Integrand size = 40, antiderivative size = 114 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a^3 (B+3 C) x+\frac {a^3 (7 B+6 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {5 a^3 B \sin (c+d x)}{2 d}+\frac {(2 B+C) \left (a^3+a^3 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {a B (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \]
a^3*(B+3*C)*x+1/2*a^3*(7*B+6*C)*arctanh(sin(d*x+c))/d-5/2*a^3*B*sin(d*x+c) /d+(2*B+C)*(a^3+a^3*cos(d*x+c))*tan(d*x+c)/d+1/2*a*B*(a+a*cos(d*x+c))^2*se c(d*x+c)*tan(d*x+c)/d
Time = 4.38 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.82 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^3 \left (4 B c+12 c C+4 B d x+12 C d x-14 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+14 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {B}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+4 C \sin (c+d x)+4 (3 B+C) \tan (c+d x)\right )}{4 d} \]
(a^3*(4*B*c + 12*c*C + 4*B*d*x + 12*C*d*x - 14*B*Log[Cos[(c + d*x)/2] - Si n[(c + d*x)/2]] - 12*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 14*B*Log [Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 12*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + B/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - B/(Cos[(c + d*x) /2] + Sin[(c + d*x)/2])^2 + 4*C*Sin[c + d*x] + 4*(3*B + C)*Tan[c + d*x]))/ (4*d)
Time = 1.09 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3454, 3042, 3454, 3042, 3447, 3042, 3502, 3042, 3214, 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.
| \(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^3 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{2} \int (\cos (c+d x) a+a)^2 (2 a (2 B+C)-a (B-2 C) \cos (c+d x)) \sec ^2(c+d x)dx+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a (2 B+C)-a (B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{2} \left (\int (\cos (c+d x) a+a) \left (a^2 (7 B+6 C)-5 a^2 B \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (7 B+6 C)-5 a^2 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{2} \left (\int \left (-5 B \cos ^2(c+d x) a^3+(7 B+6 C) a^3+\left (a^3 (7 B+6 C)-5 a^3 B\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {-5 B \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(7 B+6 C) a^3+\left (a^3 (7 B+6 C)-5 a^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{2} \left (\int \left ((7 B+6 C) a^3+2 (B+3 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}-\frac {5 a^3 B \sin (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \frac {(7 B+6 C) a^3+2 (B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}-\frac {5 a^3 B \sin (c+d x)}{d}\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{2} \left (a^3 (7 B+6 C) \int \sec (c+d x)dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}-\frac {5 a^3 B \sin (c+d x)}{d}+2 a^3 x (B+3 C)\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (a^3 (7 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}-\frac {5 a^3 B \sin (c+d x)}{d}+2 a^3 x (B+3 C)\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{2} \left (\frac {a^3 (7 B+6 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 (2 B+C) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}-\frac {5 a^3 B \sin (c+d x)}{d}+2 a^3 x (B+3 C)\right )+\frac {a B \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
(a*B*(a + a*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*a^3*(B + 3*C)*x + (a^3*(7*B + 6*C)*ArcTanh[Sin[c + d*x]])/d - (5*a^3*B*Sin[c + d*x ])/d + (2*(2*B + C)*(a^3 + a^3*Cos[c + d*x])*Tan[c + d*x])/d)/2
3.3.49.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, 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_)]), 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[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 6.78 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.13
| method | result | size |
| parts | \(\frac {\left (B \,a^{3}+3 C \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{3}+C \,a^{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 B \,a^{3}+3 C \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {a^{3} C \sin \left (d x +c \right )}{d}\) | \(129\) |
| derivativedivides | \(\frac {C \,a^{3} \sin \left (d x +c \right )+B \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+C \,a^{3} \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(137\) |
| default | \(\frac {C \,a^{3} \sin \left (d x +c \right )+B \,a^{3} \left (d x +c \right )+3 C \,a^{3} \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+C \,a^{3} \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(137\) |
| parallelrisch | \(-\frac {7 \left (\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {6 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (1+\cos \left (2 d x +2 c \right )\right ) \left (B +\frac {6 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {2 d x \left (B +3 C \right ) \cos \left (2 d x +2 c \right )}{7}+\frac {2 \left (-3 B -C \right ) \sin \left (2 d x +2 c \right )}{7}-\frac {\sin \left (3 d x +3 c \right ) C}{7}+\frac {\left (-2 B -C \right ) \sin \left (d x +c \right )}{7}-\frac {2 d x \left (B +3 C \right )}{7}\right ) a^{3}}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(154\) |
| risch | \(a^{3} B x +3 a^{3} C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{2 d}-\frac {i a^{3} \left (B \,{\mathrm e}^{3 i \left (d x +c \right )}-6 B \,{\mathrm e}^{2 i \left (d x +c \right )}-2 C \,{\mathrm e}^{2 i \left (d x +c \right )}-B \,{\mathrm e}^{i \left (d x +c \right )}-6 B -2 C \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {7 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {7 B \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(217\) |
| norman | \(\frac {\left (-B \,a^{3}-3 C \,a^{3}\right ) x +\left (-6 B \,a^{3}-18 C \,a^{3}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 B \,a^{3}-6 C \,a^{3}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 B \,a^{3}-6 C \,a^{3}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (B \,a^{3}+3 C \,a^{3}\right ) x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,a^{3}+6 C \,a^{3}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,a^{3}+6 C \,a^{3}\right ) x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 B \,a^{3}+18 C \,a^{3}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{3} \left (3 B -8 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{3} \left (25 B +16 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{3} \left (35 B +4 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {5 B \,a^{3} \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (7 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {a^{3} \left (13 B +4 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (15 B -4 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{3} \left (23 B +8 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a^{3} \left (7 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (7 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(481\) |
(B*a^3+3*C*a^3)/d*(d*x+c)+(3*B*a^3+C*a^3)/d*tan(d*x+c)+(3*B*a^3+3*C*a^3)/d *ln(sec(d*x+c)+tan(d*x+c))+B*a^3/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d *x+c)+tan(d*x+c)))+a^3*C*sin(d*x+c)/d
Time = 0.28 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.20 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (B + 3 \, C\right )} a^{3} d x \cos \left (d x + c\right )^{2} + {\left (7 \, B + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (7 \, B + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{3} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right ) + B a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")
1/4*(4*(B + 3*C)*a^3*d*x*cos(d*x + c)^2 + (7*B + 6*C)*a^3*cos(d*x + c)^2*l og(sin(d*x + c) + 1) - (7*B + 6*C)*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*C*a^3*cos(d*x + c)^2 + 2*(3*B + C)*a^3*cos(d*x + c) + B*a^3)*sin (d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
Time = 0.22 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.45 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} B a^{3} + 12 \, {\left (d x + c\right )} C a^{3} - B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{3} \sin \left (d x + c\right ) + 12 \, B a^{3} \tan \left (d x + c\right ) + 4 \, C a^{3} \tan \left (d x + c\right )}{4 \, d} \]
integrate((a+a*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")
1/4*(4*(d*x + c)*B*a^3 + 12*(d*x + c)*C*a^3 - 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)) + 6*B*a^3* (log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*C*a^3*(log(sin(d*x + c ) + 1) - log(sin(d*x + c) - 1)) + 4*C*a^3*sin(d*x + c) + 12*B*a^3*tan(d*x + c) + 4*C*a^3*tan(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.68 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {\frac {4 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (B a^{3} + 3 \, C a^{3}\right )} {\left (d x + c\right )} + {\left (7 \, B a^{3} + 6 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (7 \, B a^{3} + 6 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(4*C*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(B*a^3 + 3*C*a^3)*(d*x + c) + (7*B*a^3 + 6*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (7*B*a^3 + 6*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*B*a^3* tan(1/2*d*x + 1/2*c)^3 + 2*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*B*a^3*tan(1/2* d*x + 1/2*c) - 2*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^ 2)/d
Time = 1.58 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.82 \[ \int (a+a \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]
(C*a^3*sin(c + d*x))/d + (2*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 2)))/d + (7*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*C*a ^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (6*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (3*B*a^3*sin(c + d*x))/(d*cos(c + d*x )) + (B*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (C*a^3*sin(c + d*x))/(d*c os(c + d*x))