Integrand size = 31, antiderivative size = 218 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 A-23 B) x}{2 a^3}-\frac {4 (19 A-34 B) \sin (c+d x)}{5 a^3 d}+\frac {(13 A-23 B) \cos (c+d x) \sin (c+d x)}{2 a^3 d}+\frac {(A-B) \cos ^5(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(8 A-13 B) \cos ^4(c+d x) \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(13 A-23 B) \cos ^3(c+d x) \sin (c+d x)}{3 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {4 (19 A-34 B) \sin ^3(c+d x)}{15 a^3 d} \] Output:
1/2*(13*A-23*B)*x/a^3-4/5*(19*A-34*B)*sin(d*x+c)/a^3/d+1/2*(13*A-23*B)*cos (d*x+c)*sin(d*x+c)/a^3/d+1/5*(A-B)*cos(d*x+c)^5*sin(d*x+c)/d/(a+a*cos(d*x+ c))^3+1/15*(8*A-13*B)*cos(d*x+c)^4*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+1/3*( 13*A-23*B)*cos(d*x+c)^3*sin(d*x+c)/d/(a^3+a^3*cos(d*x+c))+4/15*(19*A-34*B) *sin(d*x+c)^3/a^3/d
Leaf count is larger than twice the leaf count of optimal. \(491\) vs. \(2(218)=436\).
Time = 3.48 (sec) , antiderivative size = 491, normalized size of antiderivative = 2.25 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (600 (13 A-23 B) d x \cos \left (\frac {d x}{2}\right )+600 (13 A-23 B) d x \cos \left (c+\frac {d x}{2}\right )+3900 A d x \cos \left (c+\frac {3 d x}{2}\right )-6900 B d x \cos \left (c+\frac {3 d x}{2}\right )+3900 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-6900 B d x \cos \left (2 c+\frac {3 d x}{2}\right )+780 A d x \cos \left (2 c+\frac {5 d x}{2}\right )-1380 B d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 A d x \cos \left (3 c+\frac {5 d x}{2}\right )-1380 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-12760 A \sin \left (\frac {d x}{2}\right )+20410 B \sin \left (\frac {d x}{2}\right )+7560 A \sin \left (c+\frac {d x}{2}\right )-11110 B \sin \left (c+\frac {d x}{2}\right )-9230 A \sin \left (c+\frac {3 d x}{2}\right )+15380 B \sin \left (c+\frac {3 d x}{2}\right )+930 A \sin \left (2 c+\frac {3 d x}{2}\right )-380 B \sin \left (2 c+\frac {3 d x}{2}\right )-2782 A \sin \left (2 c+\frac {5 d x}{2}\right )+4777 B \sin \left (2 c+\frac {5 d x}{2}\right )-750 A \sin \left (3 c+\frac {5 d x}{2}\right )+1625 B \sin \left (3 c+\frac {5 d x}{2}\right )-105 A \sin \left (3 c+\frac {7 d x}{2}\right )+230 B \sin \left (3 c+\frac {7 d x}{2}\right )-105 A \sin \left (4 c+\frac {7 d x}{2}\right )+230 B \sin \left (4 c+\frac {7 d x}{2}\right )+15 A \sin \left (4 c+\frac {9 d x}{2}\right )-20 B \sin \left (4 c+\frac {9 d x}{2}\right )+15 A \sin \left (5 c+\frac {9 d x}{2}\right )-20 B \sin \left (5 c+\frac {9 d x}{2}\right )+5 B \sin \left (5 c+\frac {11 d x}{2}\right )+5 B \sin \left (6 c+\frac {11 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[(Cos[c + d*x]^5*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]
Output:
(Cos[(c + d*x)/2]*Sec[c/2]*(600*(13*A - 23*B)*d*x*Cos[(d*x)/2] + 600*(13*A - 23*B)*d*x*Cos[c + (d*x)/2] + 3900*A*d*x*Cos[c + (3*d*x)/2] - 6900*B*d*x *Cos[c + (3*d*x)/2] + 3900*A*d*x*Cos[2*c + (3*d*x)/2] - 6900*B*d*x*Cos[2*c + (3*d*x)/2] + 780*A*d*x*Cos[2*c + (5*d*x)/2] - 1380*B*d*x*Cos[2*c + (5*d *x)/2] + 780*A*d*x*Cos[3*c + (5*d*x)/2] - 1380*B*d*x*Cos[3*c + (5*d*x)/2] - 12760*A*Sin[(d*x)/2] + 20410*B*Sin[(d*x)/2] + 7560*A*Sin[c + (d*x)/2] - 11110*B*Sin[c + (d*x)/2] - 9230*A*Sin[c + (3*d*x)/2] + 15380*B*Sin[c + (3* d*x)/2] + 930*A*Sin[2*c + (3*d*x)/2] - 380*B*Sin[2*c + (3*d*x)/2] - 2782*A *Sin[2*c + (5*d*x)/2] + 4777*B*Sin[2*c + (5*d*x)/2] - 750*A*Sin[3*c + (5*d *x)/2] + 1625*B*Sin[3*c + (5*d*x)/2] - 105*A*Sin[3*c + (7*d*x)/2] + 230*B* Sin[3*c + (7*d*x)/2] - 105*A*Sin[4*c + (7*d*x)/2] + 230*B*Sin[4*c + (7*d*x )/2] + 15*A*Sin[4*c + (9*d*x)/2] - 20*B*Sin[4*c + (9*d*x)/2] + 15*A*Sin[5* c + (9*d*x)/2] - 20*B*Sin[5*c + (9*d*x)/2] + 5*B*Sin[5*c + (11*d*x)/2] + 5 *B*Sin[6*c + (11*d*x)/2]))/(480*a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.17 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 3456, 3042, 3456, 3042, 3456, 27, 3042, 3227, 3042, 3113, 2009, 3115, 24}
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.
| \(\displaystyle \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^5 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int \frac {\cos ^4(c+d x) (5 a (A-B)-a (3 A-8 B) \cos (c+d x))}{(\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (5 a (A-B)-a (3 A-8 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\int \frac {\cos ^3(c+d x) \left (4 a^2 (8 A-13 B)-3 a^2 (11 A-21 B) \cos (c+d x)\right )}{\cos (c+d x) a+a}dx}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (4 a^2 (8 A-13 B)-3 a^2 (11 A-21 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\frac {\frac {\int 3 \cos ^2(c+d x) \left (5 a^3 (13 A-23 B)-4 a^3 (19 A-34 B) \cos (c+d x)\right )dx}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 \int \cos ^2(c+d x) \left (5 a^3 (13 A-23 B)-4 a^3 (19 A-34 B) \cos (c+d x)\right )dx}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (5 a^3 (13 A-23 B)-4 a^3 (19 A-34 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (13 A-23 B) \int \cos ^2(c+d x)dx-4 a^3 (19 A-34 B) \int \cos ^3(c+d x)dx\right )}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (13 A-23 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-4 a^3 (19 A-34 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {\frac {\frac {3 \left (\frac {4 a^3 (19 A-34 B) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}+5 a^3 (13 A-23 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (13 A-23 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\frac {4 a^3 (19 A-34 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (13 A-23 B) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {4 a^3 (19 A-34 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )}{a^2}+\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {5 a^2 (13 A-23 B) \sin (c+d x) \cos ^3(c+d x)}{d (a \cos (c+d x)+a)}+\frac {3 \left (\frac {4 a^3 (19 A-34 B) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+5 a^3 (13 A-23 B) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )}{a^2}}{3 a^2}+\frac {a (8 A-13 B) \sin (c+d x) \cos ^4(c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}+\frac {(A-B) \sin (c+d x) \cos ^5(c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[(Cos[c + d*x]^5*(A + B*Cos[c + d*x]))/(a + a*Cos[c + d*x])^3,x]
Output:
((A - B)*Cos[c + d*x]^5*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ((a*( 8*A - 13*B)*Cos[c + d*x]^4*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2) + (( 5*a^2*(13*A - 23*B)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])) + (3*(5*a^3*(13*A - 23*B)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) + (4*a ^3*(19*A - 34*B)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d))/a^2)/(3*a^2))/(5* a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] & & NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
Time = 1.59 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.57
| method | result | size |
| parallelrisch | \(\frac {\left (\frac {8 \left (-232 A +427 B \right ) \cos \left (2 d x +2 c \right )}{15}+\left (-6 A +\frac {43 B}{3}\right ) \cos \left (3 d x +3 c \right )+\left (A -B \right ) \cos \left (4 d x +4 c \right )+\frac {B \cos \left (5 d x +5 c \right )}{3}+\frac {2 \left (-1001 A +\frac {5458 B}{3}\right ) \cos \left (d x +c \right )}{5}-\frac {4303 A}{15}+\frac {7783 B}{15}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+416 x \left (A -\frac {23 B}{13}\right ) d}{64 a^{3} d}\) | \(124\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {16 \left (-\frac {7 A}{4}+\frac {17 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+16 \left (-3 A +\frac {19 B}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+16 \left (-\frac {5 A}{4}+\frac {11 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+4 \left (13 A -23 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(182\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}-\frac {10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}-31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +49 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +\frac {16 \left (-\frac {7 A}{4}+\frac {17 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+16 \left (-3 A +\frac {19 B}{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+16 \left (-\frac {5 A}{4}+\frac {11 B}{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+4 \left (13 A -23 B \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \,a^{3}}\) | \(182\) |
| risch | \(\frac {13 A x}{2 a^{3}}-\frac {23 B x}{2 a^{3}}-\frac {i B \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 a^{3} d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} A}{8 a^{3} d}+\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} B}{8 a^{3} d}+\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A}{2 a^{3} d}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} B}{8 a^{3} d}-\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A}{2 a^{3} d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} B}{8 a^{3} d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A}{8 a^{3} d}-\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} B}{8 a^{3} d}+\frac {i B \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 a^{3} d}-\frac {2 i \left (150 A \,{\mathrm e}^{4 i \left (d x +c \right )}-225 B \,{\mathrm e}^{4 i \left (d x +c \right )}+525 A \,{\mathrm e}^{3 i \left (d x +c \right )}-810 B \,{\mathrm e}^{3 i \left (d x +c \right )}+745 A \,{\mathrm e}^{2 i \left (d x +c \right )}-1160 B \,{\mathrm e}^{2 i \left (d x +c \right )}+485 A \,{\mathrm e}^{i \left (d x +c \right )}-760 B \,{\mathrm e}^{i \left (d x +c \right )}+127 A -197 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(331\) |
| norman | \(\frac {\frac {\left (13 A -23 B \right ) x}{2 a}-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{17}}{20 a d}-\frac {\left (9 A -16 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{2 a d}+\frac {\left (11 A -16 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{30 a d}+\frac {3 \left (13 A -23 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a}+\frac {15 \left (13 A -23 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2 a}+\frac {10 \left (13 A -23 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}+\frac {15 \left (13 A -23 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}+\frac {3 \left (13 A -23 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{a}+\frac {\left (13 A -23 B \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2 a}-\frac {3 \left (17 A -31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (89 A -158 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{2 a d}-\frac {5 \left (164 A -293 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{6 a d}-\frac {\left (437 A -786 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {3 \left (691 A -1236 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}-\frac {\left (1703 A -3048 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6} a^{2}}\) | \(408\) |
Input:
int(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
1/64*((8/15*(-232*A+427*B)*cos(2*d*x+2*c)+(-6*A+43/3*B)*cos(3*d*x+3*c)+(A- B)*cos(4*d*x+4*c)+1/3*B*cos(5*d*x+5*c)+2/5*(-1001*A+5458/3*B)*cos(d*x+c)-4 303/15*A+7783/15*B)*tan(1/2*d*x+1/2*c)*sec(1/2*d*x+1/2*c)^4+416*x*(A-23/13 *B)*d)/a^3/d
Time = 0.09 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.94 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left (13 \, A - 23 \, B\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (13 \, A - 23 \, B\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (13 \, A - 23 \, B\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (13 \, A - 23 \, B\right )} d x + {\left (10 \, B \cos \left (d x + c\right )^{5} + 15 \, {\left (A - B\right )} \cos \left (d x + c\right )^{4} - 5 \, {\left (9 \, A - 19 \, B\right )} \cos \left (d x + c\right )^{3} - {\left (479 \, A - 869 \, B\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (239 \, A - 429 \, B\right )} \cos \left (d x + c\right ) - 304 \, A + 544 \, B\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="f ricas")
Output:
1/30*(15*(13*A - 23*B)*d*x*cos(d*x + c)^3 + 45*(13*A - 23*B)*d*x*cos(d*x + c)^2 + 45*(13*A - 23*B)*d*x*cos(d*x + c) + 15*(13*A - 23*B)*d*x + (10*B*c os(d*x + c)^5 + 15*(A - B)*cos(d*x + c)^4 - 5*(9*A - 19*B)*cos(d*x + c)^3 - (479*A - 869*B)*cos(d*x + c)^2 - 3*(239*A - 429*B)*cos(d*x + c) - 304*A + 544*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)
Leaf count of result is larger than twice the leaf count of optimal. 1584 vs. \(2 (206) = 412\).
Time = 5.98 (sec) , antiderivative size = 1584, normalized size of antiderivative = 7.27 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)**5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)
Output:
Piecewise((390*A*d*x*tan(c/2 + d*x/2)**6/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3* d) + 1170*A*d*x*tan(c/2 + d*x/2)**4/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a **3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 1170*A*d*x*tan(c/2 + d*x/2)**2/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d *tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) + 390*A *d*x/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180 *a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 3*A*tan(c/2 + d*x/2)**11/(60*a* *3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan (c/2 + d*x/2)**2 + 60*a**3*d) + 31*A*tan(c/2 + d*x/2)**9/(60*a**3*d*tan(c/ 2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/ 2)**2 + 60*a**3*d) - 354*A*tan(c/2 + d*x/2)**7/(60*a**3*d*tan(c/2 + d*x/2) **6 + 180*a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60 *a**3*d) - 1698*A*tan(c/2 + d*x/2)**5/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180 *a**3*d*tan(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 2075*A*tan(c/2 + d*x/2)**3/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*t an(c/2 + d*x/2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 765*A*t an(c/2 + d*x/2)/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/ 2)**4 + 180*a**3*d*tan(c/2 + d*x/2)**2 + 60*a**3*d) - 690*B*d*x*tan(c/2 + d*x/2)**6/(60*a**3*d*tan(c/2 + d*x/2)**6 + 180*a**3*d*tan(c/2 + d*x/2)*...
Leaf count of result is larger than twice the leaf count of optimal. 412 vs. \(2 (204) = 408\).
Time = 0.12 (sec) , antiderivative size = 412, normalized size of antiderivative = 1.89 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {B {\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 )} - A {\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} \] Input:
integrate(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="m axima")
Output:
1/60*(B*(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) - A*(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*si n(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
Time = 0.31 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.05 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (d x + c\right )} {\left (13 \, A - 23 \, B\right )}}{a^{3}} - \frac {20 \, {\left (21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 51 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 76 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 33 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 50 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 735 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="g iac")
Output:
1/60*(30*(d*x + c)*(13*A - 23*B)/a^3 - 20*(21*A*tan(1/2*d*x + 1/2*c)^5 - 5 1*B*tan(1/2*d*x + 1/2*c)^5 + 36*A*tan(1/2*d*x + 1/2*c)^3 - 76*B*tan(1/2*d* x + 1/2*c)^3 + 15*A*tan(1/2*d*x + 1/2*c) - 33*B*tan(1/2*d*x + 1/2*c))/((ta n(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 - 40*A*a^12*tan(1/2*d*x + 1/2*c)^3 + 50*B*a^12 *tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) - 735*B*a^12*tan (1/2*d*x + 1/2*c))/a^15)/d
Time = 42.05 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.09 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {x\,\left (13\,A-23\,B\right )}{2\,a^3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A-B\right )}{2\,a^3}+\frac {4\,A-6\,B}{a^3}+\frac {5\,A-15\,B}{4\,a^3}\right )}{d}-\frac {\left (7\,A-17\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (12\,A-\frac {76\,B}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (5\,A-11\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{3\,a^3}+\frac {4\,A-6\,B}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d} \] Input:
int((cos(c + d*x)^5*(A + B*cos(c + d*x)))/(a + a*cos(c + d*x))^3,x)
Output:
(x*(13*A - 23*B))/(2*a^3) - (tan(c/2 + (d*x)/2)*((5*(A - B))/(2*a^3) + (4* A - 6*B)/a^3 + (5*A - 15*B)/(4*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(7*A - 17* B) + tan(c/2 + (d*x)/2)^3*(12*A - (76*B)/3) + tan(c/2 + (d*x)/2)*(5*A - 11 *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)^3*((A - B)/(3*a^3) + (4*A - 6*B)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d)
Time = 0.26 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^5(c+d x) (A+B \cos (c+d x))}{(a+a \cos (c+d x))^3} \, dx=\frac {-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a +25 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b -404 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +724 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b +390 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a d x -690 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b d x +6 \cos \left (d x +c \right ) a -6 \cos \left (d x +c \right ) b +10 \sin \left (d x +c \right )^{6} b +60 \sin \left (d x +c \right )^{4} a -140 \sin \left (d x +c \right )^{4} b -195 \sin \left (d x +c \right )^{3} a d x +345 \sin \left (d x +c \right )^{3} b d x -358 \sin \left (d x +c \right )^{2} a +668 \sin \left (d x +c \right )^{2} b +390 \sin \left (d x +c \right ) a d x -690 \sin \left (d x +c \right ) b d x -6 a +6 b}{30 \sin \left (d x +c \right ) a^{3} d \left (2 \cos \left (d x +c \right )-\sin \left (d x +c \right )^{2}+2\right )} \] Input:
int(cos(d*x+c)^5*(A+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x)
Output:
( - 15*cos(c + d*x)*sin(c + d*x)**4*a + 25*cos(c + d*x)*sin(c + d*x)**4*b - 404*cos(c + d*x)*sin(c + d*x)**2*a + 724*cos(c + d*x)*sin(c + d*x)**2*b + 390*cos(c + d*x)*sin(c + d*x)*a*d*x - 690*cos(c + d*x)*sin(c + d*x)*b*d* x + 6*cos(c + d*x)*a - 6*cos(c + d*x)*b + 10*sin(c + d*x)**6*b + 60*sin(c + d*x)**4*a - 140*sin(c + d*x)**4*b - 195*sin(c + d*x)**3*a*d*x + 345*sin( c + d*x)**3*b*d*x - 358*sin(c + d*x)**2*a + 668*sin(c + d*x)**2*b + 390*si n(c + d*x)*a*d*x - 690*sin(c + d*x)*b*d*x - 6*a + 6*b)/(30*sin(c + d*x)*a* *3*d*(2*cos(c + d*x) - sin(c + d*x)**2 + 2))