Integrand size = 40, antiderivative size = 196 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {(13 B-6 C) \text {arctanh}(\sin (c+d x))}{2 a^3 d}-\frac {8 (19 B-9 C) \tan (c+d x)}{15 a^3 d}+\frac {(13 B-6 C) \sec (c+d x) \tan (c+d x)}{2 a^3 d}-\frac {(B-C) \sec (c+d x) \tan (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {(11 B-6 C) \sec (c+d x) \tan (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {4 (19 B-9 C) \sec (c+d x) \tan (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
1/2*(13*B-6*C)*arctanh(sin(d*x+c))/a^3/d-8/15*(19*B-9*C)*tan(d*x+c)/a^3/d+ 1/2*(13*B-6*C)*sec(d*x+c)*tan(d*x+c)/a^3/d-1/5*(B-C)*sec(d*x+c)*tan(d*x+c) /d/(a+a*cos(d*x+c))^3-1/15*(11*B-6*C)*sec(d*x+c)*tan(d*x+c)/a/d/(a+a*cos(d *x+c))^2-4/15*(19*B-9*C)*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 = 6.27 (sec) , antiderivative size = 610, normalized size of antiderivative = 3.11 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {1920 (13 B-6 C) \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 B+870 C) \sin \left (\frac {d x}{2}\right )+5 (761 B-366 C) \sin \left (\frac {3 d x}{2}\right )-4329 B \sin \left (c-\frac {d x}{2}\right )+2094 C \sin \left (c-\frac {d x}{2}\right )+1989 B \sin \left (c+\frac {d x}{2}\right )-1314 C \sin \left (c+\frac {d x}{2}\right )-3575 B \sin \left (2 c+\frac {d x}{2}\right )+1650 C \sin \left (2 c+\frac {d x}{2}\right )-475 B \sin \left (c+\frac {3 d x}{2}\right )+450 C \sin \left (c+\frac {3 d x}{2}\right )+2005 B \sin \left (2 c+\frac {3 d x}{2}\right )-1230 C \sin \left (2 c+\frac {3 d x}{2}\right )-2275 B \sin \left (3 c+\frac {3 d x}{2}\right )+1050 C \sin \left (3 c+\frac {3 d x}{2}\right )+2673 B \sin \left (c+\frac {5 d x}{2}\right )-1278 C \sin \left (c+\frac {5 d x}{2}\right )+105 B \sin \left (2 c+\frac {5 d x}{2}\right )+90 C \sin \left (2 c+\frac {5 d x}{2}\right )+1593 B \sin \left (3 c+\frac {5 d x}{2}\right )-918 C \sin \left (3 c+\frac {5 d x}{2}\right )-975 B \sin \left (4 c+\frac {5 d x}{2}\right )+450 C \sin \left (4 c+\frac {5 d x}{2}\right )+1325 B \sin \left (2 c+\frac {7 d x}{2}\right )-630 C \sin \left (2 c+\frac {7 d x}{2}\right )+255 B \sin \left (3 c+\frac {7 d x}{2}\right )-60 C \sin \left (3 c+\frac {7 d x}{2}\right )+875 B \sin \left (4 c+\frac {7 d x}{2}\right )-480 C \sin \left (4 c+\frac {7 d x}{2}\right )-195 B \sin \left (5 c+\frac {7 d x}{2}\right )+90 C \sin \left (5 c+\frac {7 d x}{2}\right )+304 B \sin \left (3 c+\frac {9 d x}{2}\right )-144 C \sin \left (3 c+\frac {9 d x}{2}\right )+90 B \sin \left (4 c+\frac {9 d x}{2}\right )-30 C \sin \left (4 c+\frac {9 d x}{2}\right )+214 B \sin \left (5 c+\frac {9 d x}{2}\right )-114 C \sin \left (5 c+\frac {9 d x}{2}\right )\right )}{480 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[ c + d*x])^3,x]
Output:
-1/480*(1920*(13*B - 6*C)*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*B + 870*C)*Sin[(d*x)/2] + 5*(761* B - 366*C)*Sin[(3*d*x)/2] - 4329*B*Sin[c - (d*x)/2] + 2094*C*Sin[c - (d*x) /2] + 1989*B*Sin[c + (d*x)/2] - 1314*C*Sin[c + (d*x)/2] - 3575*B*Sin[2*c + (d*x)/2] + 1650*C*Sin[2*c + (d*x)/2] - 475*B*Sin[c + (3*d*x)/2] + 450*C*S in[c + (3*d*x)/2] + 2005*B*Sin[2*c + (3*d*x)/2] - 1230*C*Sin[2*c + (3*d*x) /2] - 2275*B*Sin[3*c + (3*d*x)/2] + 1050*C*Sin[3*c + (3*d*x)/2] + 2673*B*S in[c + (5*d*x)/2] - 1278*C*Sin[c + (5*d*x)/2] + 105*B*Sin[2*c + (5*d*x)/2] + 90*C*Sin[2*c + (5*d*x)/2] + 1593*B*Sin[3*c + (5*d*x)/2] - 918*C*Sin[3*c + (5*d*x)/2] - 975*B*Sin[4*c + (5*d*x)/2] + 450*C*Sin[4*c + (5*d*x)/2] + 1325*B*Sin[2*c + (7*d*x)/2] - 630*C*Sin[2*c + (7*d*x)/2] + 255*B*Sin[3*c + (7*d*x)/2] - 60*C*Sin[3*c + (7*d*x)/2] + 875*B*Sin[4*c + (7*d*x)/2] - 480 *C*Sin[4*c + (7*d*x)/2] - 195*B*Sin[5*c + (7*d*x)/2] + 90*C*Sin[5*c + (7*d *x)/2] + 304*B*Sin[3*c + (9*d*x)/2] - 144*C*Sin[3*c + (9*d*x)/2] + 90*B*Si n[4*c + (9*d*x)/2] - 30*C*Sin[4*c + (9*d*x)/2] + 214*B*Sin[5*c + (9*d*x)/2 ] - 114*C*Sin[5*c + (9*d*x)/2]))/(a^3*d*(1 + Cos[c + d*x])^3)
Time = 1.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3508, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \frac {\sec ^3(c+d x) (B+C \cos (c+d x))}{(a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {B+C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {(a (7 B-2 C)-4 a (B-C) \cos (c+d x)) \sec ^3(c+d x)}{(\cos (c+d x) a+a)^2}dx}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 B-2 C)-4 a (B-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {\left (a^2 (43 B-18 C)-3 a^2 (11 B-6 C) \cos (c+d x)\right ) \sec ^3(c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (43 B-18 C)-3 a^2 (11 B-6 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \left (15 a^3 (13 B-6 C)-8 a^3 (19 B-9 C) \cos (c+d x)\right ) \sec ^3(c+d x)dx}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {15 a^3 (13 B-6 C)-8 a^3 (19 B-9 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 B-6 C) \int \sec ^3(c+d x)dx-8 a^3 (19 B-9 C) \int \sec ^2(c+d x)dx}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 B-6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-8 a^3 (19 B-9 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 a^3 (19 B-9 C) \int 1d(-\tan (c+d x))}{d}+15 a^3 (13 B-6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 B-6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {8 a^3 (19 B-9 C) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 B-6 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 B-9 C) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 B-6 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 B-9 C) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\frac {15 a^3 (13 B-6 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {8 a^3 (19 B-9 C) \tan (c+d x)}{d}}{a^2}-\frac {4 a^2 (19 B-9 C) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)}}{3 a^2}-\frac {a (11 B-6 C) \tan (c+d x) \sec (c+d x)}{3 d (a \cos (c+d x)+a)^2}}{5 a^2}-\frac {(B-C) \tan (c+d x) \sec (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[((B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^4)/(a + a*Cos[c + d* x])^3,x]
Output:
-1/5*((B - C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^3) + (-1/ 3*(a*(11*B - 6*C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + ((-4*a^2*(19*B - 9*C)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x])) + ((-8*a^3*(19*B - 9*C)*Tan[c + d*x])/d + 15*a^3*(13*B - 6*C)*(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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*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 = 0.65 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.89
| method | result | size |
| parallelrisch | \(\frac {-1560 \left (B -\frac {6 C}{13}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+1560 \left (B -\frac {6 C}{13}\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-152 \sec \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (\frac {783 B}{76}-\frac {189 C}{38}\right ) \cos \left (2 d x +2 c \right )+\left (\frac {717 B}{152}-\frac {9 C}{4}\right ) \cos \left (3 d x +3 c \right )+\left (B -\frac {9 C}{19}\right ) \cos \left (4 d x +4 c \right )+\left (\frac {2331 B}{152}-\frac {573 C}{76}\right ) \cos \left (d x +c \right )+\frac {677 B}{76}-\frac {9 C}{2}\right )}{240 d \,a^{3} \left (\cos \left (2 d x +2 c \right )+1\right )}\) | \(175\) |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} B}{5}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} B}{3}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C -31 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B +17 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (26 B -12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-26 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {-14 B +4 C}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\frac {2 B}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}}{4 d \,a^{3}}\) | \(206\) |
| norman | \(\frac {-\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{20 a d}-\frac {\left (37 B -27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{60 d a}+\frac {\left (51 B -25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {\left (109 B -45 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d a}-\frac {7 \left (241 B -111 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d a}-\frac {\left (419 B -219 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{60 d a}+\frac {\left (959 B -429 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d a}+\frac {\left (967 B -447 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{60 d a}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a^{2}}-\frac {\left (13 B -6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a^{3} d}+\frac {\left (13 B -6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a^{3} d}\) | \(295\) |
| risch | \(-\frac {i \left (195 \,{\mathrm e}^{8 i \left (d x +c \right )} B -90 C \,{\mathrm e}^{8 i \left (d x +c \right )}+975 B \,{\mathrm e}^{7 i \left (d x +c \right )}-450 C \,{\mathrm e}^{7 i \left (d x +c \right )}+2275 B \,{\mathrm e}^{6 i \left (d x +c \right )}-1050 C \,{\mathrm e}^{6 i \left (d x +c \right )}+3575 B \,{\mathrm e}^{5 i \left (d x +c \right )}-1650 C \,{\mathrm e}^{5 i \left (d x +c \right )}+4329 B \,{\mathrm e}^{4 i \left (d x +c \right )}-2094 C \,{\mathrm e}^{4 i \left (d x +c \right )}+3805 B \,{\mathrm e}^{3 i \left (d x +c \right )}-1830 C \,{\mathrm e}^{3 i \left (d x +c \right )}+2673 B \,{\mathrm e}^{2 i \left (d x +c \right )}-1278 C \,{\mathrm e}^{2 i \left (d x +c \right )}+1325 B \,{\mathrm e}^{i \left (d x +c \right )}-630 C \,{\mathrm e}^{i \left (d x +c \right )}+304 B -144 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {13 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a^{3} d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}+\frac {13 B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a^{3} d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}\) | \(324\) |
Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x,method =_RETURNVERBOSE)
Output:
1/240*(-1560*(B-6/13*C)*(cos(2*d*x+2*c)+1)*ln(tan(1/2*d*x+1/2*c)-1)+1560*( B-6/13*C)*(cos(2*d*x+2*c)+1)*ln(tan(1/2*d*x+1/2*c)+1)-152*sec(1/2*d*x+1/2* c)^4*tan(1/2*d*x+1/2*c)*((783/76*B-189/38*C)*cos(2*d*x+2*c)+(717/152*B-9/4 *C)*cos(3*d*x+3*c)+(B-9/19*C)*cos(4*d*x+4*c)+(2331/152*B-573/76*C)*cos(d*x +c)+677/76*B-9/2*C))/d/a^3/(cos(2*d*x+2*c)+1)
Time = 0.08 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.51 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {15 \, {\left ({\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left ({\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{5} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (13 \, B - 6 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (16 \, {\left (19 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (239 \, B - 114 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (479 \, B - 234 \, C\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right ) - 15 \, B\right )} \sin \left (d x + c\right )}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{5} + 3 \, a^{3} d \cos \left (d x + c\right )^{4} + 3 \, a^{3} d \cos \left (d x + c\right )^{3} + a^{3} d \cos \left (d x + c\right )^{2}\right )}} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
Output:
1/60*(15*((13*B - 6*C)*cos(d*x + c)^5 + 3*(13*B - 6*C)*cos(d*x + c)^4 + 3* (13*B - 6*C)*cos(d*x + c)^3 + (13*B - 6*C)*cos(d*x + c)^2)*log(sin(d*x + c ) + 1) - 15*((13*B - 6*C)*cos(d*x + c)^5 + 3*(13*B - 6*C)*cos(d*x + c)^4 + 3*(13*B - 6*C)*cos(d*x + c)^3 + (13*B - 6*C)*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*(16*(19*B - 9*C)*cos(d*x + c)^4 + 3*(239*B - 114*C)*cos(d*x + c)^3 + (479*B - 234*C)*cos(d*x + c)^2 + 15*(3*B - 2*C)*cos(d*x + c) - 1 5*B)*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)
Timed out. \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**4/(a+a*cos(d*x+c))**3 ,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 377 vs. \(2 (184) = 368\).
Time = 0.07 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.92 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {B {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} - \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {390 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - 3 \, C {\left (\frac {40 \, \sin \left (d x + c\right )}{{\left (a^{3} - \frac {a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )}}{60 \, d} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
Output:
-1/60*(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 - 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*C*(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 = 0.16 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.19 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {30 \, {\left (13 \, B - 6 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {30 \, {\left (13 \, B - 6 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} + \frac {60 \, {\left (7 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3}} - \frac {3 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, B a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \] Input:
integrate((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(30*(13*B - 6*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 30*(13*B - 6*C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 + 60*(7*B*tan(1/2*d*x + 1/2*c) ^3 - 2*C*tan(1/2*d*x + 1/2*c)^3 - 5*B*tan(1/2*d*x + 1/2*c) + 2*C*tan(1/2*d *x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 1)^2*a^3) - (3*B*a^12*tan(1/2*d*x + 1/2*c)^5 - 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^12*tan(1/2*d*x + 1/2 *c)^3 - 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*B*a^12*tan(1/2*d*x + 1/2*c) - 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 0.27 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.10 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,B-2\,C\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,B-2\,C\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (B-C\right )}{2\,a^3}+\frac {3\,\left (5\,B-3\,C\right )}{4\,a^3}+\frac {10\,B-2\,C}{4\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {B-C}{4\,a^3}+\frac {5\,B-3\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (B-C\right )}{20\,a^3\,d}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (13\,B-6\,C\right )}{a^3\,d} \] Input:
int((B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^4*(a + a*cos(c + d*x ))^3),x)
Output:
(tan(c/2 + (d*x)/2)^3*(7*B - 2*C) - tan(c/2 + (d*x)/2)*(5*B - 2*C))/(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*(B - C))/(2*a^3) + (3*(5*B - 3*C))/(4*a^3) + (10*B - 2*C)/(4*a ^3)))/d - (tan(c/2 + (d*x)/2)^3*((B - C)/(4*a^3) + (5*B - 3*C)/(12*a^3)))/ d - (tan(c/2 + (d*x)/2)^5*(B - C))/(20*a^3*d) + (atanh(tan(c/2 + (d*x)/2)) *(13*B - 6*C))/(a^3*d)
Time = 0.17 (sec) , antiderivative size = 441, normalized size of antiderivative = 2.25 \[ \int \frac {\left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x)}{(a+a \cos (c+d x))^3} \, dx =\text {Too large to display} \] Input:
int((B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4/(a+a*cos(d*x+c))^3,x)
Output:
( - 390*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*b + 180*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**4*c + 780*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*b - 360*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**2*c - 39 0*log(tan((c + d*x)/2) - 1)*b + 180*log(tan((c + d*x)/2) - 1)*c + 390*log( tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**4*b - 180*log(tan((c + d*x)/2) + 1 )*tan((c + d*x)/2)**4*c - 780*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)** 2*b + 360*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**2*c + 390*log(tan((c + d*x)/2) + 1)*b - 180*log(tan((c + d*x)/2) + 1)*c - 3*tan((c + d*x)/2)** 9*b + 3*tan((c + d*x)/2)**9*c - 34*tan((c + d*x)/2)**7*b + 24*tan((c + d*x )/2)**7*c - 388*tan((c + d*x)/2)**5*b + 198*tan((c + d*x)/2)**5*c + 1310*t an((c + d*x)/2)**3*b - 600*tan((c + d*x)/2)**3*c - 765*tan((c + d*x)/2)*b + 375*tan((c + d*x)/2)*c)/(60*a**3*d*(tan((c + d*x)/2)**4 - 2*tan((c + d*x )/2)**2 + 1))