Integrand size = 31, antiderivative size = 196 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 A-6 B) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {8 (19 A-9 B) \tan (c+d x)}{15 a^3 d}+\frac {(13 A-6 B) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 A-6 B) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 A-9 B) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/2*(13*A-6*B)*arctanh(sin(d*x+c))/a^3/d-8/15*(19*A-9*B)*tan(d*x+c)/a^3/d+ 1/2*(13*A-6*B)*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*(A-B)*sec(d*x+c)*tan(d*x+c) /d/(a+a*cos(d*x+c))^3-1/15*(11*A-6*B)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d *x+c))^2-4/15*(19*A-9*B)*sec(d*x+c)*tan(d*x+c)/d/(a^3+a^3*cos(d*x+c))
Leaf count is larger than twice the leaf count of optimal. \(610\) vs. \(2(196)=392\).
Time = 6.51 (sec) , antiderivative size = 610, normalized size of antiderivative = 3.11 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {1920 (13 A-6 B) \cos ^6\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sec (c) \sec ^2(c+d x) \left ((-1235 A+870 B) \sin \left (\frac {d x}{2}\right )+5 (761 A-366 B) \sin \left (\frac {3 d x}{2}\right )-4329 A \sin \left (c-\frac {d x}{2}\right )+2094 B \sin \left (c-\frac {d x}{2}\right )+1989 A \sin \left (c+\frac {d x}{2}\right )-1314 B \sin \left (c+\frac {d x}{2}\right )-3575 A \sin \left (2 c+\frac {d x}{2}\right )+1650 B \sin \left (2 c+\frac {d x}{2}\right )-475 A \sin \left (c+\frac {3 d x}{2}\right )+450 B \sin \left (c+\frac {3 d x}{2}\right )+2005 A \sin \left (2 c+\frac {3 d x}{2}\right )-1230 B \sin \left (2 c+\frac {3 d x}{2}\right )-2275 A \sin \left (3 c+\frac {3 d x}{2}\right )+1050 B \sin \left (3 c+\frac {3 d x}{2}\right )+2673 A \sin \left (c+\frac {5 d x}{2}\right )-1278 B \sin \left (c+\frac {5 d x}{2}\right )+105 A \sin \left (2 c+\frac {5 d x}{2}\right )+90 B \sin \left (2 c+\frac {5 d x}{2}\right )+1593 A \sin \left (3 c+\frac {5 d x}{2}\right )-918 B \sin \left (3 c+\frac {5 d x}{2}\right )-975 A \sin \left (4 c+\frac {5 d x}{2}\right )+450 B \sin \left (4 c+\frac {5 d x}{2}\right )+1325 A \sin \left (2 c+\frac {7 d x}{2}\right )-630 B \sin \left (2 c+\frac {7 d x}{2}\right )+255 A \sin \left (3 c+\frac {7 d x}{2}\right )-60 B \sin \left (3 c+\frac {7 d x}{2}\right )+875 A \sin \left (4 c+\frac {7 d x}{2}\right )-480 B \sin \left (4 c+\frac {7 d x}{2}\right )-195 A \sin \left (5 c+\frac {7 d x}{2}\right )+90 B \sin \left (5 c+\frac {7 d x}{2}\right )+304 A \sin \left (3 c+\frac {9 d x}{2}\right )-144 B \sin \left (3 c+\frac {9 d x}{2}\right )+90 A \sin \left (4 c+\frac {9 d x}{2}\right )-30 B \sin \left (4 c+\frac {9 d x}{2}\right )+214 A \sin \left (5 c+\frac {9 d x}{2}\right )-114 B \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]
Output:
-1/480*(1920*(13*A - 6*B)*Cos[(c + d*x)/2]^6*(Log[Cos[(c + d*x)/2] - Sin[( c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Cos[(c + d*x)/2 ]*Sec[c/2]*Sec[c]*Sec[c + d*x]^2*((-1235*A + 870*B)*Sin[(d*x)/2] + 5*(761* A - 366*B)*Sin[(3*d*x)/2] - 4329*A*Sin[c - (d*x)/2] + 2094*B*Sin[c - (d*x) /2] + 1989*A*Sin[c + (d*x)/2] - 1314*B*Sin[c + (d*x)/2] - 3575*A*Sin[2*c + (d*x)/2] + 1650*B*Sin[2*c + (d*x)/2] - 475*A*Sin[c + (3*d*x)/2] + 450*B*S in[c + (3*d*x)/2] + 2005*A*Sin[2*c + (3*d*x)/2] - 1230*B*Sin[2*c + (3*d*x) /2] - 2275*A*Sin[3*c + (3*d*x)/2] + 1050*B*Sin[3*c + (3*d*x)/2] + 2673*A*S in[c + (5*d*x)/2] - 1278*B*Sin[c + (5*d*x)/2] + 105*A*Sin[2*c + (5*d*x)/2] + 90*B*Sin[2*c + (5*d*x)/2] + 1593*A*Sin[3*c + (5*d*x)/2] - 918*B*Sin[3*c + (5*d*x)/2] - 975*A*Sin[4*c + (5*d*x)/2] + 450*B*Sin[4*c + (5*d*x)/2] + 1325*A*Sin[2*c + (7*d*x)/2] - 630*B*Sin[2*c + (7*d*x)/2] + 255*A*Sin[3*c + (7*d*x)/2] - 60*B*Sin[3*c + (7*d*x)/2] + 875*A*Sin[4*c + (7*d*x)/2] - 480 *B*Sin[4*c + (7*d*x)/2] - 195*A*Sin[5*c + (7*d*x)/2] + 90*B*Sin[5*c + (7*d *x)/2] + 304*A*Sin[3*c + (9*d*x)/2] - 144*B*Sin[3*c + (9*d*x)/2] + 90*A*Si n[4*c + (9*d*x)/2] - 30*B*Sin[4*c + (9*d*x)/2] + 214*A*Sin[5*c + (9*d*x)/2 ] - 114*B*Sin[5*c + (9*d*x)/2]))/(a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.30 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.04, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 3457, 3042, 3457, 3042, 3457, 3042, 3227, 3042, 4254, 24, 4255, 3042, 4257}
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.
| \(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \cos (c+d x))}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(a (7 A-2 B)-4 a (A-B) \cos (c+d x)) \sec ^3(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 A-2 B)-4 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2 (43 A-18 B)-3 a^2 (11 A-6 B) \cos (c+d x)\right ) \sec ^3(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (43 A-18 B)-3 a^2 (11 A-6 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \left (15 a^3 (13 A-6 B)-8 a^3 (19 A-9 B) \cos (c+d x)\right ) \sec ^3(c+d x)dx}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^3 (13 A-6 B)-8 a^3 (19 A-9 B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B) \int \sec ^3(c+d x)dx-8 a^3 (19 A-9 B) \int \sec ^2(c+d x)dx}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-8 a^3 (19 A-9 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (19 A-9 B) \int 1d(-\tan (c+d x))}{d}+15 a^3 (13 A-6 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {8 a^3 (19 A-9 B) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 A-9 B) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 A-9 B) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 A-6 B) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 A-9 B) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 A-9 B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 A-6 B) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^3)/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*((A - B)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/ 3*(a*(11*A - 6*B)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((-4*a^2*(19*A - 9*B)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-8*a^3*(19*A - 9*B)*Tan[c + d*x])/d + 15*a^3*(13*A - 6*B)*(ArcTanh[Sin [c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/a^2)/(3*a^2))/(5*a^ 2)
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[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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] && (IntegerQ[2*n] || EqQ[c, 0])
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.87 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.89
| method | result | size |
| parallelrisch | \(\frac {-1560 \left (A -\frac {6 B}{13}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+1560 \left (A -\frac {6 B}{13}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-152 \left (\left (\frac {783 A}{76}-\frac {189 B}{38}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {717 A}{152}-\frac {9 B}{4}\right ) \cos \left (3 d x +3 c \right )+\left (A -\frac {9 B}{19}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {2331 A}{152}-\frac {573 B}{76}\right ) \cos \left (d x +c \right )+\frac {677 A}{76}-\frac {9 B}{2}\right ) \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{240 d \,a^{3} \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(175\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 A -12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 A +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} A}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 A -12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 A +12 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 A +4 B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 A}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| norman | \(\frac {-\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {\left (37 A -27 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}-\frac {\left (51 A -25 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\left (109 A -45 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}-\frac {\left (211 A -111 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 a d}+\frac {\left (461 A -201 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{2}}-\frac {\left (13 A -6 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (13 A -6 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(243\) |
| risch | \(-\frac {i \left (195 A \,{\mathrm e}^{8 i \left (d x +c \right )}-90 B \,{\mathrm e}^{8 i \left (d x +c \right )}+975 A \,{\mathrm e}^{7 i \left (d x +c \right )}-450 B \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 A \,{\mathrm e}^{6 i \left (d x +c \right )}-1050 B \,{\mathrm e}^{6 i \left (d x +c \right )}+3575 A \,{\mathrm e}^{5 i \left (d x +c \right )}-1650 B \,{\mathrm e}^{5 i \left (d x +c \right )}+4329 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2094 B \,{\mathrm e}^{4 i \left (d x +c \right )}+3805 A \,{\mathrm e}^{3 i \left (d x +c \right )}-1830 B \,{\mathrm e}^{3 i \left (d x +c \right )}+2673 A \,{\mathrm e}^{2 i \left (d x +c \right )}-1278 B \,{\mathrm e}^{2 i \left (d x +c \right )}+1325 A \,{\mathrm e}^{i \left (d x +c \right )}-630 B \,{\mathrm e}^{i \left (d x +c \right )}+304 A -144 B \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a^{3} d}-\frac {13 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a^{3} d}\) | \(324\) |
Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBO SE)
Output:
1/240*(-1560*(A-6/13*B)*(cos(2*d*x+2*c)+1)*ln(tan(1/2*d*x+1/2*c)-1)+1560*( A-6/13*B)*(cos(2*d*x+2*c)+1)*ln(tan(1/2*d*x+1/2*c)+1)-152*((783/76*A-189/3 8*B)*cos(2*d*x+2*c)+(717/152*A-9/4*B)*cos(3*d*x+3*c)+(A-9/19*B)*cos(4*d*x+ 4*c)+(2331/152*A-573/76*B)*cos(d*x+c)+677/76*A-9/2*B)*sec(1/2*d*x+1/2*c)^4 *tan(1/2*d*x+1/2*c))/d/a^3/(cos(2*d*x+2*c)+1)
Time = 0.09 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.51 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (19 \, A - 9 \, B\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, A - 114 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, A - 234 \, B\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right ) - 15 \, A\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="f ricas")
Output:
1/60*(15*((13*A - 6*B)*cos(d*x + c)^5 + 3*(13*A - 6*B)*cos(d*x + c)^4 + 3* (13*A - 6*B)*cos(d*x + c)^3 + (13*A - 6*B)*cos(d*x + c)^2)*log(sin(d*x + c ) + 1) - 15*((13*A - 6*B)*cos(d*x + c)^5 + 3*(13*A - 6*B)*cos(d*x + c)^4 + 3*(13*A - 6*B)*cos(d*x + c)^3 + (13*A - 6*B)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(19*A - 9*B)*cos(d*x + c)^4 + 3*(239*A - 114*B)*cos(d*x + c)^3 + (479*A - 234*B)*cos(d*x + c)^2 + 15*(3*A - 2*B)*cos(d*x + c) - 1 5*A)*sin(d*x + c))/(a^3*d*cos(d*x + c)^5 + 3*a^3*d*cos(d*x + c)^4 + 3*a^3* d*cos(d*x + c)^3 + a^3*d*cos(d*x + c)^2)
\[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos ^{3}{\left (c + d x \right )} + 3 \cos ^{2}{\left (c + d x \right )} + 3 \cos {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)**3/(a+a*cos(d*x+c))**3,x)
Output:
(Integral(A*sec(c + d*x)**3/(cos(c + d*x)**3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)**3/(cos(c + d*x)** 3 + 3*cos(c + d*x)**2 + 3*cos(c + d*x) + 1), x))/a**3
Leaf count of result is larger than twice the leaf count of optimal. 377 vs. \(2 (184) = 368\).
Time = 0.04 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.92 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {A {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, B {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="m axima")
Output:
-1/60*(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*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 - 390*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 390*log(sin (d*x + c)/(cos(d*x + c) + 1) - 1)/a^3) - 3*B*(40*sin(d*x + c)/((a^3 - a^3* sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (85*sin(d*x + c )/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3))/d
Time = 0.35 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.19 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (13 \, A - 6 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, A - 6 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="g iac")
Output:
1/60*(30*(13*A - 6*B)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*A - 6*B)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*A*tan(1/2*d*x + 1/2*c) ^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 - 5*A*tan(1/2*d*x + 1/2*c) + 2*B*tan(1/2*d *x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*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 - 30*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) - 255*B*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 42.36 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.10 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A-2\,B\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A-B\right )}{2\,a^3}+\frac {3\,\left (5\,A-3\,B\right )}{4\,a^3}+\frac {10\,A-2\,B}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A-B}{4\,a^3}+\frac {5\,A-3\,B}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-B\right )}{20\,a^3\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,A-6\,B\right )}{a^3\,d} \] Input:
int((A + B*cos(c + d*x))/(cos(c + d*x)^3*(a + a*cos(c + d*x))^3),x)
Output:
(tan(c/2 + (d*x)/2)^3*(7*A - 2*B) - tan(c/2 + (d*x)/2)*(5*A - 2*B))/(d*(a^ 3*tan(c/2 + (d*x)/2)^4 - 2*a^3*tan(c/2 + (d*x)/2)^2 + a^3)) - (tan(c/2 + ( d*x)/2)*((3*(A - B))/(2*a^3) + (3*(5*A - 3*B))/(4*a^3) + (10*A - 2*B)/(4*a ^3)))/d - (tan(c/2 + (d*x)/2)^3*((A - B)/(4*a^3) + (5*A - 3*B)/(12*a^3)))/ d - (tan(c/2 + (d*x)/2)^5*(A - B))/(20*a^3*d) + (atanh(tan(c/2 + (d*x)/2)) *(13*A - 6*B))/(a^3*d)
Time = 0.20 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.25 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int((A+B*cos(d*x+c))*sec(d*x+c)^3/(a+a*cos(d*x+c))^3,x)
Output:
( - 390*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*a + 180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*b + 780*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*a - 360*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b - 39 0*log(tan((c + d*x)/2) - 1)*a + 180*log(tan((c + d*x)/2) - 1)*b + 390*log( tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*a - 180*log(tan((c + d*x)/2) + 1 )*tan((c + d*x)/2)**4*b - 780*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)** 2*a + 360*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*b + 390*log(tan((c + d*x)/2) + 1)*a - 180*log(tan((c + d*x)/2) + 1)*b - 3*tan((c + d*x)/2)** 9*a + 3*tan((c + d*x)/2)**9*b - 34*tan((c + d*x)/2)**7*a + 24*tan((c + d*x )/2)**7*b - 388*tan((c + d*x)/2)**5*a + 198*tan((c + d*x)/2)**5*b + 1310*t an((c + d*x)/2)**3*a - 600*tan((c + d*x)/2)**3*b - 765*tan((c + d*x)/2)*a + 375*tan((c + d*x)/2)*b)/(60*a**3*d*(tan((c + d*x)/2)**4 - 2*tan((c + d*x )/2)**2 + 1))