Integrand size = 40, antiderivative size = 169 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a^2 (6 B+7 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (9 B+10 C) \tan (c+d x)}{5 d}+\frac {a^2 (6 B+7 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (6 B+5 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a^2 (9 B+10 C) \tan ^3(c+d x)}{15 d} \]
1/8*a^2*(6*B+7*C)*arctanh(sin(d*x+c))/d+1/5*a^2*(9*B+10*C)*tan(d*x+c)/d+1/ 8*a^2*(6*B+7*C)*sec(d*x+c)*tan(d*x+c)/d+1/20*a^2*(6*B+5*C)*sec(d*x+c)^3*ta n(d*x+c)/d+1/5*B*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^4*tan(d*x+c)/d+1/15*a^2*( 9*B+10*C)*tan(d*x+c)^3/d
Time = 1.78 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a^2 \left (15 (6 B+7 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 (6 B+7 C) \sec (c+d x)+30 (2 B+C) \sec ^3(c+d x)+8 \left (30 (B+C)+5 (3 B+2 C) \tan ^2(c+d x)+3 B \tan ^4(c+d x)\right )\right )\right )}{120 d} \]
(a^2*(15*(6*B + 7*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(6*B + 7*C)* Sec[c + d*x] + 30*(2*B + C)*Sec[c + d*x]^3 + 8*(30*(B + C) + 5*(3*B + 2*C) *Tan[c + d*x]^2 + 3*B*Tan[c + d*x]^4))))/(120*d)
Time = 1.16 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.95, 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, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 2009, 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.
| \(\displaystyle \int \sec ^7(c+d x) (a \cos (c+d x)+a)^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^7}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^6(c+d x) (a \cos (c+d x)+a)^2 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{5} \int (\cos (c+d x) a+a) (a (6 B+5 C)+a (3 B+5 C) \cos (c+d x)) \sec ^5(c+d x)dx+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (6 B+5 C)+a (3 B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{5} \int \left ((3 B+5 C) \cos ^2(c+d x) a^2+(6 B+5 C) a^2+\left ((3 B+5 C) a^2+(6 B+5 C) a^2\right ) \cos (c+d x)\right ) \sec ^5(c+d x)dx+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {(3 B+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(6 B+5 C) a^2+\left ((3 B+5 C) a^2+(6 B+5 C) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \left (4 (9 B+10 C) a^2+5 (6 B+7 C) \cos (c+d x) a^2\right ) \sec ^4(c+d x)dx+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {4 (9 B+10 C) a^2+5 (6 B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (4 a^2 (9 B+10 C) \int \sec ^4(c+d x)dx+5 a^2 (6 B+7 C) \int \sec ^3(c+d x)dx\right )+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+4 a^2 (9 B+10 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx\right )+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a^2 (9 B+10 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\right )+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 a^2 (9 B+10 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (9 B+10 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (9 B+10 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {4 a^2 (9 B+10 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\right )+\frac {a^2 (6 B+5 C) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {B \tan (c+d x) \sec ^4(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
(B*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a^2*(6* B + 5*C)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (5*a^2*(6*B + 7*C)*(ArcTanh[ Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) - (4*a^2*(9*B + 1 0*C)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/4)/5
3.3.43.3.1 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_)]*(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 = 8.10 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02
| method | result | size |
| parts | \(-\frac {\left (B \,a^{2}+2 a^{2} C \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 B \,a^{2}+a^{2} C \right ) \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {B \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(173\) |
| parallelrisch | \(\frac {8 \left (-\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (B +\frac {7 C}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16}+\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (B +\frac {7 C}{6}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16}+\left (\frac {7 B}{8}+\frac {11 C}{16}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {3 B}{4}+\frac {5 C}{6}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 B}{16}+\frac {7 C}{32}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {3 B}{20}+\frac {C}{6}\right ) \sin \left (5 d x +5 c \right )+\left (B +\frac {2 C}{3}\right ) \sin \left (d x +c \right )\right ) a^{2}}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(217\) |
| derivativedivides | \(\frac {a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-2 a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(223\) |
| default | \(\frac {a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-2 a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} C \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}\) | \(223\) |
| risch | \(-\frac {i a^{2} \left (90 B \,{\mathrm e}^{9 i \left (d x +c \right )}+105 C \,{\mathrm e}^{9 i \left (d x +c \right )}+420 B \,{\mathrm e}^{7 i \left (d x +c \right )}+330 C \,{\mathrm e}^{7 i \left (d x +c \right )}-240 B \,{\mathrm e}^{6 i \left (d x +c \right )}-480 C \,{\mathrm e}^{6 i \left (d x +c \right )}-1200 B \,{\mathrm e}^{4 i \left (d x +c \right )}-1120 C \,{\mathrm e}^{4 i \left (d x +c \right )}-420 B \,{\mathrm e}^{3 i \left (d x +c \right )}-330 C \,{\mathrm e}^{3 i \left (d x +c \right )}-720 B \,{\mathrm e}^{2 i \left (d x +c \right )}-800 C \,{\mathrm e}^{2 i \left (d x +c \right )}-90 B \,{\mathrm e}^{i \left (d x +c \right )}-105 C \,{\mathrm e}^{i \left (d x +c \right )}-144 B -160 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{4 d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{4 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\) | \(287\) |
-(B*a^2+2*C*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(2*B*a^2+C*a^2)/d*(- (-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c) ))-B*a^2/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+a^2*C/d*( 1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))
Time = 0.28 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.98 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + 24 \, B a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]
integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="fricas")
1/240*(15*(6*B + 7*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(6*B + 7*C)*a^2*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(9*B + 10*C)*a^2*c os(d*x + c)^4 + 15*(6*B + 7*C)*a^2*cos(d*x + c)^3 + 8*(9*B + 10*C)*a^2*cos (d*x + c)^2 + 30*(2*B + C)*a^2*cos(d*x + c) + 24*B*a^2)*sin(d*x + c))/(d*c os(d*x + c)^5)
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.64 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 30 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, C a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]
integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^7,x, algorithm="maxima")
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 160*(tan(d*x + c)^3 + 3*tan( d*x + c))*C*a^2 - 30*B*a^2*(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)) - 15*C*a^2*(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)) - 60*C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)))/d
Time = 0.35 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.46 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {15 \, {\left (6 \, B a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (6 \, B a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (90 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 420 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 490 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 864 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 540 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 790 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 390 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]
1/120*(15*(6*B*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(6*B *a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(90*B*a^2*tan(1/2*d *x + 1/2*c)^9 + 105*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 420*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 490*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 864*B*a^2*tan(1/2*d*x + 1/2 *c)^5 + 800*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 540*B*a^2*tan(1/2*d*x + 1/2*c)^ 3 - 790*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 390*B*a^2*tan(1/2*d*x + 1/2*c) + 37 5*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d
Time = 3.99 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.33 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^7(c+d x) \, dx=\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (6\,B+7\,C\right )}{4\,d}-\frac {\left (\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-7\,B\,a^2-\frac {49\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,B\,a^2}{5}+\frac {40\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-9\,B\,a^2-\frac {79\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,B\,a^2}{2}+\frac {25\,C\,a^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(a^2*atanh(tan(c/2 + (d*x)/2))*(6*B + 7*C))/(4*d) - (tan(c/2 + (d*x)/2)*(( 13*B*a^2)/2 + (25*C*a^2)/4) + tan(c/2 + (d*x)/2)^9*((3*B*a^2)/2 + (7*C*a^2 )/4) - tan(c/2 + (d*x)/2)^7*(7*B*a^2 + (49*C*a^2)/6) - tan(c/2 + (d*x)/2)^ 3*(9*B*a^2 + (79*C*a^2)/6) + tan(c/2 + (d*x)/2)^5*((72*B*a^2)/5 + (40*C*a^ 2)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))