac {3 d x}{2}\right )-2800 A \sin \left (2 c+\frac {5 d x}{2}\right )+294 B \sin \left (2 c+\frac {5 d x}{2}\right )+182 C \sin \left (2 c+\frac {5 d x}{2}\right )+840 A \sin \left (3 c+\frac {5 d x}{2}\right )-210 B \sin \left (3 c+\frac {5 d x}{2}\right )-520 A \sin \left (3 c+\frac {7 d x}{2}\right )+72 B \sin \left (3 c+\frac {7 d x}{2}\right )+26 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{13440 a^4 d} \] Input:

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^4,x 
]
 

Output:

(Sec[c/2]*Sec[(c + d*x)/2]^7*(3675*A*d*x*Cos[(d*x)/2] + 3675*A*d*x*Cos[c + 
 (d*x)/2] + 2205*A*d*x*Cos[c + (3*d*x)/2] + 2205*A*d*x*Cos[2*c + (3*d*x)/2 
] + 735*A*d*x*Cos[2*c + (5*d*x)/2] + 735*A*d*x*Cos[3*c + (5*d*x)/2] + 105* 
A*d*x*Cos[3*c + (7*d*x)/2] + 105*A*d*x*Cos[4*c + (7*d*x)/2] - 9940*A*Sin[( 
d*x)/2] + 1260*B*Sin[(d*x)/2] + 560*C*Sin[(d*x)/2] + 8260*A*Sin[c + (d*x)/ 
2] - 1260*B*Sin[c + (d*x)/2] - 350*C*Sin[c + (d*x)/2] - 7140*A*Sin[c + (3* 
d*x)/2] + 882*B*Sin[c + (3*d*x)/2] + 336*C*Sin[c + (3*d*x)/2] + 3780*A*Sin 
[2*c + (3*d*x)/2] - 630*B*Sin[2*c + (3*d*x)/2] - 210*C*Sin[2*c + (3*d*x)/2 
] - 2800*A*Sin[2*c + (5*d*x)/2] + 294*B*Sin[2*c + (5*d*x)/2] + 182*C*Sin[2 
*c + (5*d*x)/2] + 840*A*Sin[3*c + (5*d*x)/2] - 210*B*Sin[3*c + (5*d*x)/2] 
- 520*A*Sin[3*c + (7*d*x)/2] + 72*B*Sin[3*c + (7*d*x)/2] + 26*C*Sin[3*c + 
(7*d*x)/2]))/(13440*a^4*d)
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 4540, 25, 3042, 4410, 25, 3042, 4410, 25, 3042, 4407, 3042, 4281}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4540

\(\displaystyle -\frac {\int -\frac {7 a A-a (3 A-3 B-4 C) \sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {7 a A-a (3 A-3 B-4 C) \sec (c+d x)}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4410

\(\displaystyle \frac {-\frac {\int -\frac {35 a^2 A-2 a^2 (10 A-3 B-4 C) \sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {35 a^2 A-2 a^2 (10 A-3 B-4 C) \sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {35 a^2 A-2 a^2 (10 A-3 B-4 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4410

\(\displaystyle \frac {\frac {-\frac {\int -\frac {105 a^3 A-a^3 (55 A-6 B-8 C) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(55 A-6 B-8 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\frac {\int \frac {105 a^3 A-a^3 (55 A-6 B-8 C) \sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(55 A-6 B-8 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {105 a^3 A-a^3 (55 A-6 B-8 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(55 A-6 B-8 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4407

\(\displaystyle \frac {\frac {\frac {105 a^2 A x-2 a^3 (80 A-3 B-4 C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(55 A-6 B-8 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {105 a^2 A x-2 a^3 (80 A-3 B-4 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {(55 A-6 B-8 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {\frac {105 a^2 A x-\frac {2 a^3 (80 A-3 B-4 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(55 A-6 B-8 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (10 A-3 B-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B+C) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

Input:

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^4,x]
 

Output:

-1/7*((A - B + C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + (-1/5*(a*(10* 
A - 3*B - 4*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (-1/3*((55*A - 6 
*B - 8*C)*Tan[c + d*x])/(d*(1 + Sec[c + d*x])^2) + (105*a^2*A*x - (2*a^3*( 
80*A - 3*B - 4*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2) 
)/(7*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 4281

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} 
, x] && EqQ[a^2 - b^2, 0]
 

rule 4407

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

rule 4410

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

rule 4540

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

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66

method	result	size
parallelrisch	\(\frac {15 \left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+21 \left (-5 A +3 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+35 \left (11 A -3 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+105 \left (-15 A +B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+840 d x A}{840 a^{4} d}\)	\(97\)
derivativedivides	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A +\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +16 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\)	\(183\)
default	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} C}{7}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A +\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C +16 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\)	\(183\)
norman	\(\frac {\frac {A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}-\frac {A x}{a}+\frac {\left (A -B +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{56 a d}+\frac {\left (15 A -B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (20 A -13 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{140 a d}-\frac {\left (28 A -3 B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {\left (35 A -12 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) a^{3}}\)	\(184\)
risch	\(\frac {A x}{a^{4}}-\frac {2 i \left (420 A \,{\mathrm e}^{6 i \left (d x +c \right )}-105 B \,{\mathrm e}^{6 i \left (d x +c \right )}+1890 A \,{\mathrm e}^{5 i \left (d x +c \right )}-315 B \,{\mathrm e}^{5 i \left (d x +c \right )}-105 C \,{\mathrm e}^{5 i \left (d x +c \right )}+4130 A \,{\mathrm e}^{4 i \left (d x +c \right )}-630 B \,{\mathrm e}^{4 i \left (d x +c \right )}-175 C \,{\mathrm e}^{4 i \left (d x +c \right )}+4970 A \,{\mathrm e}^{3 i \left (d x +c \right )}-630 B \,{\mathrm e}^{3 i \left (d x +c \right )}-280 C \,{\mathrm e}^{3 i \left (d x +c \right )}+3570 A \,{\mathrm e}^{2 i \left (d x +c \right )}-441 B \,{\mathrm e}^{2 i \left (d x +c \right )}-168 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1400 A \,{\mathrm e}^{i \left (d x +c \right )}-147 B \,{\mathrm e}^{i \left (d x +c \right )}-91 C \,{\mathrm e}^{i \left (d x +c \right )}+260 A -36 B -13 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(244\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVER 
BOSE)
 

Output:

1/840*(15*(A-B+C)*tan(1/2*d*x+1/2*c)^7+21*(-5*A+3*B-C)*tan(1/2*d*x+1/2*c)^ 
5+35*(11*A-3*B-C)*tan(1/2*d*x+1/2*c)^3+105*(-15*A+B+C)*tan(1/2*d*x+1/2*c)+ 
840*d*x*A)/a^4/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.30 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {105 \, A d x \cos \left (d x + c\right )^{4} + 420 \, A d x \cos \left (d x + c\right )^{3} + 630 \, A d x \cos \left (d x + c\right )^{2} + 420 \, A d x \cos \left (d x + c\right ) + 105 \, A d x - {\left ({\left (260 \, A - 36 \, B - 13 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (620 \, A - 39 \, B - 52 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (535 \, A - 24 \, B - 32 \, C\right )} \cos \left (d x + c\right ) + 160 \, A - 6 \, B - 8 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm= 
"fricas")
 

Output:

1/105*(105*A*d*x*cos(d*x + c)^4 + 420*A*d*x*cos(d*x + c)^3 + 630*A*d*x*cos 
(d*x + c)^2 + 420*A*d*x*cos(d*x + c) + 105*A*d*x - ((260*A - 36*B - 13*C)* 
cos(d*x + c)^3 + (620*A - 39*B - 52*C)*cos(d*x + c)^2 + (535*A - 24*B - 32 
*C)*cos(d*x + c) + 160*A - 6*B - 8*C)*sin(d*x + c))/(a^4*d*cos(d*x + c)^4 
+ 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + 
 a^4*d)
 

Sympy [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**4,x)
 

Output:

(Integral(A/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*s 
ec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c 
+ d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) + Integral(C*sec(c 
 + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*se 
c(c + d*x) + 1), x))/a**4
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 286 vs. \(2 (140) = 280\).

Time = 0.13 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.93 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=-\frac {5 \, A {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - \frac {C {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} - \frac {3 \, B {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm= 
"maxima")
                                                                                    
                                                                                    
 

Output:

-1/840*(5*A*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*sin(d*x + c)^3/(cos 
(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c) 
^7/(cos(d*x + c) + 1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1)) 
/a^4) - C*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*sin(d*x + c)^3/(cos(d* 
x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 15*sin(d*x + c)^7 
/(cos(d*x + c) + 1)^7)/a^4 - 3*B*(35*sin(d*x + c)/(cos(d*x + c) + 1) - 35* 
sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1) 
^5 - 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d
 

Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.49 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {840 \, {\left (d x + c\right )} A}{a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 105 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 385 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 35 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1575 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x, algorithm= 
"giac")
 

Output:

1/840*(840*(d*x + c)*A/a^4 + (15*A*a^24*tan(1/2*d*x + 1/2*c)^7 - 15*B*a^24 
*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 105*A*a^24*ta 
n(1/2*d*x + 1/2*c)^5 + 63*B*a^24*tan(1/2*d*x + 1/2*c)^5 - 21*C*a^24*tan(1/ 
2*d*x + 1/2*c)^5 + 385*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 105*B*a^24*tan(1/2* 
d*x + 1/2*c)^3 - 35*C*a^24*tan(1/2*d*x + 1/2*c)^3 - 1575*A*a^24*tan(1/2*d* 
x + 1/2*c) + 105*B*a^24*tan(1/2*d*x + 1/2*c) + 105*C*a^24*tan(1/2*d*x + 1/ 
2*c))/a^28)/d
 

Mupad [B] (verification not implemented)

Time = 14.02 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.55 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {A\,x}{a^4}+\frac {{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}-\frac {15\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}+\frac {C\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{8}-\frac {11\,A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}+\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24}\right )-{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{8}-\frac {3\,B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}+\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{40}\right )+\frac {A\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}-\frac {B\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}+\frac {C\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7} \] Input:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^4,x)
 

Output:

(A*x)/a^4 + (cos(c/2 + (d*x)/2)^6*((B*sin(c/2 + (d*x)/2))/8 - (15*A*sin(c/ 
2 + (d*x)/2))/8 + (C*sin(c/2 + (d*x)/2))/8) - cos(c/2 + (d*x)/2)^4*((B*sin 
(c/2 + (d*x)/2)^3)/8 - (11*A*sin(c/2 + (d*x)/2)^3)/24 + (C*sin(c/2 + (d*x) 
/2)^3)/24) - cos(c/2 + (d*x)/2)^2*((A*sin(c/2 + (d*x)/2)^5)/8 - (3*B*sin(c 
/2 + (d*x)/2)^5)/40 + (C*sin(c/2 + (d*x)/2)^5)/40) + (A*sin(c/2 + (d*x)/2) 
^7)/56 - (B*sin(c/2 + (d*x)/2)^7)/56 + (C*sin(c/2 + (d*x)/2)^7)/56)/(a^4*d 
*cos(c/2 + (d*x)/2)^7)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.19 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^4} \, dx=\frac {15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} c -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +63 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b -21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} c +385 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} c -1575 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) c +840 a d x}{840 a^{4} d} \] Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x)
 

Output:

(15*tan((c + d*x)/2)**7*a - 15*tan((c + d*x)/2)**7*b + 15*tan((c + d*x)/2) 
**7*c - 105*tan((c + d*x)/2)**5*a + 63*tan((c + d*x)/2)**5*b - 21*tan((c + 
 d*x)/2)**5*c + 385*tan((c + d*x)/2)**3*a - 105*tan((c + d*x)/2)**3*b - 35 
*tan((c + d*x)/2)**3*c - 1575*tan((c + d*x)/2)*a + 105*tan((c + d*x)/2)*b 
+ 105*tan((c + d*x)/2)*c + 840*a*d*x)/(840*a**4*d)