Integrand size = 33, antiderivative size = 169 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^3 (15 A+28 C) x+\frac {a^3 C \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d} \]
1/8*a^3*(15*A+28*C)*x+a^3*C*arctanh(sin(d*x+c))/d+5/8*a^3*(3*A+4*C)*sin(d* x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/4*A*cos(d*x+c) ^2*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d+1/8*(5*A+4*C)*cos(d*x+c)*(a^3+a^3 *sec(d*x+c))*sin(d*x+c)/d
Time = 1.78 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^3 \left (60 A d x+112 C d x-32 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 (13 A+12 C) \sin (c+d x)+8 (4 A+C) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+A \sin (4 (c+d x))\right )}{32 d} \]
(a^3*(60*A*d*x + 112*C*d*x - 32*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*(13*A + 12*C)*Sin[c + d*x] + 8*(4*A + C)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c + d*x)] + A*Sin[4*(c + d*x)]))/(32*d)
Time = 0.93 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4575, 3042, 4505, 27, 3042, 4505, 3042, 4484, 2009}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^3 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int \cos ^3(c+d x) (\sec (c+d x) a+a)^3 (3 a A+4 a C \sec (c+d x))dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (3 a A+4 a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{3} \int 3 \cos ^2(c+d x) (\sec (c+d x) a+a)^2 \left ((5 A+4 C) a^2+4 C \sec (c+d x) a^2\right )dx+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \cos ^2(c+d x) (\sec (c+d x) a+a)^2 \left ((5 A+4 C) a^2+4 C \sec (c+d x) a^2\right )dx+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left ((5 A+4 C) a^2+4 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{2} \int \cos (c+d x) (\sec (c+d x) a+a) \left (5 (3 A+4 C) a^3+8 C \sec (c+d x) a^3\right )dx+\frac {(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 (3 A+4 C) a^3+8 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {5 a^4 (3 A+4 C) \sin (c+d x)}{d}-\int \left (-\left ((15 A+28 C) a^4\right )-8 C \sec (c+d x) a^4\right )dx\right )+\frac {(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {5 a^4 (3 A+4 C) \sin (c+d x)}{d}+a^4 x (15 A+28 C)+\frac {8 a^4 C \text {arctanh}(\sin (c+d x))}{d}\right )+\frac {(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{2 d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d}\) |
(A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + ((A*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/d + ((5*A + 4*C)*Cos[c + d*x]*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(2*d) + (a^4*(15*A + 28*C)*x + (8*a^4*C*ArcTanh[Sin[c + d*x]])/d + (5*a^4*(3*A + 4*C)*Sin[c + d*x])/d)/2 )/(4*a)
3.2.8.3.1 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[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n) Int[(d*Csc[e + f*x])^( n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
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 = 0.38 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.61
| method | result | size |
| parallelrisch | \(\frac {a^{3} \left (-32 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+32 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (32 A +8 C \right ) \sin \left (2 d x +2 c \right )+8 A \sin \left (3 d x +3 c \right )+A \sin \left (4 d x +4 c \right )+\left (104 A +96 C \right ) \sin \left (d x +c \right )+60 x d \left (A +\frac {28 C}{15}\right )\right )}{32 d}\) | \(103\) |
| derivativedivides | \(\frac {a^{3} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 a^{3} C \sin \left (d x +c \right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+a^{3} A \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(173\) |
| default | \(\frac {a^{3} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+3 a^{3} C \sin \left (d x +c \right )+3 a^{3} A \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+a^{3} A \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(173\) |
| risch | \(\frac {15 a^{3} A x}{8}+\frac {7 a^{3} x C}{2}-\frac {13 i a^{3} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{3} C}{2 d}+\frac {13 i a^{3} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} C}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {a^{3} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{3} A \sin \left (3 d x +3 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} A}{d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} C}{4 d}\) | \(206\) |
| norman | \(\frac {\left (\frac {15}{8} a^{3} A +\frac {7}{2} a^{3} C \right ) x +\left (-\frac {15}{2} a^{3} A -14 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {15}{2} a^{3} A -14 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {15}{8} a^{3} A +\frac {7}{2} a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (\frac {45}{4} a^{3} A +21 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {19 a^{3} \left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}+\frac {5 a^{3} \left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {a^{3} \left (5 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {7 a^{3} \left (7 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (27 A +52 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}+\frac {a^{3} \left (37 A +92 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {3 a^{3} \left (41 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 d}+\frac {a^{3} \left (-68 C +57 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{4 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(400\) |
1/32*a^3*(-32*C*ln(tan(1/2*d*x+1/2*c)-1)+32*C*ln(tan(1/2*d*x+1/2*c)+1)+(32 *A+8*C)*sin(2*d*x+2*c)+8*A*sin(3*d*x+3*c)+A*sin(4*d*x+4*c)+(104*A+96*C)*si n(d*x+c)+60*x*d*(A+28/15*C))/d
Time = 0.28 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (15 \, A + 28 \, C\right )} a^{3} d x + 4 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, A a^{3} \cos \left (d x + c\right )^{2} + {\left (15 \, A + 4 \, C\right )} a^{3} \cos \left (d x + c\right ) + 24 \, {\left (A + C\right )} a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/8*((15*A + 28*C)*a^3*d*x + 4*C*a^3*log(sin(d*x + c) + 1) - 4*C*a^3*log(- sin(d*x + c) + 1) + (2*A*a^3*cos(d*x + c)^3 + 8*A*a^3*cos(d*x + c)^2 + (15 *A + 4*C)*a^3*cos(d*x + c) + 24*(A + C)*a^3)*sin(d*x + c))/d
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.01 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 96 \, {\left (d x + c\right )} C a^{3} - 16 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 32 \, A a^{3} \sin \left (d x + c\right ) - 96 \, C a^{3} \sin \left (d x + c\right )}{32 \, d} \]
-1/32*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - (12*d*x + 12*c + sin(4 *d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2* c))*A*a^3 - 8*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 96*(d*x + c)*C*a^3 - 16*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 32*A*a^3*sin( d*x + c) - 96*C*a^3*sin(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.26 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {8 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (15 \, A a^{3} + 28 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 20 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 55 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 68 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 73 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 76 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 49 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 28 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{8 \, d} \]
1/8*(8*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*C*a^3*log(abs(tan(1/2* d*x + 1/2*c) - 1)) + (15*A*a^3 + 28*C*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2 *d*x + 1/2*c)^7 + 20*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 55*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 68*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 73*A*a^3*tan(1/2*d*x + 1/2*c )^3 + 76*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 49*A*a^3*tan(1/2*d*x + 1/2*c) + 28 *C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 15.62 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.15 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {13\,A\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {15\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {7\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]
(13*A*a^3*sin(c + d*x))/(4*d) + (3*C*a^3*sin(c + d*x))/d + (15*A*a^3*atan( sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(4*d) + (7*C*a^3*atan(sin(c/2 + (d *x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (A*a^3*sin(2*c + 2*d*x))/d + (A*a^3*sin(3*c + 3*d*x))/(4* d) + (A*a^3*sin(4*c + 4*d*x))/(32*d) + (C*a^3*sin(2*c + 2*d*x))/(4*d)