Integrand size = 31, antiderivative size = 145 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=3 a^3 A x+\frac {a^3 (6 A+5 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+a \sec (c+d x))^3 \sin (c+d x)}{d}+\frac {5 a^3 C \tan (c+d x)}{2 d}-\frac {(3 A-C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{3 a d}-\frac {(6 A-5 C) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d} \]
3*a^3*A*x+1/2*a^3*(6*A+5*C)*arctanh(sin(d*x+c))/d+A*(a+a*sec(d*x+c))^3*sin (d*x+c)/d+5/2*a^3*C*tan(d*x+c)/d-1/3*(3*A-C)*(a^2+a^2*sec(d*x+c))^2*tan(d* x+c)/a/d-1/6*(6*A-5*C)*(a^3+a^3*sec(d*x+c))*tan(d*x+c)/d
Time = 1.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=3 a^3 A x+\frac {3 a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 C \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^3 A \sin (c+d x)}{d}+\frac {4 a^3 C \tan (c+d x)}{d}+\frac {3 a^3 C \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 A \sec ^2(c+d x) \tan (c+d x)}{d}-\frac {a^3 A \tan ^3(c+d x)}{d}+\frac {a^3 C \tan ^3(c+d x)}{3 d} \]
3*a^3*A*x + (3*a^3*A*ArcTanh[Sin[c + d*x]])/d + (5*a^3*C*ArcTanh[Sin[c + d *x]])/(2*d) + (a^3*A*Sin[c + d*x])/d + (4*a^3*C*Tan[c + d*x])/d + (3*a^3*C *Sec[c + d*x]*Tan[c + d*x])/(2*d) + (a^3*A*Sec[c + d*x]^2*Tan[c + d*x])/d - (a^3*A*Tan[c + d*x]^3)/d + (a^3*C*Tan[c + d*x]^3)/(3*d)
Time = 0.91 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4575, 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.
| \(\displaystyle \int \cos (c+d x) (a \sec (c+d x)+a)^3 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int (\sec (c+d x) a+a)^3 (3 a A-a (3 A-C) \sec (c+d x))dx}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a A-a (3 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {1}{3} \int (\sec (c+d x) a+a)^2 \left (9 a^2 A-a^2 (6 A-5 C) \sec (c+d x)\right )dx-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (9 a^2 A-a^2 (6 A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 4405 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int 3 (\sec (c+d x) a+a) \left (6 A a^3+5 C \sec (c+d x) a^3\right )dx-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int (\sec (c+d x) a+a) \left (6 A a^3+5 C \sec (c+d x) a^3\right )dx-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (6 A a^3+5 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )dx-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 4402 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (a^4 (6 A+5 C) \int \sec (c+d x)dx+5 a^4 C \int \sec ^2(c+d x)dx+6 a^4 A x\right )-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (a^4 (6 A+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+5 a^4 C \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+6 a^4 A x\right )-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (a^4 (6 A+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {5 a^4 C \int 1d(-\tan (c+d x))}{d}+6 a^4 A x\right )-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (a^4 (6 A+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+6 a^4 A x+\frac {5 a^4 C \tan (c+d x)}{d}\right )-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {3}{2} \left (\frac {a^4 (6 A+5 C) \text {arctanh}(\sin (c+d x))}{d}+6 a^4 A x+\frac {5 a^4 C \tan (c+d x)}{d}\right )-\frac {(6 A-5 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}\right )-\frac {(3 A-C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{3 d}}{a}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^3}{d}\) |
(A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/d + (-1/3*((3*A - C)*(a^2 + a^2*Se c[c + d*x])^2*Tan[c + d*x])/d + (-1/2*((6*A - 5*C)*(a^4 + a^4*Sec[c + d*x] )*Tan[c + d*x])/d + (3*(6*a^4*A*x + (a^4*(6*A + 5*C)*ArcTanh[Sin[c + d*x]] )/d + (5*a^4*C*Tan[c + d*x])/d))/2)/3)/a
3.2.5.3.1 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 = 0.49 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {a^{3} A \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \left (d x +c \right )+3 a^{3} C \tan \left (d x +c \right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \tan \left (d x +c \right )-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(151\) |
| default | \(\frac {a^{3} A \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \left (d x +c \right )+3 a^{3} C \tan \left (d x +c \right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \tan \left (d x +c \right )-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(151\) |
| parallelrisch | \(\frac {a^{3} \left (-9 \left (A +\frac {5 C}{6}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+9 \left (A +\frac {5 C}{6}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 d x A \cos \left (3 d x +3 c \right )+\left (A +3 C \right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {11 C}{3}\right ) \sin \left (3 d x +3 c \right )+\frac {A \sin \left (4 d x +4 c \right )}{2}+9 d x A \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (A +5 C \right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(182\) |
| risch | \(3 a^{3} A x -\frac {i a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{3} \left (9 C \,{\mathrm e}^{5 i \left (d x +c \right )}-6 A \,{\mathrm e}^{4 i \left (d x +c \right )}-18 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-48 C \,{\mathrm e}^{2 i \left (d x +c \right )}-9 C \,{\mathrm e}^{i \left (d x +c \right )}-6 A -22 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\) | \(236\) |
| norman | \(\frac {\frac {a^{3} \left (4 A +11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+3 a^{3} A x -9 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+6 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+6 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-9 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+3 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-\frac {5 a^{3} C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}+\frac {6 a^{3} \left (2 A -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {4 a^{3} \left (3 A -10 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {4 a^{3} \left (9 A +10 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a^{3} \left (6 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (6 A +5 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(306\) |
1/d*(a^3*A*sin(d*x+c)+a^3*C*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*A*(d*x+c)+3*a^ 3*C*tan(d*x+c)+3*a^3*A*ln(sec(d*x+c)+tan(d*x+c))+3*a^3*C*(1/2*sec(d*x+c)*t an(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+a^3*A*tan(d*x+c)-a^3*C*(-2/3-1/3* sec(d*x+c)^2)*tan(d*x+c))
Time = 0.27 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.04 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {36 \, A a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (6 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (6 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, A a^{3} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 11 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 9 \, C a^{3} \cos \left (d x + c\right ) + 2 \, C a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
1/12*(36*A*a^3*d*x*cos(d*x + c)^3 + 3*(6*A + 5*C)*a^3*cos(d*x + c)^3*log(s in(d*x + c) + 1) - 3*(6*A + 5*C)*a^3*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(6*A*a^3*cos(d*x + c)^3 + 2*(3*A + 11*C)*a^3*cos(d*x + c)^2 + 9*C*a^3 *cos(d*x + c) + 2*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)
\[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{3} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 C \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 3 C \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(A*cos(c + d*x), x) + Integral(3*A*cos(c + d*x)*sec(c + d*x) , x) + Integral(3*A*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**2, x) + I ntegral(3*C*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(3*C*cos(c + d*x)*s ec(c + d*x)**4, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**5, x))
Time = 0.21 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.22 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {36 \, {\left (d x + c\right )} A a^{3} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 9 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, A a^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right ) + 36 \, C a^{3} \tan \left (d x + c\right )}{12 \, d} \]
1/12*(36*(d*x + c)*A*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 9*C *a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(si n(d*x + c) - 1)) + 18*A*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)) + 12*A*a^3*sin (d*x + c) + 12*A*a^3*tan(d*x + c) + 36*C*a^3*tan(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.51 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {18 \, {\left (d x + c\right )} A a^{3} + \frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 3 \, {\left (6 \, A a^{3} + 5 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (6 \, A a^{3} + 5 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
1/6*(18*(d*x + c)*A*a^3 + 12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2 *c)^2 + 1) + 3*(6*A*a^3 + 5*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3* (6*A*a^3 + 5*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*A*a^3*tan(1/ 2*d*x + 1/2*c)^5 + 15*C*a^3*tan(1/2*d*x + 1/2*c)^5 - 12*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^3*tan(1/2*d*x + 1/2*c ) + 33*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 15.61 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.37 \[ \int \cos (c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {6\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {6\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {11\,C\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {C\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \]
(A*a^3*sin(c + d*x))/d + (6*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/ 2)))/d + (6*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (5*C*a ^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^3*sin(c + d*x))/ (d*cos(c + d*x)) + (11*C*a^3*sin(c + d*x))/(3*d*cos(c + d*x)) + (3*C*a^3*s in(c + d*x))/(2*d*cos(c + d*x)^2) + (C*a^3*sin(c + d*x))/(3*d*cos(c + d*x) ^3)