Integrand size = 31, antiderivative size = 181 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=4 a^4 A x+\frac {a^4 (52 A+35 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}+\frac {5 a^4 (4 A+7 C) \tan (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac {(12 A-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d} \]
4*a^4*A*x+1/8*a^4*(52*A+35*C)*arctanh(sin(d*x+c))/d+A*(a+a*sec(d*x+c))^4*s in(d*x+c)/d+5/8*a^4*(4*A+7*C)*tan(d*x+c)/d-1/4*a*(4*A-C)*(a+a*sec(d*x+c))^ 3*tan(d*x+c)/d-1/12*(12*A-7*C)*(a^2+a^2*sec(d*x+c))^2*tan(d*x+c)/d-1/24*(1 2*A-35*C)*(a^4+a^4*sec(d*x+c))*tan(d*x+c)/d
Time = 5.21 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.12 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=4 a^4 A x+\frac {13 a^4 A \text {arctanh}(\sin (c+d x))}{2 d}+\frac {35 a^4 C \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^4 A \sin (c+d x)}{d}+\frac {8 a^4 C \tan (c+d x)}{d}+\frac {a^4 A \sec (c+d x) \tan (c+d x)}{2 d}+\frac {27 a^4 C \sec (c+d x) \tan (c+d x)}{8 d}+\frac {4 a^4 A \sec ^2(c+d x) \tan (c+d x)}{d}+\frac {a^4 C \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {4 a^4 A \tan ^3(c+d x)}{d}+\frac {4 a^4 C \tan ^3(c+d x)}{3 d} \]
4*a^4*A*x + (13*a^4*A*ArcTanh[Sin[c + d*x]])/(2*d) + (35*a^4*C*ArcTanh[Sin [c + d*x]])/(8*d) + (a^4*A*Sin[c + d*x])/d + (8*a^4*C*Tan[c + d*x])/d + (a ^4*A*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (27*a^4*C*Sec[c + d*x]*Tan[c + d*x ])/(8*d) + (4*a^4*A*Sec[c + d*x]^2*Tan[c + d*x])/d + (a^4*C*Sec[c + d*x]^3 *Tan[c + d*x])/(4*d) - (4*a^4*A*Tan[c + d*x]^3)/d + (4*a^4*C*Tan[c + d*x]^ 3)/(3*d)
Time = 1.18 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.09, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 4575, 3042, 4405, 3042, 4405, 3042, 4405, 27, 3042, 4402, 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.
| \(\displaystyle \int \cos (c+d x) (a \sec (c+d x)+a)^4 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^4 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^4 (4 a A-a (4 A-C) \sec (c+d x))dx}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^4 \left (4 a A-a (4 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {1}{4} \int (\sec (c+d x) a+a)^3 \left (16 a^2 A-a^2 (12 A-7 C) \sec (c+d x)\right )dx-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (16 a^2 A-a^2 (12 A-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int (\sec (c+d x) a+a)^2 \left (48 a^3 A-a^3 (12 A-35 C) \sec (c+d x)\right )dx-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (48 a^3 A-a^3 (12 A-35 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int 3 (\sec (c+d x) a+a) \left (32 A a^4+5 (4 A+7 C) \sec (c+d x) a^4\right )dx-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int (\sec (c+d x) a+a) \left (32 A a^4+5 (4 A+7 C) \sec (c+d x) a^4\right )dx-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (32 A a^4+5 (4 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4\right )dx-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 4402 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (5 a^5 (4 A+7 C) \int \sec ^2(c+d x)dx+a^5 (52 A+35 C) \int \sec (c+d x)dx+32 a^5 A x\right )-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (a^5 (52 A+35 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+5 a^5 (4 A+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+32 a^5 A x\right )-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (-\frac {5 a^5 (4 A+7 C) \int 1d(-\tan (c+d x))}{d}+a^5 (52 A+35 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+32 a^5 A x\right )-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (a^5 (52 A+35 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^5 (4 A+7 C) \tan (c+d x)}{d}+32 a^5 A x\right )-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {3}{2} \left (\frac {a^5 (52 A+35 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^5 (4 A+7 C) \tan (c+d x)}{d}+32 a^5 A x\right )-\frac {(12 A-35 C) \tan (c+d x) \left (a^5 \sec (c+d x)+a^5\right )}{2 d}\right )-\frac {a^3 (12 A-7 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\right )-\frac {a^2 (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d}\) |
(A*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/d + (-1/4*(a^2*(4*A - C)*(a + a*Se c[c + d*x])^3*Tan[c + d*x])/d + (-1/3*(a^3*(12*A - 7*C)*(a + a*Sec[c + d*x ])^2*Tan[c + d*x])/d + (-1/2*((12*A - 35*C)*(a^5 + a^5*Sec[c + d*x])*Tan[c + d*x])/d + (3*(32*a^5*A*x + (a^5*(52*A + 35*C)*ArcTanh[Sin[c + d*x]])/d + (5*a^5*(4*A + 7*C)*Tan[c + d*x])/d))/2)/3)/4)/a
3.2.14.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x] + (Simp[b*d Int[Csc[e + f*x]^2, x], x ] + Simp[(b*c + a*d) Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f }, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m - 1)/(f*m)), x] + Simp[1/m Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 *m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[A*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 0.68 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.23
| method | result | size |
| parallelrisch | \(\frac {8 a^{4} \left (-\frac {13 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {35 C}{52}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {13 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {35 C}{52}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+2 d x A \cos \left (2 d x +2 c \right )+\frac {d x A \cos \left (4 d x +4 c \right )}{2}+\left (A +\frac {7 C}{3}\right ) \sin \left (2 d x +2 c \right )+\frac {\left (5 A +\frac {27 C}{2}\right ) \sin \left (3 d x +3 c \right )}{16}+\frac {\left (A +\frac {5 C}{3}\right ) \sin \left (4 d x +4 c \right )}{2}+\frac {A \sin \left (5 d x +5 c \right )}{16}+\frac {\left (A +\frac {35 C}{8}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x A}{2}\right )}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(223\) |
| derivativedivides | \(\frac {a^{4} A \sin \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (d x +c \right )+4 a^{4} C \tan \left (d x +c \right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \tan \left (d x +c \right )-4 a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(237\) |
| default | \(\frac {a^{4} A \sin \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} A \left (d x +c \right )+4 a^{4} C \tan \left (d x +c \right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \tan \left (d x +c \right )-4 a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{4} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(237\) |
| risch | \(4 a^{4} A x -\frac {i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{4} \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+81 C \,{\mathrm e}^{7 i \left (d x +c \right )}-96 A \,{\mathrm e}^{6 i \left (d x +c \right )}-96 C \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+105 C \,{\mathrm e}^{5 i \left (d x +c \right )}-288 A \,{\mathrm e}^{4 i \left (d x +c \right )}-480 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-105 C \,{\mathrm e}^{3 i \left (d x +c \right )}-288 A \,{\mathrm e}^{2 i \left (d x +c \right )}-544 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,{\mathrm e}^{i \left (d x +c \right )}-81 C \,{\mathrm e}^{i \left (d x +c \right )}-96 A -160 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\) | \(332\) |
| norman | \(\frac {-4 a^{4} A x +16 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-20 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+20 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-16 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+4 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}-\frac {5 a^{4} \left (4 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}-\frac {a^{4} \left (44 A +93 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {7 a^{4} \left (60 A +73 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {a^{4} \left (-203 C +12 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {a^{4} \left (53 C +204 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}+\frac {a^{4} \left (385 C +156 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{4} \left (52 A +35 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{4} \left (52 A +35 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(339\) |
8*a^4*(-13/4*(3/4+1/4*cos(4*d*x+4*c)+cos(2*d*x+2*c))*(A+35/52*C)*ln(tan(1/ 2*d*x+1/2*c)-1)+13/4*(3/4+1/4*cos(4*d*x+4*c)+cos(2*d*x+2*c))*(A+35/52*C)*l n(tan(1/2*d*x+1/2*c)+1)+2*d*x*A*cos(2*d*x+2*c)+1/2*d*x*A*cos(4*d*x+4*c)+(A +7/3*C)*sin(2*d*x+2*c)+1/16*(5*A+27/2*C)*sin(3*d*x+3*c)+1/2*(A+5/3*C)*sin( 4*d*x+4*c)+1/16*A*sin(5*d*x+5*c)+1/4*(A+35/8*C)*sin(d*x+c)+3/2*d*x*A)/d/(c os(4*d*x+4*c)+4*cos(2*d*x+2*c)+3)
Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.94 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {192 \, A a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (52 \, A + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (52 \, A + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 32 \, {\left (3 \, A + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 27 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 32 \, C a^{4} \cos \left (d x + c\right ) + 6 \, C a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(192*A*a^4*d*x*cos(d*x + c)^4 + 3*(52*A + 35*C)*a^4*cos(d*x + c)^4*lo g(sin(d*x + c) + 1) - 3*(52*A + 35*C)*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*A*a^4*cos(d*x + c)^4 + 32*(3*A + 5*C)*a^4*cos(d*x + c)^3 + 3 *(4*A + 27*C)*a^4*cos(d*x + c)^2 + 32*C*a^4*cos(d*x + c) + 6*C*a^4)*sin(d* x + c))/(d*cos(d*x + c)^4)
\[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{4} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 C \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 C \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 C \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]
a**4*(Integral(A*cos(c + d*x), x) + Integral(4*A*cos(c + d*x)*sec(c + d*x) , x) + Integral(6*A*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(4*A*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(4*C*cos(c + d*x)*s ec(c + d*x)**3, x) + Integral(6*C*cos(c + d*x)*sec(c + d*x)**4, x) + Integ ral(4*C*cos(c + d*x)*sec(c + d*x)**5, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**6, x))
Time = 0.21 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.64 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {192 \, {\left (d x + c\right )} A a^{4} + 64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 3 \, C a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 192 \, A a^{4} \tan \left (d x + c\right ) + 192 \, C a^{4} \tan \left (d x + c\right )}{48 \, d} \]
1/48*(192*(d*x + c)*A*a^4 + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 3 *C*a^4*(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)) - 12*A*a^ 4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d *x + c) - 1)) - 72*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d* x + c) + 1) + log(sin(d*x + c) - 1)) + 144*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c ) - 1)) + 48*A*a^4*sin(d*x + c) + 192*A*a^4*tan(d*x + c) + 192*C*a^4*tan(d *x + c))/d
Time = 0.36 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.40 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {96 \, {\left (d x + c\right )} A a^{4} + \frac {48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 3 \, {\left (52 \, A a^{4} + 35 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (52 \, A a^{4} + 35 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (84 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 276 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 385 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 300 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 511 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 279 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
1/24*(96*(d*x + c)*A*a^4 + 48*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/ 2*c)^2 + 1) + 3*(52*A*a^4 + 35*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(52*A*a^4 + 35*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(84*A*a^4* tan(1/2*d*x + 1/2*c)^7 + 105*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 276*A*a^4*tan( 1/2*d*x + 1/2*c)^5 - 385*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 300*A*a^4*tan(1/2* d*x + 1/2*c)^3 + 511*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 108*A*a^4*tan(1/2*d*x + 1/2*c) - 279*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4) /d
Time = 15.74 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.36 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,a^4\,\sin \left (c+d\,x\right )}{d}+\frac {8\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {13\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {35\,C\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {4\,A\,a^4\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^4\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {20\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {27\,C\,a^4\,\sin \left (c+d\,x\right )}{8\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {4\,C\,a^4\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^4\,\sin \left (c+d\,x\right )}{4\,d\,{\cos \left (c+d\,x\right )}^4} \]
(A*a^4*sin(c + d*x))/d + (8*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 2)))/d + (13*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (35*C *a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (4*A*a^4*sin(c + d*x))/(d*cos(c + d*x)) + (A*a^4*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (20 *C*a^4*sin(c + d*x))/(3*d*cos(c + d*x)) + (27*C*a^4*sin(c + d*x))/(8*d*cos (c + d*x)^2) + (4*C*a^4*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (C*a^4*sin(c + d*x))/(4*d*cos(c + d*x)^4)