Integrand size = 40, antiderivative size = 154 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {5 a^3 (3 B+4 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (9 B+11 C) \tan (c+d x)}{3 d}+\frac {a^3 (27 B+28 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 B+2 C) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
5/8*a^3*(3*B+4*C)*arctanh(sin(d*x+c))/d+1/3*a^3*(9*B+11*C)*tan(d*x+c)/d+1/ 24*a^3*(27*B+28*C)*sec(d*x+c)*tan(d*x+c)/d+1/6*(3*B+2*C)*(a^3+a^3*cos(d*x+ c))*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a*B*(a+a*cos(d*x+c))^2*sec(d*x+c)^3*tan( d*x+c)/d
Time = 1.55 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 a^3 B \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^3 C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 B \tan (c+d x)}{d}+\frac {4 a^3 C \tan (c+d x)}{d}+\frac {15 a^3 B \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 B \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 B \tan ^3(c+d x)}{d}+\frac {a^3 C \tan ^3(c+d x)}{3 d} \]
(15*a^3*B*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*C*ArcTanh[Sin[c + d*x]])/( 2*d) + (4*a^3*B*Tan[c + d*x])/d + (4*a^3*C*Tan[c + d*x])/d + (15*a^3*B*Sec [c + d*x]*Tan[c + d*x])/(8*d) + (3*a^3*C*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^3*B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*B*Tan[c + d*x]^3)/d + ( a^3*C*Tan[c + d*x]^3)/(3*d)
Time = 1.31 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.07, 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, 3454, 3042, 3454, 3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^3 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{4} \int (\cos (c+d x) a+a)^2 (2 a (3 B+2 C)+a (B+4 C) \cos (c+d x)) \sec ^4(c+d x)dx+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a (3 B+2 C)+a (B+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int (\cos (c+d x) a+a) \left ((27 B+28 C) a^2+(9 B+16 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((27 B+28 C) a^2+(9 B+16 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left ((9 B+16 C) \cos ^2(c+d x) a^3+(27 B+28 C) a^3+\left ((9 B+16 C) a^3+(27 B+28 C) a^3\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {(9 B+16 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(27 B+28 C) a^3+\left ((9 B+16 C) a^3+(27 B+28 C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (8 (9 B+11 C) a^3+15 (3 B+4 C) \cos (c+d x) a^3\right ) \sec ^2(c+d x)dx+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {8 (9 B+11 C) a^3+15 (3 B+4 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (8 a^3 (9 B+11 C) \int \sec ^2(c+d x)dx+15 a^3 (3 B+4 C) \int \sec (c+d x)dx\right )+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a^3 (3 B+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 a^3 (9 B+11 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a^3 (3 B+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 a^3 (9 B+11 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a^3 (3 B+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {8 a^3 (9 B+11 C) \tan (c+d x)}{d}\right )+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {15 a^3 (3 B+4 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a^3 (9 B+11 C) \tan (c+d x)}{d}\right )+\frac {a^3 (27 B+28 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (3 B+2 C) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}\right )+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
(a*B*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*(3*B + 2*C)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^3 *(27*B + 28*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((15*a^3*(3*B + 4*C)*Arc Tanh[Sin[c + d*x]])/d + (8*a^3*(9*B + 11*C)*Tan[c + d*x])/d)/2)/3)/4
3.3.51.3.1 Defintions of rubi rules used
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_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
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^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 9.12 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.16
| method | result | size |
| parallelrisch | \(\frac {10 a^{3} \left (-\frac {3 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {4 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {3 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {4 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\left (B +\frac {13 C}{15}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {3 B}{8}+\frac {3 C}{10}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 B}{10}+\frac {11 C}{30}\right ) \sin \left (4 d x +4 c \right )+\frac {23 \left (B +\frac {12 C}{23}\right ) \sin \left (d x +c \right )}{40}\right )}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(178\) |
| parts | \(\frac {\left (B \,a^{3}+3 C \,a^{3}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (3 B \,a^{3}+C \,a^{3}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{3}+3 C \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{3}}{d}\) | \(181\) |
| derivativedivides | \(\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )+3 C \,a^{3} \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-C \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(219\) |
| default | \(\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \tan \left (d x +c \right )+3 C \,a^{3} \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-C \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(219\) |
| risch | \(-\frac {i a^{3} \left (45 B \,{\mathrm e}^{7 i \left (d x +c \right )}+36 C \,{\mathrm e}^{7 i \left (d x +c \right )}-24 B \,{\mathrm e}^{6 i \left (d x +c \right )}-72 C \,{\mathrm e}^{6 i \left (d x +c \right )}+69 B \,{\mathrm e}^{5 i \left (d x +c \right )}+36 C \,{\mathrm e}^{5 i \left (d x +c \right )}-216 B \,{\mathrm e}^{4 i \left (d x +c \right )}-264 C \,{\mathrm e}^{4 i \left (d x +c \right )}-69 B \,{\mathrm e}^{3 i \left (d x +c \right )}-36 C \,{\mathrm e}^{3 i \left (d x +c \right )}-264 B \,{\mathrm e}^{2 i \left (d x +c \right )}-280 C \,{\mathrm e}^{2 i \left (d x +c \right )}-45 B \,{\mathrm e}^{i \left (d x +c \right )}-36 C \,{\mathrm e}^{i \left (d x +c \right )}-72 B -88 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {15 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {15 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) | \(287\) |
10*a^3*(-3/4*(3/4+1/4*cos(4*d*x+4*c)+cos(2*d*x+2*c))*(B+4/3*C)*ln(tan(1/2* d*x+1/2*c)-1)+3/4*(3/4+1/4*cos(4*d*x+4*c)+cos(2*d*x+2*c))*(B+4/3*C)*ln(tan (1/2*d*x+1/2*c)+1)+(B+13/15*C)*sin(2*d*x+2*c)+(3/8*B+3/10*C)*sin(3*d*x+3*c )+(3/10*B+11/30*C)*sin(4*d*x+4*c)+23/40*(B+12/23*C)*sin(d*x+c))/d/(cos(4*d *x+4*c)+4*cos(2*d*x+2*c)+3)
Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.94 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (9 \, B + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (5 \, B + 4 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + 6 \, B a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")
1/48*(15*(3*B + 4*C)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*(3*B + 4*C)*a^3*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(9*B + 11*C)*a^3*cos (d*x + c)^3 + 9*(5*B + 4*C)*a^3*cos(d*x + c)^2 + 8*(3*B + C)*a^3*cos(d*x + c) + 6*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.75 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 3 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{3} \tan \left (d x + c\right ) + 144 \, C a^{3} \tan \left (d x + c\right )}{48 \, d} \]
integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")
1/48*(48*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 16*(tan(d*x + c)^3 + 3* tan(d*x + c))*C*a^3 - 3*B*a^3*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin( d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d *x + c) - 1)) - 36*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d* x + c) + 1) + log(sin(d*x + c) - 1)) - 36*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*a^3*(lo g(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*B*a^3*tan(d*x + c) + 144 *C*a^3*tan(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.38 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 4 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, B a^{3} + 4 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 165 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 220 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 219 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 147 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 132 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(15*(3*B*a^3 + 4*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B* a^3 + 4*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(45*B*a^3*tan(1/2*d* x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 165*B*a^3*tan(1/2*d*x + 1 /2*c)^5 - 220*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 219*B*a^3*tan(1/2*d*x + 1/2*c )^3 + 292*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 147*B*a^3*tan(1/2*d*x + 1/2*c) - 132*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 3.78 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.20 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\left (-\frac {15\,B\,a^3}{4}-5\,C\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {55\,B\,a^3}{4}+\frac {55\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {73\,B\,a^3}{4}-\frac {73\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,B\,a^3}{4}+11\,C\,a^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,B+4\,C\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*((49*B*a^3)/4 + 11*C*a^3) - tan(c/2 + (d*x)/2)^7*((15* B*a^3)/4 + 5*C*a^3) + tan(c/2 + (d*x)/2)^5*((55*B*a^3)/4 + (55*C*a^3)/3) - tan(c/2 + (d*x)/2)^3*((73*B*a^3)/4 + (73*C*a^3)/3))/(d*(6*tan(c/2 + (d*x) /2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/ 2)^8 + 1)) + (5*a^3*atanh(tan(c/2 + (d*x)/2))*(3*B + 4*C))/(4*d)