\(\int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [107]

   Optimal result
   Rubi [A] (verified)
   Mathematica [B] (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)

Optimal result

Integrand size = 31, antiderivative size = 218 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {(23 A-13 B) x}{2 a^3}+\frac {4 (34 A-19 B) \sin (c+d x)}{5 a^3 d}-\frac {(23 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {4 (34 A-19 B) \sin ^3(c+d x)}{15 a^3 d} \]

[Out]

-1/2*(23*A-13*B)*x/a^3+4/5*(34*A-19*B)*sin(d*x+c)/a^3/d-1/2*(23*A-13*B)*cos(d*x+c)*sin(d*x+c)/a^3/d-1/5*(A-B)*
cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(13*A-8*B)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/
3*(23*A-13*B)*cos(d*x+c)^2*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))-4/15*(34*A-19*B)*sin(d*x+c)^3/a^3/d

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4105, 3872, 2713, 2715, 8} \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {4 (34 A-19 B) \sin ^3(c+d x)}{15 a^3 d}+\frac {4 (34 A-19 B) \sin (c+d x)}{5 a^3 d}-\frac {(23 A-13 B) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {(23 A-13 B) \sin (c+d x) \cos ^2(c+d x)}{3 d \left (a^3 \sec (c+d x)+a^3\right )}-\frac {x (23 A-13 B)}{2 a^3}-\frac {(13 A-8 B) \sin (c+d x) \cos ^2(c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A-B) \sin (c+d x) \cos ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[In]

Int[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

-1/2*((23*A - 13*B)*x)/a^3 + (4*(34*A - 19*B)*Sin[c + d*x])/(5*a^3*d) - ((23*A - 13*B)*Cos[c + d*x]*Sin[c + d*
x])/(2*a^3*d) - ((A - B)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((13*A - 8*B)*Cos[c + d*x
]^2*Sin[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((23*A - 13*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*d*(a^3 + a^
3*Sec[c + d*x])) - (4*(34*A - 19*B)*Sin[c + d*x]^3)/(15*a^3*d)

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3872

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

Rule 4105

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] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos ^3(c+d x) (a (8 A-3 B)-5 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2} \\ & = -\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {\int \frac {\cos ^3(c+d x) \left (3 a^2 (21 A-11 B)-4 a^2 (13 A-8 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4} \\ & = -\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {\int \cos ^3(c+d x) \left (12 a^3 (34 A-19 B)-15 a^3 (23 A-13 B) \sec (c+d x)\right ) \, dx}{15 a^6} \\ & = -\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(4 (34 A-19 B)) \int \cos ^3(c+d x) \, dx}{5 a^3}-\frac {(23 A-13 B) \int \cos ^2(c+d x) \, dx}{a^3} \\ & = -\frac {(23 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {(23 A-13 B) \int 1 \, dx}{2 a^3}-\frac {(4 (34 A-19 B)) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 a^3 d} \\ & = -\frac {(23 A-13 B) x}{2 a^3}+\frac {4 (34 A-19 B) \sin (c+d x)}{5 a^3 d}-\frac {(23 A-13 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A-B) \cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(13 A-8 B) \cos ^2(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(23 A-13 B) \cos ^2(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac {4 (34 A-19 B) \sin ^3(c+d x)}{15 a^3 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(491\) vs. \(2(218)=436\).

Time = 2.78 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.25 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-600 (23 A-13 B) d x \cos \left (\frac {d x}{2}\right )-600 (23 A-13 B) d x \cos \left (c+\frac {d x}{2}\right )-6900 A d x \cos \left (c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (c+\frac {3 d x}{2}\right )-6900 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-1380 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (2 c+\frac {5 d x}{2}\right )-1380 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac {5 d x}{2}\right )+20410 A \sin \left (\frac {d x}{2}\right )-12760 B \sin \left (\frac {d x}{2}\right )-11110 A \sin \left (c+\frac {d x}{2}\right )+7560 B \sin \left (c+\frac {d x}{2}\right )+15380 A \sin \left (c+\frac {3 d x}{2}\right )-9230 B \sin \left (c+\frac {3 d x}{2}\right )-380 A \sin \left (2 c+\frac {3 d x}{2}\right )+930 B \sin \left (2 c+\frac {3 d x}{2}\right )+4777 A \sin \left (2 c+\frac {5 d x}{2}\right )-2782 B \sin \left (2 c+\frac {5 d x}{2}\right )+1625 A \sin \left (3 c+\frac {5 d x}{2}\right )-750 B \sin \left (3 c+\frac {5 d x}{2}\right )+230 A \sin \left (3 c+\frac {7 d x}{2}\right )-105 B \sin \left (3 c+\frac {7 d x}{2}\right )+230 A \sin \left (4 c+\frac {7 d x}{2}\right )-105 B \sin \left (4 c+\frac {7 d x}{2}\right )-20 A \sin \left (4 c+\frac {9 d x}{2}\right )+15 B \sin \left (4 c+\frac {9 d x}{2}\right )-20 A \sin \left (5 c+\frac {9 d x}{2}\right )+15 B \sin \left (5 c+\frac {9 d x}{2}\right )+5 A \sin \left (5 c+\frac {11 d x}{2}\right )+5 A \sin \left (6 c+\frac {11 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \]

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-600*(23*A - 13*B)*d*x*Cos[(d*x)/2] - 600*(23*A - 13*B)*d*x*Cos[c + (d*x)/2] - 690
0*A*d*x*Cos[c + (3*d*x)/2] + 3900*B*d*x*Cos[c + (3*d*x)/2] - 6900*A*d*x*Cos[2*c + (3*d*x)/2] + 3900*B*d*x*Cos[
2*c + (3*d*x)/2] - 1380*A*d*x*Cos[2*c + (5*d*x)/2] + 780*B*d*x*Cos[2*c + (5*d*x)/2] - 1380*A*d*x*Cos[3*c + (5*
d*x)/2] + 780*B*d*x*Cos[3*c + (5*d*x)/2] + 20410*A*Sin[(d*x)/2] - 12760*B*Sin[(d*x)/2] - 11110*A*Sin[c + (d*x)
/2] + 7560*B*Sin[c + (d*x)/2] + 15380*A*Sin[c + (3*d*x)/2] - 9230*B*Sin[c + (3*d*x)/2] - 380*A*Sin[2*c + (3*d*
x)/2] + 930*B*Sin[2*c + (3*d*x)/2] + 4777*A*Sin[2*c + (5*d*x)/2] - 2782*B*Sin[2*c + (5*d*x)/2] + 1625*A*Sin[3*
c + (5*d*x)/2] - 750*B*Sin[3*c + (5*d*x)/2] + 230*A*Sin[3*c + (7*d*x)/2] - 105*B*Sin[3*c + (7*d*x)/2] + 230*A*
Sin[4*c + (7*d*x)/2] - 105*B*Sin[4*c + (7*d*x)/2] - 20*A*Sin[4*c + (9*d*x)/2] + 15*B*Sin[4*c + (9*d*x)/2] - 20
*A*Sin[5*c + (9*d*x)/2] + 15*B*Sin[5*c + (9*d*x)/2] + 5*A*Sin[5*c + (11*d*x)/2] + 5*A*Sin[6*c + (11*d*x)/2]))/
(480*a^3*d*(1 + Cos[c + d*x])^3)

Maple [A] (verified)

Time = 1.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.57

method result size
parallelrisch \(\frac {-\left (\frac {8 \left (-427 A +232 B \right ) \cos \left (2 d x +2 c \right )}{15}+\left (-\frac {43 A}{3}+6 B \right ) \cos \left (3 d x +3 c \right )+\left (A -B \right ) \cos \left (4 d x +4 c \right )-\frac {A \cos \left (5 d x +5 c \right )}{3}+\frac {2 \left (-\frac {5458 A}{3}+1001 B \right ) \cos \left (d x +c \right )}{5}-\frac {7783 A}{15}+\frac {4303 B}{15}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-736 d \left (A -\frac {13 B}{23}\right ) x}{64 a^{3} d}\) \(125\)
derivativedivides \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {16 \left (\left (-\frac {17 A}{4}+\frac {7 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {19 A}{3}+3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {11 A}{4}+\frac {5 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-4 \left (23 A -13 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) \(182\)
default \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {16 \left (\left (-\frac {17 A}{4}+\frac {7 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {19 A}{3}+3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-\frac {11 A}{4}+\frac {5 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-4 \left (23 A -13 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) \(182\)
norman \(\frac {-\frac {\left (23 A -13 B \right ) x}{2 a}+\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {3 \left (23 A -13 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}-\frac {3 \left (23 A -13 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}-\frac {\left (23 A -13 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2 a}+\frac {3 \left (31 A -17 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (41 A -31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (99 A -59 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}+\frac {5 \left (147 A -83 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {\left (513 A -283 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2}}\) \(258\)
risch \(-\frac {23 A x}{2 a^{3}}+\frac {13 x B}{2 a^{3}}-\frac {i A \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 a^{3} d}+\frac {3 i A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 a^{3} d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B}{8 a^{3} d}-\frac {27 i A \,{\mathrm e}^{i \left (d x +c \right )}}{8 a^{3} d}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B}{2 a^{3} d}+\frac {27 i A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a^{3} d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B}{2 a^{3} d}-\frac {3 i A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 a^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B}{8 a^{3} d}+\frac {i A \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 a^{3} d}+\frac {2 i \left (225 A \,{\mathrm e}^{4 i \left (d x +c \right )}-150 B \,{\mathrm e}^{4 i \left (d x +c \right )}+810 A \,{\mathrm e}^{3 i \left (d x +c \right )}-525 B \,{\mathrm e}^{3 i \left (d x +c \right )}+1160 A \,{\mathrm e}^{2 i \left (d x +c \right )}-745 B \,{\mathrm e}^{2 i \left (d x +c \right )}+760 \,{\mathrm e}^{i \left (d x +c \right )} A -485 B \,{\mathrm e}^{i \left (d x +c \right )}+197 A -127 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) \(331\)

[In]

int(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/64*(-(8/15*(-427*A+232*B)*cos(2*d*x+2*c)+(-43/3*A+6*B)*cos(3*d*x+3*c)+(A-B)*cos(4*d*x+4*c)-1/3*A*cos(5*d*x+5
*c)+2/5*(-5458/3*A+1001*B)*cos(d*x+c)-7783/15*A+4303/15*B)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4-736*d*(A-13
/23*B)*x)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {15 \, {\left (23 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (23 \, A - 13 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (23 \, A - 13 \, B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (23 \, A - 13 \, B\right )} d x - {\left (10 \, A \cos \left (d x + c\right )^{5} - 15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} + 5 \, {\left (19 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (869 \, A - 479 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (429 \, A - 239 \, B\right )} \cos \left (d x + c\right ) + 544 \, A - 304 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/30*(15*(23*A - 13*B)*d*x*cos(d*x + c)^3 + 45*(23*A - 13*B)*d*x*cos(d*x + c)^2 + 45*(23*A - 13*B)*d*x*cos(d*
x + c) + 15*(23*A - 13*B)*d*x - (10*A*cos(d*x + c)^5 - 15*(A - B)*cos(d*x + c)^4 + 5*(19*A - 9*B)*cos(d*x + c)
^3 + (869*A - 479*B)*cos(d*x + c)^2 + 3*(429*A - 239*B)*cos(d*x + c) + 544*A - 304*B)*sin(d*x + c))/(a^3*d*cos
(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

Sympy [F]

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

[In]

integrate(cos(d*x+c)**3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**3,x)

[Out]

(Integral(A*cos(c + d*x)**3/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x) + Integral(B*cos(c
+ d*x)**3*sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x))/a**3

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (204) = 408\).

Time = 0.31 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {A {\left (\frac {20 \, {\left (\frac {33 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {76 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {51 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {735 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {50 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {1380 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(A*(20*(33*sin(d*x + c)/(cos(d*x + c) + 1) + 76*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 51*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5)/(a^3 + 3*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1
)^4 + a^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (735*sin(d*x + c)/(cos(d*x + c) + 1) - 50*sin(d*x + c)^3/(cos
(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 1380*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a
^3) - B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x +
 c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1) -
 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 780*arctan(sin(d*x + c)
/(cos(d*x + c) + 1))/a^3))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {30 \, {\left (d x + c\right )} {\left (23 \, A - 13 \, B\right )}}{a^{3}} - \frac {20 \, {\left (51 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 76 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 50 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 735 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-1/60*(30*(d*x + c)*(23*A - 13*B)/a^3 - 20*(51*A*tan(1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/2*c)^5 + 76*A*t
an(1/2*d*x + 1/2*c)^3 - 36*B*tan(1/2*d*x + 1/2*c)^3 + 33*A*tan(1/2*d*x + 1/2*c) - 15*B*tan(1/2*d*x + 1/2*c))/(
(tan(1/2*d*x + 1/2*c)^2 + 1)^3*a^3) - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 50*
A*a^12*tan(1/2*d*x + 1/2*c)^3 + 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 735*A*a^12*tan(1/2*d*x + 1/2*c) - 465*B*a^1
2*tan(1/2*d*x + 1/2*c))/a^15)/d

Mupad [B] (verification not implemented)

Time = 14.02 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^3(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx=\frac {\left (17\,A-7\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {76\,A}{3}-12\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (11\,A-5\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A-B\right )}{2\,a^3}+\frac {6\,A-4\,B}{a^3}+\frac {15\,A-5\,B}{4\,a^3}\right )}{d}-\frac {x\,\left (23\,A-13\,B\right )}{2\,a^3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{3\,a^3}+\frac {6\,A-4\,B}{12\,a^3}\right )}{d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^5*(17*A - 7*B) + tan(c/2 + (d*x)/2)^3*((76*A)/3 - 12*B) + tan(c/2 + (d*x)/2)*(11*A - 5*B))
/(d*(3*a^3*tan(c/2 + (d*x)/2)^2 + 3*a^3*tan(c/2 + (d*x)/2)^4 + a^3*tan(c/2 + (d*x)/2)^6 + a^3)) + (tan(c/2 + (
d*x)/2)*((5*(A - B))/(2*a^3) + (6*A - 4*B)/a^3 + (15*A - 5*B)/(4*a^3)))/d - (x*(23*A - 13*B))/(2*a^3) - (tan(c
/2 + (d*x)/2)^3*((A - B)/(3*a^3) + (6*A - 4*B)/(12*a^3)))/d + (tan(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d)