Integrand size = 41, antiderivative size = 278 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {(3 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d} \] Output:
1/8*(3*B*a^3+12*B*a*b^2+4*b^3*(A+2*C)+3*a^2*b*(3*A+4*C))*arctanh(sin(d*x+c ))/d+1/15*(30*B*a^2*b+15*B*b^3+15*a*b^2*(2*A+3*C)+2*a^3*(4*A+5*C))*tan(d*x +c)/d+1/40*(6*A*b^3+15*B*a^3+50*B*a*b^2+15*a^2*b*(3*A+4*C))*sec(d*x+c)*tan (d*x+c)/d+1/30*a*(3*A*b^2+15*B*a*b+2*a^2*(4*A+5*C))*sec(d*x+c)^2*tan(d*x+c )/d+1/20*(3*A*b+5*B*a)*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A* (a+b*cos(d*x+c))^3*sec(d*x+c)^4*tan(d*x+c)/d
Time = 6.19 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.98 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {b^3 C \coth ^{-1}(\sin (c+d x))}{d}+\frac {3 a^2 (3 A b+a B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b \left (A b^2+3 a (b B+a C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {b^2 (b B+3 a C) \tan (c+d x)}{d}+\frac {3 a^2 (3 A b+a B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \left (A b^2+3 a (b B+a C)\right ) \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^2 (3 A b+a B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \left (3 A b^2+a (3 b B+a C)\right ) \left (3 \tan (c+d x)+\tan ^3(c+d x)\right )}{3 d}+\frac {a^3 A \left (15 \tan (c+d x)+10 \tan ^3(c+d x)+3 \tan ^5(c+d x)\right )}{15 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^6,x]
Output:
(b^3*C*ArcCoth[Sin[c + d*x]])/d + (3*a^2*(3*A*b + a*B)*ArcTanh[Sin[c + d*x ]])/(8*d) + (b*(A*b^2 + 3*a*(b*B + a*C))*ArcTanh[Sin[c + d*x]])/(2*d) + (b ^2*(b*B + 3*a*C)*Tan[c + d*x])/d + (3*a^2*(3*A*b + a*B)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*(A*b^2 + 3*a*(b*B + a*C))*Sec[c + d*x]*Tan[c + d*x])/( 2*d) + (a^2*(3*A*b + a*B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*(3*A*b^2 + a*(3*b*B + a*C))*(3*Tan[c + d*x] + Tan[c + d*x]^3))/(3*d) + (a^3*A*(15* Tan[c + d*x] + 10*Tan[c + d*x]^3 + 3*Tan[c + d*x]^5))/(15*d)
Time = 1.85 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 3526, 3042, 3526, 3042, 3510, 25, 3042, 3500, 3042, 3227, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{5} \int (a+b \cos (c+d x))^2 \left (b (A+5 C) \cos ^2(c+d x)+(4 a A+5 b B+5 a C) \cos (c+d x)+3 A b+5 a B\right ) \sec ^5(c+d x)dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b (A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(4 a A+5 b B+5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b+5 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int (a+b \cos (c+d x)) \left (b (7 A b+20 C b+5 a B) \cos ^2(c+d x)+\left (15 B a^2+b (29 A+40 C) a+20 b^2 B\right ) \cos (c+d x)+2 \left (2 (4 A+5 C) a^2+15 b B a+3 A b^2\right )\right ) \sec ^4(c+d x)dx+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (7 A b+20 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (15 B a^2+b (29 A+40 C) a+20 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left (2 (4 A+5 C) a^2+15 b B a+3 A b^2\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}-\frac {1}{3} \int -\left (\left (3 b^2 (7 A b+20 C b+5 a B) \cos ^2(c+d x)+4 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right ) \cos (c+d x)+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )\right ) \sec ^3(c+d x)\right )dx\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \left (3 b^2 (7 A b+20 C b+5 a B) \cos ^2(c+d x)+4 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right ) \cos (c+d x)+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )\right ) \sec ^3(c+d x)dx+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {3 b^2 (7 A b+20 C b+5 a B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (8 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right )+15 \left (3 B a^3+3 b (3 A+4 C) a^2+12 b^2 B a+4 b^3 (A+2 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {3 \tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {8 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right )+15 \left (3 B a^3+3 b (3 A+4 C) a^2+12 b^2 B a+4 b^3 (A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 \tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (8 \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right ) \int \sec ^2(c+d x)dx+15 \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \int \sec (c+d x)dx\right )+\frac {3 \tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 \tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 \tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {8 \tan (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{d}\right )+\frac {3 \tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )+\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \tan (c+d x) \sec ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}+\frac {1}{3} \left (\frac {1}{2} \left (\frac {15 \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {8 \tan (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{d}\right )+\frac {3 \tan (c+d x) \sec (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )\right )+\frac {(5 a B+3 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^3*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
Output:
(A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (((3*A*b + 5*a*B)*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*a*( 3*A*b^2 + 15*a*b*B + 2*a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*(6*A*b^3 + 15*a^3*B + 50*a*b^2*B + 15*a^2*b*(3*A + 4*C))*Sec[c + d* x]*Tan[c + d*x])/(2*d) + ((15*(3*a^3*B + 12*a*b^2*B + 4*b^3*(A + 2*C) + 3* a^2*b*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/d + (8*(30*a^2*b*B + 15*b^3*B + 15*a*b^2*(2*A + 3*C) + 2*a^3*(4*A + 5*C))*Tan[c + d*x])/d)/2)/3)/4)/5
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_.)*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_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && 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[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
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]
Time = 2.42 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.87
| method | result | size |
| parts | \(-\frac {A \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (B \,b^{3}+3 C a \,b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{3}+3 B a \,b^{2}+3 a^{2} b C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (3 a A \,b^{2}+3 B \,a^{2} b +a^{3} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(243\) |
| derivativedivides | \(\frac {-A \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 A \,a^{2} b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \tan \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right ) b^{3}+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(369\) |
| default | \(\frac {-A \,a^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{3} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 A \,a^{2} b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a^{2} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 a A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C \tan \left (d x +c \right ) a \,b^{2}+A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right ) b^{3}+C \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(369\) |
| parallelrisch | \(\frac {-675 \left (\frac {4 b^{3} \left (A +2 C \right )}{9}+\frac {4 B a \,b^{2}}{3}+\left (A +\frac {4 C}{3}\right ) a^{2} b +\frac {B \,a^{3}}{3}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+675 \left (\frac {4 b^{3} \left (A +2 C \right )}{9}+\frac {4 B a \,b^{2}}{3}+\left (A +\frac {4 C}{3}\right ) a^{2} b +\frac {B \,a^{3}}{3}\right ) \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+40 \left (9 B \,b^{3}+30 \left (A +\frac {9 C}{10}\right ) a \,b^{2}+30 B \,a^{2} b +8 \left (A +\frac {5 C}{4}\right ) a^{3}\right ) \sin \left (3 d x +3 c \right )+8 \left (15 B \,b^{3}+30 a \left (A +\frac {3 C}{2}\right ) b^{2}+30 B \,a^{2} b +8 \left (A +\frac {5 C}{4}\right ) a^{3}\right ) \sin \left (5 d x +5 c \right )+60 \left (4 A \,b^{3}+12 B a \,b^{2}+21 \left (A +\frac {4 C}{7}\right ) a^{2} b +7 B \,a^{3}\right ) \sin \left (2 d x +2 c \right )+30 \left (4 A \,b^{3}+12 B a \,b^{2}+9 \left (A +\frac {4 C}{3}\right ) a^{2} b +3 B \,a^{3}\right ) \sin \left (4 d x +4 c \right )+640 \sin \left (d x +c \right ) \left (\frac {3 B \,b^{3}}{8}+\frac {3 a \left (A +\frac {3 C}{4}\right ) b^{2}}{2}+\frac {3 B \,a^{2} b}{2}+a^{3} \left (A +\frac {C}{2}\right )\right )}{600 \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) d}\) | \(410\) |
| risch | \(\text {Expression too large to display}\) | \(937\) |
Input:
int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,meth od=_RETURNVERBOSE)
Output:
-A*a^3/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(3*A*a^2*b+ B*a^3)/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*b^3+3*C*a*b^2)/d*tan(d*x+c)+(A*b^3+3*B*a*b^2+3*C*a^2*b)/d *(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-(3*A*a*b^2+3*B* a^2*b+C*a^3)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C*b^3/d*ln(sec(d*x+c)+ta n(d*x+c))
Time = 0.10 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 30 \, B a^{2} b + 15 \, {\left (2 \, A + 3 \, C\right )} a b^{2} + 15 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} + 15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6, x, algorithm="fricas")
Output:
1/240*(15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*(A + 2*C)*b^3)*c os(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*(A + 2*C)*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8 *(2*(4*A + 5*C)*a^3 + 30*B*a^2*b + 15*(2*A + 3*C)*a*b^2 + 15*B*b^3)*cos(d* x + c)^4 + 24*A*a^3 + 15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*A *b^3)*cos(d*x + c)^3 + 8*((4*A + 5*C)*a^3 + 15*B*a^2*b + 15*A*a*b^2)*cos(d *x + c)^2 + 30*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)* *6,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.63 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(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 )} A a^{3} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} - 15 \, B a^{3} {\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 )} - 45 \, A a^{2} b {\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 )} - 180 \, C a^{2} b {\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 )} - 180 \, B a b^{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 )} - 60 \, A b^{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 )} + 120 \, C b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a b^{2} \tan \left (d x + c\right ) + 240 \, B b^{3} \tan \left (d x + c\right )}{240 \, d} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6, x, algorithm="maxima")
Output:
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 + 240*(tan(d*x + c)^3 + 3*tan( d*x + c))*B*a^2*b + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 - 15*B*a ^3*(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)) - 45*A*a^2*b* (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)) - 180*C*a^2*b*(2 *sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 180*B*a*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*A*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*C*b^3*(lo g(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 720*C*a*b^2*tan(d*x + c) + 240*B*b^3*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 989 vs. \(2 (266) = 532\).
Time = 0.20 (sec) , antiderivative size = 989, normalized size of antiderivative = 3.56 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6, x, algorithm="giac")
Output:
1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*C*a^2*b + 12*B*a*b^2 + 4*A*b^3 + 8*C*b ^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a^3 + 9*A*a^2*b + 12*C*a^ 2*b + 12*B*a*b^2 + 4*A*b^3 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 1 20*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360 *B*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 180*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360 *A*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 180*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 360 *C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*B* b^3*tan(1/2*d*x + 1/2*c)^9 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 30*B*a^3*t an(1/2*d*x + 1/2*c)^7 - 320*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 90*A*a^2*b*tan( 1/2*d*x + 1/2*c)^7 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 360*C*a^2*b*tan( 1/2*d*x + 1/2*c)^7 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 360*B*a*b^2*tan( 1/2*d*x + 1/2*c)^7 - 1440*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*b^3*tan(1 /2*d*x + 1/2*c)^7 - 480*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^3*tan(1/2*d *x + 1/2*c)^5 + 400*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a^2*b*tan(1/2*d* x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 2160*C*a*b^2*tan(1/2* d*x + 1/2*c)^5 + 720*B*b^3*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 320*C*a^3*tan(1/2*d*x + 1/2 *c)^3 - 90*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*B*a^2*b*tan(1/2*d*x + 1/2* c)^3 - 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*A*a*b^2*tan(1/2*d*x + 1...
Time = 1.99 (sec) , antiderivative size = 601, normalized size of antiderivative = 2.16 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A\,b^3}{2}+\frac {3\,B\,a^3}{8}+C\,b^3+\frac {9\,A\,a^2\,b}{8}+\frac {3\,B\,a\,b^2}{2}+\frac {3\,C\,a^2\,b}{2}\right )}{2\,A\,b^3+\frac {3\,B\,a^3}{2}+4\,C\,b^3+\frac {9\,A\,a^2\,b}{2}+6\,B\,a\,b^2+6\,C\,a^2\,b}\right )\,\left (A\,b^3+\frac {3\,B\,a^3}{4}+2\,C\,b^3+\frac {9\,A\,a^2\,b}{4}+3\,B\,a\,b^2+3\,C\,a^2\,b\right )}{d}-\frac {\left (2\,A\,a^3-A\,b^3-\frac {5\,B\,a^3}{4}+2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}-3\,B\,a\,b^2+6\,B\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,b^3-\frac {8\,A\,a^3}{3}+\frac {B\,a^3}{2}-8\,B\,b^3-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}+6\,B\,a\,b^2-16\,B\,a^2\,b-24\,C\,a\,b^2+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+12\,B\,b^3+\frac {20\,C\,a^3}{3}+20\,A\,a\,b^2+20\,B\,a^2\,b+36\,C\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^3}{3}-2\,A\,b^3-\frac {B\,a^3}{2}-8\,B\,b^3-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}-6\,B\,a\,b^2-16\,B\,a^2\,b-24\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+\frac {5\,B\,a^3}{4}+2\,B\,b^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+3\,B\,a\,b^2+6\,B\,a^2\,b+6\,C\,a\,b^2+3\,C\,a^2\,b\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 )} \] Input:
int(((a + b*cos(c + d*x))^3*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^6,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((A*b^3)/2 + (3*B*a^3)/8 + C*b^3 + (9*A*a^2*b )/8 + (3*B*a*b^2)/2 + (3*C*a^2*b)/2))/(2*A*b^3 + (3*B*a^3)/2 + 4*C*b^3 + ( 9*A*a^2*b)/2 + 6*B*a*b^2 + 6*C*a^2*b))*(A*b^3 + (3*B*a^3)/4 + 2*C*b^3 + (9 *A*a^2*b)/4 + 3*B*a*b^2 + 3*C*a^2*b))/d - (tan(c/2 + (d*x)/2)*(2*A*a^3 + A *b^3 + (5*B*a^3)/4 + 2*B*b^3 + 2*C*a^3 + 6*A*a*b^2 + (15*A*a^2*b)/4 + 3*B* a*b^2 + 6*B*a^2*b + 6*C*a*b^2 + 3*C*a^2*b) + tan(c/2 + (d*x)/2)^5*((116*A* a^3)/15 + 12*B*b^3 + (20*C*a^3)/3 + 20*A*a*b^2 + 20*B*a^2*b + 36*C*a*b^2) + tan(c/2 + (d*x)/2)^9*(2*A*a^3 - A*b^3 - (5*B*a^3)/4 + 2*B*b^3 + 2*C*a^3 + 6*A*a*b^2 - (15*A*a^2*b)/4 - 3*B*a*b^2 + 6*B*a^2*b + 6*C*a*b^2 - 3*C*a^2 *b) - tan(c/2 + (d*x)/2)^3*((8*A*a^3)/3 + 2*A*b^3 + (B*a^3)/2 + 8*B*b^3 + (16*C*a^3)/3 + 16*A*a*b^2 + (3*A*a^2*b)/2 + 6*B*a*b^2 + 16*B*a^2*b + 24*C* a*b^2 + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^7*((8*A*a^3)/3 - 2*A*b^3 - (B*a^3) /2 + 8*B*b^3 + (16*C*a^3)/3 + 16*A*a*b^2 - (3*A*a^2*b)/2 - 6*B*a*b^2 + 16* B*a^2*b + 24*C*a*b^2 - 6*C*a^2*b))/(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))
Time = 0.21 (sec) , antiderivative size = 1065, normalized size of antiderivative = 3.83 \[ \int (a+b \cos (c+d x))^3 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
Output:
( - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*b - 45* cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b*c - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**3 - 30*cos(c + d*x) *log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**3*c + 90*cos(c + d*x)*log(ta n((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b + 90*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b*c + 120*cos(c + d*x)*log(tan((c + d*x) /2) - 1)*sin(c + d*x)**2*a*b**3 + 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1 )*sin(c + d*x)**2*b**3*c - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3* b - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b*c - 60*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*a*b**3 - 30*cos(c + d*x)*log(tan((c + d*x)/2) - 1 )*b**3*c + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3* b + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b*c + 6 0*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**3 + 30*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**3*c - 90*cos(c + d*x) *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b - 90*cos(c + d*x)*log(ta n((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b*c - 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3 - 60*cos(c + d*x)*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**2*b**3*c + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1 )*a**3*b + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**3 + 30*cos(c + d*x)*log(tan((c + d...