\(\int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [115]

Optimal result

Integrand size = 29, antiderivative size = 166 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {(4 A-B) x}{a^4}+\frac {8 (83 A-20 B) \sin (c+d x)}{105 a^4 d}-\frac {(88 A-25 B) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(4 A-B) \sin (c+d x)}{a^4 d (1+\sec (c+d x))}-\frac {(A-B) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(12 A-5 B) \sin (c+d x)}{35 a d (a+a \sec (c+d x))^3} \] Output:

-(4*A-B)*x/a^4+8/105*(83*A-20*B)*sin(d*x+c)/a^4/d-1/105*(88*A-25*B)*sin(d* 
x+c)/a^4/d/(1+sec(d*x+c))^2-(4*A-B)*sin(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A 
-B)*sin(d*x+c)/d/(a+a*sec(d*x+c))^4-1/35*(12*A-5*B)*sin(d*x+c)/a/d/(a+a*se 
c(d*x+c))^3
 

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(485\) vs. \(2(166)=332\).

Time = 5.21 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.92 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-7350 (4 A-B) d x \cos \left (\frac {d x}{2}\right )-7350 (4 A-B) d x \cos \left (c+\frac {d x}{2}\right )-17640 A d x \cos \left (c+\frac {3 d x}{2}\right )+4410 B d x \cos \left (c+\frac {3 d x}{2}\right )-17640 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+4410 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-5880 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+1470 B d x \cos \left (2 c+\frac {5 d x}{2}\right )-5880 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+1470 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-840 A d x \cos \left (3 c+\frac {7 d x}{2}\right )+210 B d x \cos \left (3 c+\frac {7 d x}{2}\right )-840 A d x \cos \left (4 c+\frac {7 d x}{2}\right )+210 B d x \cos \left (4 c+\frac {7 d x}{2}\right )+60830 A \sin \left (\frac {d x}{2}\right )-19880 B \sin \left (\frac {d x}{2}\right )-46130 A \sin \left (c+\frac {d x}{2}\right )+16520 B \sin \left (c+\frac {d x}{2}\right )+46116 A \sin \left (c+\frac {3 d x}{2}\right )-14280 B \sin \left (c+\frac {3 d x}{2}\right )-18060 A \sin \left (2 c+\frac {3 d x}{2}\right )+7560 B \sin \left (2 c+\frac {3 d x}{2}\right )+19292 A \sin \left (2 c+\frac {5 d x}{2}\right )-5600 B \sin \left (2 c+\frac {5 d x}{2}\right )-2100 A \sin \left (3 c+\frac {5 d x}{2}\right )+1680 B \sin \left (3 c+\frac {5 d x}{2}\right )+3791 A \sin \left (3 c+\frac {7 d x}{2}\right )-1040 B \sin \left (3 c+\frac {7 d x}{2}\right )+735 A \sin \left (4 c+\frac {7 d x}{2}\right )+105 A \sin \left (4 c+\frac {9 d x}{2}\right )+105 A \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{1680 a^4 d (1+\cos (c+d x))^4} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sec[c/2]*(-7350*(4*A - B)*d*x*Cos[(d*x)/2] - 7350*(4*A - 
 B)*d*x*Cos[c + (d*x)/2] - 17640*A*d*x*Cos[c + (3*d*x)/2] + 4410*B*d*x*Cos 
[c + (3*d*x)/2] - 17640*A*d*x*Cos[2*c + (3*d*x)/2] + 4410*B*d*x*Cos[2*c + 
(3*d*x)/2] - 5880*A*d*x*Cos[2*c + (5*d*x)/2] + 1470*B*d*x*Cos[2*c + (5*d*x 
)/2] - 5880*A*d*x*Cos[3*c + (5*d*x)/2] + 1470*B*d*x*Cos[3*c + (5*d*x)/2] - 
 840*A*d*x*Cos[3*c + (7*d*x)/2] + 210*B*d*x*Cos[3*c + (7*d*x)/2] - 840*A*d 
*x*Cos[4*c + (7*d*x)/2] + 210*B*d*x*Cos[4*c + (7*d*x)/2] + 60830*A*Sin[(d* 
x)/2] - 19880*B*Sin[(d*x)/2] - 46130*A*Sin[c + (d*x)/2] + 16520*B*Sin[c + 
(d*x)/2] + 46116*A*Sin[c + (3*d*x)/2] - 14280*B*Sin[c + (3*d*x)/2] - 18060 
*A*Sin[2*c + (3*d*x)/2] + 7560*B*Sin[2*c + (3*d*x)/2] + 19292*A*Sin[2*c + 
(5*d*x)/2] - 5600*B*Sin[2*c + (5*d*x)/2] - 2100*A*Sin[3*c + (5*d*x)/2] + 1 
680*B*Sin[3*c + (5*d*x)/2] + 3791*A*Sin[3*c + (7*d*x)/2] - 1040*B*Sin[3*c 
+ (7*d*x)/2] + 735*A*Sin[4*c + (7*d*x)/2] + 105*A*Sin[4*c + (9*d*x)/2] + 1 
05*A*Sin[5*c + (9*d*x)/2]))/(1680*a^4*d*(1 + Cos[c + d*x])^4)
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.448, Rules used = {3042, 4508, 3042, 4508, 3042, 4508, 3042, 4508, 3042, 4274, 24, 3042, 3117}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\int \frac {\cos (c+d x) (a (8 A-B)-4 a (A-B) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (8 A-B)-4 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {\cos (c+d x) \left (2 a^2 (26 A-5 B)-3 a^2 (12 A-5 B) \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {2 a^2 (26 A-5 B)-3 a^2 (12 A-5 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\frac {\int \frac {\cos (c+d x) \left (a^3 (244 A-55 B)-2 a^3 (88 A-25 B) \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {a^3 (244 A-55 B)-2 a^3 (88 A-25 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\frac {\frac {\int \cos (c+d x) \left (8 a^4 (83 A-20 B)-105 a^4 (4 A-B) \sec (c+d x)\right )dx}{a^2}-\frac {105 a^3 (4 A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {8 a^4 (83 A-20 B)-105 a^4 (4 A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {105 a^3 (4 A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {\frac {\frac {8 a^4 (83 A-20 B) \int \cos (c+d x)dx-105 a^4 (4 A-B) \int 1dx}{a^2}-\frac {105 a^3 (4 A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {\frac {\frac {8 a^4 (83 A-20 B) \int \cos (c+d x)dx-105 a^4 x (4 A-B)}{a^2}-\frac {105 a^3 (4 A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\frac {8 a^4 (83 A-20 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx-105 a^4 x (4 A-B)}{a^2}-\frac {105 a^3 (4 A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3117

\(\displaystyle \frac {\frac {\frac {\frac {\frac {8 a^4 (83 A-20 B) \sin (c+d x)}{d}-105 a^4 x (4 A-B)}{a^2}-\frac {105 a^3 (4 A-B) \sin (c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {(88 A-25 B) \sin (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {a (12 A-5 B) \sin (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \sin (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

Input:

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

Output:

-1/7*((A - B)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + (-1/5*(a*(12*A - 
5*B)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (-1/3*((88*A - 25*B)*Sin[c 
 + d*x])/(d*(1 + Sec[c + d*x])^2) + ((-105*a^3*(4*A - B)*Sin[c + d*x])/(d* 
(a + a*Sec[c + d*x])) + (-105*a^4*(4*A - B)*x + (8*a^4*(83*A - 20*B)*Sin[c 
 + d*x])/d)/a^2)/(3*a^2))/(5*a^2))/(7*a^2)
 

Defintions of rubi rules used

rule 24

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

rule 3042

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

rule 3117

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; 
 FreeQ[{c, d}, x]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4508

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

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(\frac {105 \left (\left (\frac {10964 A}{105}-\frac {496 B}{21}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {2368 A}{105}-\frac {104 B}{21}\right ) \cos \left (3 d x +3 c \right )+A \cos \left (4 d x +4 c \right )+\left (\frac {24992 A}{105}-\frac {1168 B}{21}\right ) \cos \left (d x +c \right )+\frac {16171 A}{105}-\frac {752 B}{21}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-26880 d \left (A -\frac {B}{4}\right ) x}{6720 a^{4} d}\)	\(107\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B -\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {16 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-16 \left (4 A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\)	\(164\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} A}{7}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} B}{7}+\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B -\frac {23 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {16 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}-16 \left (4 A -B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}\)	\(164\)
norman	\(\frac {-\frac {\left (4 A -B \right ) x}{a}-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{56 a d}-\frac {\left (4 A -B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {5 \left (13 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (22 A -15 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{140 a d}-\frac {\left (47 A -20 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {\left (62 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{3}}\)	\(184\)
risch	\(-\frac {4 A x}{a^{4}}+\frac {x B}{a^{4}}-\frac {i A \,{\mathrm e}^{i \left (d x +c \right )}}{2 a^{4} d}+\frac {i A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{4} d}+\frac {4 i \left (525 A \,{\mathrm e}^{6 i \left (d x +c \right )}-210 B \,{\mathrm e}^{6 i \left (d x +c \right )}+2625 A \,{\mathrm e}^{5 i \left (d x +c \right )}-945 B \,{\mathrm e}^{5 i \left (d x +c \right )}+5950 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2065 B \,{\mathrm e}^{4 i \left (d x +c \right )}+7420 A \,{\mathrm e}^{3 i \left (d x +c \right )}-2485 B \,{\mathrm e}^{3 i \left (d x +c \right )}+5397 A \,{\mathrm e}^{2 i \left (d x +c \right )}-1785 B \,{\mathrm e}^{2 i \left (d x +c \right )}+2149 \,{\mathrm e}^{i \left (d x +c \right )} A -700 B \,{\mathrm e}^{i \left (d x +c \right )}+382 A -130 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(226\)

Input:

int(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBOSE 
)
 

Output:

1/6720*(105*((10964/105*A-496/21*B)*cos(2*d*x+2*c)+(2368/105*A-104/21*B)*c 
os(3*d*x+3*c)+A*cos(4*d*x+4*c)+(24992/105*A-1168/21*B)*cos(d*x+c)+16171/10 
5*A-752/21*B)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^6-26880*d*(A-1/4*B)*x) 
/a^4/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.34 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {105 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (4 \, A - B\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (4 \, A - B\right )} d x - {\left (105 \, A \cos \left (d x + c\right )^{4} + 4 \, {\left (296 \, A - 65 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (659 \, A - 155 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (2236 \, A - 535 \, B\right )} \cos \left (d x + c\right ) + 664 \, A - 160 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:

integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="fri 
cas")
 

Output:

-1/105*(105*(4*A - B)*d*x*cos(d*x + c)^4 + 420*(4*A - B)*d*x*cos(d*x + c)^ 
3 + 630*(4*A - B)*d*x*cos(d*x + c)^2 + 420*(4*A - B)*d*x*cos(d*x + c) + 10 
5*(4*A - B)*d*x - (105*A*cos(d*x + c)^4 + 4*(296*A - 65*B)*cos(d*x + c)^3 
+ 4*(659*A - 155*B)*cos(d*x + c)^2 + (2236*A - 535*B)*cos(d*x + c) + 664*A 
 - 160*B)*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]

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

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

Output:

(Integral(A*cos(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(B*cos(c + d*x)*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))/a**4
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.63 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {A {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - 5 \, B {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \] Input:

integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="max 
ima")
 

Output:

1/840*(A*(1680*sin(d*x + c)/((a^4 + a^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^ 
2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) - 805*sin(d 
*x + c)^3/(cos(d*x + c) + 1)^3 + 147*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 - 6720*arctan(sin(d*x + c)/(c 
os(d*x + c) + 1))/a^4) - 5*B*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*si 
n(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)/(c 
os(d*x + c) + 1))/a^4))/d
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.14 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {840 \, {\left (d x + c\right )} {\left (4 \, A - B\right )}}{a^{4}} - \frac {1680 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 147 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 105 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 805 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 385 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5145 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1575 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \] Input:

integrate(cos(d*x+c)*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="gia 
c")
 

Output:

-1/840*(840*(d*x + c)*(4*A - B)/a^4 - 1680*A*tan(1/2*d*x + 1/2*c)/((tan(1/ 
2*d*x + 1/2*c)^2 + 1)*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 - 147*A*a^24*tan(1/2*d*x + 1/2*c)^5 + 105*B*a^24*t 
an(1/2*d*x + 1/2*c)^5 + 805*A*a^24*tan(1/2*d*x + 1/2*c)^3 - 385*B*a^24*tan 
(1/2*d*x + 1/2*c)^3 - 5145*A*a^24*tan(1/2*d*x + 1/2*c) + 1575*B*a^24*tan(1 
/2*d*x + 1/2*c))/a^28)/d
 

Mupad [B] (verification not implemented)

Time = 11.08 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.22 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\left (\frac {764\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}-\frac {52\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {16\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{21}-\frac {143\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{105}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {8\,A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{35}-\frac {5\,B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{28}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\frac {A\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}+\frac {B\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{56}}{a^4\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}-\frac {4\,A\,d\,x-B\,d\,x}{a^4\,d}+\frac {2\,A\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d} \] Input:

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

Output:

((B*sin(c/2 + (d*x)/2))/56 - (A*sin(c/2 + (d*x)/2))/56 + cos(c/2 + (d*x)/2 
)^2*((8*A*sin(c/2 + (d*x)/2))/35 - (5*B*sin(c/2 + (d*x)/2))/28) - cos(c/2 
+ (d*x)/2)^4*((143*A*sin(c/2 + (d*x)/2))/105 - (16*B*sin(c/2 + (d*x)/2))/2 
1) + cos(c/2 + (d*x)/2)^6*((764*A*sin(c/2 + (d*x)/2))/105 - (52*B*sin(c/2 
+ (d*x)/2))/21))/(a^4*d*cos(c/2 + (d*x)/2)^7) - (4*A*d*x - B*d*x)/(a^4*d) 
+ (2*A*cos(c/2 + (d*x)/2)*sin(c/2 + (d*x)/2))/(a^4*d)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.22 \[ \int \frac {\cos (c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {-15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} a +15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9} b +132 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} a -90 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7} b -658 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a +280 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} b +4340 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} a -1190 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b -3360 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a d x +840 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b d x +6825 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -1575 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b -3360 a d x +840 b d x}{840 a^{4} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )} \] Input:

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

Output:

( - 15*tan((c + d*x)/2)**9*a + 15*tan((c + d*x)/2)**9*b + 132*tan((c + d*x 
)/2)**7*a - 90*tan((c + d*x)/2)**7*b - 658*tan((c + d*x)/2)**5*a + 280*tan 
((c + d*x)/2)**5*b + 4340*tan((c + d*x)/2)**3*a - 1190*tan((c + d*x)/2)**3 
*b - 3360*tan((c + d*x)/2)**2*a*d*x + 840*tan((c + d*x)/2)**2*b*d*x + 6825 
*tan((c + d*x)/2)*a - 1575*tan((c + d*x)/2)*b - 3360*a*d*x + 840*b*d*x)/(8 
40*a**4*d*(tan((c + d*x)/2)**2 + 1))