Integrand size = 31, antiderivative size = 267 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {\left (12 a^3 A b+16 a A b^3+3 a^4 B+24 a^2 b^2 B+8 b^4 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (8 a^4 A+60 a^2 A b^2+15 A b^4+40 a^3 b B+60 a b^3 B\right ) \tan (c+d x)}{15 d}+\frac {a \left (60 a^2 A b+56 A b^3+15 a^3 B+110 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a^2 \left (8 a^2 A+18 A b^2+25 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {a (8 A b+5 a B) (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d} \] Output:
1/8*(12*A*a^3*b+16*A*a*b^3+3*B*a^4+24*B*a^2*b^2+8*B*b^4)*arctanh(sin(d*x+c ))/d+1/15*(8*A*a^4+60*A*a^2*b^2+15*A*b^4+40*B*a^3*b+60*B*a*b^3)*tan(d*x+c) /d+1/40*a*(60*A*a^2*b+56*A*b^3+15*B*a^3+110*B*a*b^2)*sec(d*x+c)*tan(d*x+c) /d+1/30*a^2*(8*A*a^2+18*A*b^2+25*B*a*b)*sec(d*x+c)^2*tan(d*x+c)/d+1/20*a*( 8*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*(a+b*cos (d*x+c))^3*sec(d*x+c)^4*tan(d*x+c)/d
Time = 6.18 (sec) , antiderivative size = 254, normalized size of antiderivative = 0.95 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {b^4 B \coth ^{-1}(\sin (c+d x))}{d}+\frac {3 a^3 (4 A b+a B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a b^2 (2 A b+3 a B) \text {arctanh}(\sin (c+d x))}{d}+\frac {b^3 (A b+4 a B) \tan (c+d x)}{d}+\frac {3 a^3 (4 A b+a B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a b^2 (2 A b+3 a B) \sec (c+d x) \tan (c+d x)}{d}+\frac {a^3 (4 A b+a B) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {2 a^2 b (3 A b+2 a B) \left (3 \tan (c+d x)+\tan ^3(c+d x)\right )}{3 d}+\frac {a^4 A \left (15 \tan (c+d x)+10 \tan ^3(c+d x)+3 \tan ^5(c+d x)\right )}{15 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]
Output:
(b^4*B*ArcCoth[Sin[c + d*x]])/d + (3*a^3*(4*A*b + a*B)*ArcTanh[Sin[c + d*x ]])/(8*d) + (a*b^2*(2*A*b + 3*a*B)*ArcTanh[Sin[c + d*x]])/d + (b^3*(A*b + 4*a*B)*Tan[c + d*x])/d + (3*a^3*(4*A*b + a*B)*Sec[c + d*x]*Tan[c + d*x])/( 8*d) + (a*b^2*(2*A*b + 3*a*B)*Sec[c + d*x]*Tan[c + d*x])/d + (a^3*(4*A*b + a*B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (2*a^2*b*(3*A*b + 2*a*B)*(3*Tan [c + d*x] + Tan[c + d*x]^3))/(3*d) + (a^4*A*(15*Tan[c + d*x] + 10*Tan[c + d*x]^3 + 3*Tan[c + d*x]^5))/(15*d)
Time = 1.75 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.484, Rules used = {3042, 3468, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3468 |
| \(\displaystyle \frac {1}{5} \int (a+b \cos (c+d x))^2 \left (b (a A+5 b B) \cos ^2(c+d x)+\left (4 A a^2+10 b B a+5 A b^2\right ) \cos (c+d x)+a (8 A b+5 a B)\right ) \sec ^5(c+d x)dx+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b (a A+5 b B) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (4 A a^2+10 b B a+5 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+a (8 A b+5 a B)\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int (a+b \cos (c+d x)) \left (b \left (5 B a^2+12 A b a+20 b^2 B\right ) \cos ^2(c+d x)+\left (15 B a^3+44 A b a^2+60 b^2 B a+20 A b^3\right ) \cos (c+d x)+2 a \left (8 A a^2+25 b B a+18 A b^2\right )\right ) \sec ^4(c+d x)dx+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b \left (5 B a^2+12 A b a+20 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (15 B a^3+44 A b a^2+60 b^2 B a+20 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+2 a \left (8 A a^2+25 b B a+18 A b^2\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 b^2 \left (5 B a^2+12 A b a+20 b^2 B\right ) \cos ^2(c+d x)+4 \left (8 A a^4+40 b B a^3+60 A b^2 a^2+60 b^3 B a+15 A b^4\right ) \cos (c+d x)+3 a \left (15 B a^3+60 A b a^2+110 b^2 B a+56 A b^3\right )\right ) \sec ^3(c+d x)\right )dx\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \left (3 b^2 \left (5 B a^2+12 A b a+20 b^2 B\right ) \cos ^2(c+d x)+4 \left (8 A a^4+40 b B a^3+60 A b^2 a^2+60 b^3 B a+15 A b^4\right ) \cos (c+d x)+3 a \left (15 B a^3+60 A b a^2+110 b^2 B a+56 A b^3\right )\right ) \sec ^3(c+d x)dx+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {3 b^2 \left (5 B a^2+12 A b a+20 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2+4 \left (8 A a^4+40 b B a^3+60 A b^2 a^2+60 b^3 B a+15 A b^4\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+3 a \left (15 B a^3+60 A b a^2+110 b^2 B a+56 A b^3\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (8 \left (8 A a^4+40 b B a^3+60 A b^2 a^2+60 b^3 B a+15 A b^4\right )+15 \left (3 B a^4+12 A b a^3+24 b^2 B a^2+16 A b^3 a+8 b^4 B\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {3 a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {8 \left (8 A a^4+40 b B a^3+60 A b^2 a^2+60 b^3 B a+15 A b^4\right )+15 \left (3 B a^4+12 A b a^3+24 b^2 B a^2+16 A b^3 a+8 b^4 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (8 \left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \int \sec ^2(c+d x)dx+15 \left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \int \sec (c+d x)dx\right )+\frac {3 a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 \left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 \left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 \left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {8 \left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \tan (c+d x)}{d}\right )+\frac {3 a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a^2 \left (8 a^2 A+25 a b B+18 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {1}{3} \left (\frac {3 a \left (15 a^3 B+60 a^2 A b+110 a b^2 B+56 A b^3\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (\frac {15 \left (3 a^4 B+12 a^3 A b+24 a^2 b^2 B+16 a A b^3+8 b^4 B\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {8 \left (8 a^4 A+40 a^3 b B+60 a^2 A b^2+60 a b^3 B+15 A b^4\right ) \tan (c+d x)}{d}\right )\right )\right )+\frac {a (5 a B+8 A b) \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\right )+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^4*(A + B*Cos[c + d*x])*Sec[c + d*x]^6,x]
Output:
(a*A*(a + b*Cos[c + d*x])^3*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a*(8*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^2*(8*a^2*A + 18*A*b^2 + 25*a*b*B)*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*a*(60*a^2*A*b + 56*A*b^3 + 15*a^3*B + 110*a*b^2*B)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((15*(12*a^3*A*b + 16*a*A*b^3 + 3*a^4*B + 24*a^2*b^2*B + 8 *b^4*B)*ArcTanh[Sin[c + d*x]])/d + (8*(8*a^4*A + 60*a^2*A*b^2 + 15*A*b^4 + 40*a^3*b*B + 60*a*b^3*B)*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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(-(b*c - a*d))*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (A*b + a *B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2 , 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && LtQ[n, -1]
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 = 21.60 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.89
| method | result | size |
| parts | \(-\frac {a^{4} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{4}+4 B a \,b^{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 A a \,b^{3}+6 B \,a^{2} b^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (6 A \,a^{2} b^{2}+4 B \,a^{3} b \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 A \,a^{3} b +B \,a^{4}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(237\) |
| derivativedivides | \(\frac {-a^{4} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 A \,a^{3} b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 B \,a^{3} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-6 A \,a^{2} b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 B \,a^{2} b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 A a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \tan \left (d x +c \right ) a \,b^{3}+A \tan \left (d x +c \right ) b^{4}+B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(313\) |
| default | \(\frac {-a^{4} A \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{4} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 A \,a^{3} b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-4 B \,a^{3} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-6 A \,a^{2} b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 B \,a^{2} b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 A a \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 B \tan \left (d x +c \right ) a \,b^{3}+A \tan \left (d x +c \right ) b^{4}+B \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(313\) |
| parallelrisch | \(\frac {-180 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A \,a^{3} b +\frac {4}{3} A a \,b^{3}+\frac {1}{4} B \,a^{4}+2 B \,a^{2} b^{2}+\frac {2}{3} B \,b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+180 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A \,a^{3} b +\frac {4}{3} A a \,b^{3}+\frac {1}{4} B \,a^{4}+2 B \,a^{2} b^{2}+\frac {2}{3} B \,b^{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (320 a^{4} A +2400 A \,a^{2} b^{2}+360 A \,b^{4}+1600 B \,a^{3} b +1440 B a \,b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (64 a^{4} A +480 A \,a^{2} b^{2}+120 A \,b^{4}+320 B \,a^{3} b +480 B a \,b^{3}\right ) \sin \left (5 d x +5 c \right )+\left (1680 A \,a^{3} b +960 A a \,b^{3}+420 B \,a^{4}+1440 B \,a^{2} b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (360 A \,a^{3} b +480 A a \,b^{3}+90 B \,a^{4}+720 B \,a^{2} b^{2}\right ) \sin \left (4 d x +4 c \right )+640 \sin \left (d x +c \right ) \left (a^{4} A +3 A \,a^{2} b^{2}+\frac {3}{8} A \,b^{4}+2 B \,a^{3} b +\frac {3}{2} B a \,b^{3}\right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(409\) |
| risch | \(\frac {i \left (480 B a \,b^{3}+480 A \,a^{2} b^{2}+320 B \,a^{3} b +45 B \,a^{4} {\mathrm e}^{i \left (d x +c \right )}+640 A \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}+480 A \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}+320 A \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}+480 A \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}+210 B \,a^{4} {\mathrm e}^{3 i \left (d x +c \right )}+720 A \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-210 B \,a^{4} {\mathrm e}^{7 i \left (d x +c \right )}+120 A \,b^{4} {\mathrm e}^{8 i \left (d x +c \right )}-45 B \,a^{4} {\mathrm e}^{9 i \left (d x +c \right )}+120 A \,b^{4}-240 A a \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-360 B \,a^{2} b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+480 B a \,b^{3} {\mathrm e}^{8 i \left (d x +c \right )}-840 A \,a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}-720 B \,a^{2} b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+1440 A \,a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+960 B \,a^{3} b \,{\mathrm e}^{6 i \left (d x +c \right )}+1920 B a \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}+3360 A \,a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+2240 B \,a^{3} b \,{\mathrm e}^{4 i \left (d x +c \right )}+2880 B a \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+840 A \,a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-480 A a \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-180 A \,a^{3} b \,{\mathrm e}^{9 i \left (d x +c \right )}+64 a^{4} A +480 A a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+720 B \,a^{2} b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+2400 A \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+1600 B \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}+1920 B a \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+180 A \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}+240 A a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}+360 B \,a^{2} b^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{2 d}-\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{3}}{d}-\frac {3 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{4}}{d}+\frac {3 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{2 d}+\frac {2 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{3}}{d}+\frac {3 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{4}}{d}\) | \(802\) |
Input:
int((a+cos(d*x+c)*b)^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x,method=_RETURNVERBO SE)
Output:
-a^4*A/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(A*b^4+4*B* a*b^3)/d*tan(d*x+c)+(4*A*a*b^3+6*B*a^2*b^2)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1 /2*ln(sec(d*x+c)+tan(d*x+c)))-(6*A*a^2*b^2+4*B*a^3*b)/d*(-2/3-1/3*sec(d*x+ c)^2)*tan(d*x+c)+(4*A*a^3*b+B*a^4)/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^4/d*ln(sec(d*x+c)+tan(d*x+c) )
Time = 0.10 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3} + 8 \, B b^{4}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, A a^{4} + 8 \, {\left (8 \, A a^{4} + 40 \, B a^{3} b + 60 \, A a^{2} b^{2} + 60 \, B a b^{3} + 15 \, A b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, B a^{4} + 12 \, A a^{3} b + 24 \, B a^{2} b^{2} + 16 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (2 \, A a^{4} + 10 \, B a^{3} b + 15 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="f ricas")
Output:
1/240*(15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*cos (d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^ 2 + 16*A*a*b^3 + 8*B*b^4)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(24*A* a^4 + 8*(8*A*a^4 + 40*B*a^3*b + 60*A*a^2*b^2 + 60*B*a*b^3 + 15*A*b^4)*cos( d*x + c)^4 + 15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3)*cos(d*x + c)^3 + 16*(2*A*a^4 + 10*B*a^3*b + 15*A*a^2*b^2)*cos(d*x + c)^2 + 30*(B* a^4 + 4*A*a^3*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**4*(A+B*cos(d*x+c))*sec(d*x+c)**6,x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.45 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{4} + 320 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} b + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} b^{2} - 15 \, B a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{3} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 360 \, B a^{2} b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 240 \, A a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, B b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 960 \, B a b^{3} \tan \left (d x + c\right ) + 240 \, A b^{4} \tan \left (d x + c\right )}{240 \, d} \] Input:
integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="m axima")
Output:
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 320*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3*b + 480*(tan(d*x + c)^3 + 3*t an(d*x + c))*A*a^2*b^2 - 15*B*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(s in(d*x + c) - 1)) - 60*A*a^3*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)) - 360*B*a^2*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)) - 240*A*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 *B*b^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 960*B*a*b^3*tan(d *x + c) + 240*A*b^4*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 850 vs. \(2 (255) = 510\).
Time = 0.20 (sec) , antiderivative size = 850, normalized size of antiderivative = 3.18 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x, algorithm="g iac")
Output:
1/120*(15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*log (abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a^4 + 12*A*a^3*b + 24*B*a^2*b^2 + 16*A*a*b^3 + 8*B*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^4* tan(1/2*d*x + 1/2*c)^9 - 75*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 300*A*a^3*b*tan (1/2*d*x + 1/2*c)^9 + 480*B*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 720*A*a^2*b^2*t an(1/2*d*x + 1/2*c)^9 - 360*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 240*A*a*b^3 *tan(1/2*d*x + 1/2*c)^9 + 480*B*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^4*t an(1/2*d*x + 1/2*c)^9 - 160*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 30*B*a^4*tan(1/ 2*d*x + 1/2*c)^7 + 120*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 1280*B*a^3*b*tan(1 /2*d*x + 1/2*c)^7 - 1920*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*B*a^2*b^2* tan(1/2*d*x + 1/2*c)^7 + 480*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 1920*B*a*b^3 *tan(1/2*d*x + 1/2*c)^7 - 480*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^4*tan (1/2*d*x + 1/2*c)^5 + 1600*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 2400*A*a^2*b^2 *tan(1/2*d*x + 1/2*c)^5 + 2880*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 720*A*b^4* tan(1/2*d*x + 1/2*c)^5 - 160*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^4*tan(1 /2*d*x + 1/2*c)^3 - 120*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1280*B*a^3*b*tan( 1/2*d*x + 1/2*c)^3 - 1920*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 720*B*a^2*b^2 *tan(1/2*d*x + 1/2*c)^3 - 480*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 1920*B*a*b^ 3*tan(1/2*d*x + 1/2*c)^3 - 480*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^4*ta n(1/2*d*x + 1/2*c) + 75*B*a^4*tan(1/2*d*x + 1/2*c) + 300*A*a^3*b*tan(1/...
Time = 45.21 (sec) , antiderivative size = 555, normalized size of antiderivative = 2.08 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^4}{8}+\frac {3\,A\,a^3\,b}{2}+3\,B\,a^2\,b^2+2\,A\,a\,b^3+B\,b^4\right )}{\frac {3\,B\,a^4}{2}+6\,A\,a^3\,b+12\,B\,a^2\,b^2+8\,A\,a\,b^3+4\,B\,b^4}\right )\,\left (\frac {3\,B\,a^4}{4}+3\,A\,a^3\,b+6\,B\,a^2\,b^2+4\,A\,a\,b^3+2\,B\,b^4\right )}{d}-\frac {\left (2\,A\,a^4+2\,A\,b^4-\frac {5\,B\,a^4}{4}+12\,A\,a^2\,b^2-6\,B\,a^2\,b^2-4\,A\,a\,b^3-5\,A\,a^3\,b+8\,B\,a\,b^3+8\,B\,a^3\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,a^4}{2}-8\,A\,b^4-\frac {8\,A\,a^4}{3}-32\,A\,a^2\,b^2+12\,B\,a^2\,b^2+8\,A\,a\,b^3+2\,A\,a^3\,b-32\,B\,a\,b^3-\frac {64\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^4}{15}+\frac {80\,B\,a^3\,b}{3}+40\,A\,a^2\,b^2+48\,B\,a\,b^3+12\,A\,b^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^4}{3}-8\,A\,b^4-\frac {B\,a^4}{2}-32\,A\,a^2\,b^2-12\,B\,a^2\,b^2-8\,A\,a\,b^3-2\,A\,a^3\,b-32\,B\,a\,b^3-\frac {64\,B\,a^3\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^4+2\,A\,b^4+\frac {5\,B\,a^4}{4}+12\,A\,a^2\,b^2+6\,B\,a^2\,b^2+4\,A\,a\,b^3+5\,A\,a^3\,b+8\,B\,a\,b^3+8\,B\,a^3\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^4)/cos(c + d*x)^6,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((3*B*a^4)/8 + B*b^4 + 3*B*a^2*b^2 + 2*A*a*b^ 3 + (3*A*a^3*b)/2))/((3*B*a^4)/2 + 4*B*b^4 + 12*B*a^2*b^2 + 8*A*a*b^3 + 6* A*a^3*b))*((3*B*a^4)/4 + 2*B*b^4 + 6*B*a^2*b^2 + 4*A*a*b^3 + 3*A*a^3*b))/d - (tan(c/2 + (d*x)/2)*(2*A*a^4 + 2*A*b^4 + (5*B*a^4)/4 + 12*A*a^2*b^2 + 6 *B*a^2*b^2 + 4*A*a*b^3 + 5*A*a^3*b + 8*B*a*b^3 + 8*B*a^3*b) + tan(c/2 + (d *x)/2)^5*((116*A*a^4)/15 + 12*A*b^4 + 40*A*a^2*b^2 + 48*B*a*b^3 + (80*B*a^ 3*b)/3) + tan(c/2 + (d*x)/2)^9*(2*A*a^4 + 2*A*b^4 - (5*B*a^4)/4 + 12*A*a^2 *b^2 - 6*B*a^2*b^2 - 4*A*a*b^3 - 5*A*a^3*b + 8*B*a*b^3 + 8*B*a^3*b) - tan( c/2 + (d*x)/2)^3*((8*A*a^4)/3 + 8*A*b^4 + (B*a^4)/2 + 32*A*a^2*b^2 + 12*B* a^2*b^2 + 8*A*a*b^3 + 2*A*a^3*b + 32*B*a*b^3 + (64*B*a^3*b)/3) - tan(c/2 + (d*x)/2)^7*((8*A*a^4)/3 + 8*A*b^4 - (B*a^4)/2 + 32*A*a^2*b^2 - 12*B*a^2*b ^2 - 8*A*a*b^3 - 2*A*a^3*b + 32*B*a*b^3 + (64*B*a^3*b)/3))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/ 2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))
Time = 0.18 (sec) , antiderivative size = 773, normalized size of antiderivative = 2.90 \[ \int (a+b \cos (c+d x))^4 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^4*(A+B*cos(d*x+c))*sec(d*x+c)^6,x)
Output:
( - 225*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**4*b - 60 0*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**3 - 120*c os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**5 + 450*cos(c + d *x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b + 1200*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**3 + 240*cos(c + d*x)*log( tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b**5 - 225*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**4*b - 600*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b** 3 - 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**5 + 225*cos(c + d*x)*log (tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**4*b + 600*cos(c + d*x)*log(tan(( c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b**3 + 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**5 - 450*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4*b - 1200*cos(c + d*x)*log(tan((c + d*x)/2) + 1) *sin(c + d*x)**2*a**2*b**3 - 240*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**2*b**5 + 225*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**4*b + 6 00*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**3 + 120*cos(c + d*x)*log (tan((c + d*x)/2) + 1)*b**5 - 225*cos(c + d*x)*sin(c + d*x)**3*a**4*b - 60 0*cos(c + d*x)*sin(c + d*x)**3*a**2*b**3 + 375*cos(c + d*x)*sin(c + d*x)*a **4*b + 600*cos(c + d*x)*sin(c + d*x)*a**2*b**3 + 64*sin(c + d*x)**5*a**5 + 800*sin(c + d*x)**5*a**3*b**2 + 600*sin(c + d*x)**5*a*b**4 - 160*sin(c + d*x)**3*a**5 - 2000*sin(c + d*x)**3*a**3*b**2 - 1200*sin(c + d*x)**3*a...