Integrand size = 41, antiderivative size = 269 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) x+\frac {\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac {\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {(3 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d} \] 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))*x+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))*sin(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))*cos(d*x+c)*sin(d*x+c)/d+1/30*a*(3* A*b^2+15*B*a*b+2*a^2*(4*A+5*C))*cos(d*x+c)^2*sin(d*x+c)/d+1/20*(3*A*b+5*B* a)*cos(d*x+c)^3*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+b*se c(d*x+c))^3*sin(d*x+c)/d
Time = 2.64 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {540 a^2 A b c+240 A b^3 c+180 a^3 B c+720 a b^2 B c+720 a^2 b c C+480 b^3 c C+540 a^2 A b d x+240 A b^3 d x+180 a^3 B d x+720 a b^2 B d x+720 a^2 b C d x+480 b^3 C d x+60 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sin (c+d x)+120 \left (A b^3+a^3 B+3 a b^2 B+3 a^2 b (A+C)\right ) \sin (2 (c+d x))+50 a^3 A \sin (3 (c+d x))+120 a A b^2 \sin (3 (c+d x))+120 a^2 b B \sin (3 (c+d x))+40 a^3 C \sin (3 (c+d x))+45 a^2 A b \sin (4 (c+d x))+15 a^3 B \sin (4 (c+d x))+6 a^3 A \sin (5 (c+d x))}{480 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
(540*a^2*A*b*c + 240*A*b^3*c + 180*a^3*B*c + 720*a*b^2*B*c + 720*a^2*b*c*C + 480*b^3*c*C + 540*a^2*A*b*d*x + 240*A*b^3*d*x + 180*a^3*B*d*x + 720*a*b ^2*B*d*x + 720*a^2*b*C*d*x + 480*b^3*C*d*x + 60*(18*a^2*b*B + 8*b^3*B + 6* a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*Sin[c + d*x] + 120*(A*b^3 + a^3*B + 3 *a*b^2*B + 3*a^2*b*(A + C))*Sin[2*(c + d*x)] + 50*a^3*A*Sin[3*(c + d*x)] + 120*a*A*b^2*Sin[3*(c + d*x)] + 120*a^2*b*B*Sin[3*(c + d*x)] + 40*a^3*C*Si n[3*(c + d*x)] + 45*a^2*A*b*Sin[4*(c + d*x)] + 15*a^3*B*Sin[4*(c + d*x)] + 6*a^3*A*Sin[5*(c + d*x)])/(480*d)
Time = 1.62 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4582, 3042, 4582, 3042, 4562, 25, 3042, 4535, 3042, 3117, 4533, 24}
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.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (b (A+5 C) \sec ^2(c+d x)+(4 a A+5 b B+5 a C) \sec (c+d x)+3 A b+5 a B\right )dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (b (A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(4 a A+5 b B+5 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+3 A b+5 a B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (b (7 A b+20 C b+5 a B) \sec ^2(c+d x)+\left (15 B a^2+b (29 A+40 C) a+20 b^2 B\right ) \sec (c+d x)+2 \left (2 (4 A+5 C) a^2+15 b B a+3 A b^2\right )\right )dx+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (7 A b+20 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (15 B a^2+b (29 A+40 C) a+20 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 \left (2 (4 A+5 C) a^2+15 b B a+3 A b^2\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) \left (3 b^2 (7 A b+20 C b+5 a B) \sec ^2(c+d x)+4 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right ) \sec (c+d x)+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )\right )dx\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \cos ^2(c+d x) \left (3 b^2 (7 A b+20 C b+5 a B) \sec ^2(c+d x)+4 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right ) \sec (c+d x)+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )\right )dx+\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {3 b^2 (7 A b+20 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 \left (2 (4 A+5 C) a^3+30 b B a^2+15 b^2 (2 A+3 C) a+15 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (4 \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right ) \int \cos (c+d x)dx+\int \cos ^2(c+d x) \left (3 b^2 (7 A b+20 C b+5 a B) \sec ^2(c+d x)+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )\right )dx\right )+\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (4 \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+\int \frac {3 b^2 (7 A b+20 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\int \frac {3 b^2 (7 A b+20 C b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 \left (15 B a^3+15 b (3 A+4 C) a^2+50 b^2 B a+6 A b^3\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {4 \sin (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{d}\right )+\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {15}{2} \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right ) \int 1dx+\frac {4 \sin (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{d}+\frac {3 \sin (c+d x) \cos (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}\right )+\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {2 a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{3 d}+\frac {1}{3} \left (\frac {4 \sin (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{d}+\frac {3 \sin (c+d x) \cos (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{2 d}+\frac {15}{2} x \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right )\right )\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) + (((3*A*b + 5*a*B)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(4*d) + ((2*a*( 3*A*b^2 + 15*a*b*B + 2*a^2*(4*A + 5*C))*Cos[c + d*x]^2*Sin[c + d*x])/(3*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))*x)/ 2 + (4*(30*a^2*b*B + 15*b^3*B + 15*a*b^2*(2*A + 3*C) + 2*a^3*(4*A + 5*C))* Sin[c + d*x])/d + (3*(6*A*b^3 + 15*a^3*B + 50*a*b^2*B + 15*a^2*b*(3*A + 4* C))*Cos[c + d*x]*Sin[c + d*x])/(2*d))/3)/4)/5
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Si mp[1/(d*n) Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B* b) + A*a*(n + 1))*Csc[e + f*x] + b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[ {a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Time = 1.68 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.75
| method | result | size |
| parallelrisch | \(\frac {120 \left (B \,a^{3}+3 b \left (A +C \right ) a^{2}+3 B a \,b^{2}+A \,b^{3}\right ) \sin \left (2 d x +2 c \right )+10 \left (\left (5 A +4 C \right ) a^{3}+12 B \,a^{2} b +12 a A \,b^{2}\right ) \sin \left (3 d x +3 c \right )+15 \left (3 A \,a^{2} b +B \,a^{3}\right ) \sin \left (4 d x +4 c \right )+6 a^{3} A \sin \left (5 d x +5 c \right )+60 \left (a^{3} \left (5 A +6 C \right )+18 B \,a^{2} b +18 a \left (A +\frac {4 C}{3}\right ) b^{2}+8 B \,b^{3}\right ) \sin \left (d x +c \right )+540 x \left (\frac {B \,a^{3}}{3}+b \left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 B a \,b^{2}}{3}+\frac {4 b^{3} \left (A +2 C \right )}{9}\right ) d}{480 d}\) | \(203\) |
| derivativedivides | \(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{3} \sin \left (d x +c \right )+3 C a \,b^{2} \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\) | \(301\) |
| default | \(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{3} \sin \left (d x +c \right )+3 C a \,b^{2} \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\) | \(301\) |
| risch | \(\frac {3 C \,a^{2} b x}{2}+\frac {3 a^{3} B x}{8}+x C \,b^{3}+\frac {A \,b^{3} x}{2}+\frac {3 x B a \,b^{2}}{2}+\frac {9 a^{2} A b x}{8}+\frac {5 a^{3} A \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) a A \,b^{2}}{4 d}+\frac {9 \sin \left (d x +c \right ) B \,a^{2} b}{4 d}+\frac {\sin \left (d x +c \right ) B \,b^{3}}{d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{4 d}+\frac {3 \sin \left (d x +c \right ) C a \,b^{2}}{d}+\frac {a^{3} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 \sin \left (4 d x +4 c \right ) A \,a^{2} b}{32 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{3}}{32 d}+\frac {5 a^{3} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a A \,b^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2} b}{4 d}+\frac {\sin \left (3 d x +3 c \right ) a^{3} C}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) A \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B a \,b^{2}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b C}{4 d}\) | \(360\) |
Input:
int(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth od=_RETURNVERBOSE)
Output:
1/480*(120*(B*a^3+3*b*(A+C)*a^2+3*B*a*b^2+A*b^3)*sin(2*d*x+2*c)+10*((5*A+4 *C)*a^3+12*B*a^2*b+12*a*A*b^2)*sin(3*d*x+3*c)+15*(3*A*a^2*b+B*a^3)*sin(4*d *x+4*c)+6*a^3*A*sin(5*d*x+5*c)+60*(a^3*(5*A+6*C)+18*B*a^2*b+18*a*(A+4/3*C) *b^2+8*B*b^3)*sin(d*x+c)+540*x*(1/3*B*a^3+b*(A+4/3*C)*a^2+4/3*B*a*b^2+4/9* b^3*(A+2*C))*d)/d
Time = 0.09 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.77 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 16 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 240 \, B a^{2} b + 120 \, {\left (2 \, A + 3 \, C\right )} a b^{2} + 120 \, B b^{3} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
Output:
1/120*(15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*(A + 2*C)*b^3)*d *x + (24*A*a^3*cos(d*x + c)^4 + 16*(4*A + 5*C)*a^3 + 240*B*a^2*b + 120*(2* A + 3*C)*a*b^2 + 120*B*b^3 + 30*(B*a^3 + 3*A*a^2*b)*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 + 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))*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)** 2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \, {\left (d x + c\right )} C b^{3} + 1440 \, C a b^{2} \sin \left (d x + c\right ) + 480 \, B b^{3} \sin \left (d x + c\right )}{480 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
Output:
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^3 - 160*(s in(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4* c) + 8*sin(2*d*x + 2*c))*A*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*B *a^2*b + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2*b - 480*(sin(d*x + c)^ 3 - 3*sin(d*x + c))*A*a*b^2 + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^3 + 480*(d*x + c)*C*b^3 + 1440 *C*a*b^2*sin(d*x + c) + 480*B*b^3*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 926 vs. \(2 (257) = 514\).
Time = 0.35 (sec) , antiderivative size = 926, normalized size of antiderivative = 3.44 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), 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)*(d*x + c) + 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 + 120*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*tan(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 - 3 60*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 - 3 60*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*ta n(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 + 1/2*c)^3 + 360*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 1440*C*a*b^2*ta n(1/2*d*x + 1/2*c)^3 + 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*B*b^3*tan...
Time = 13.26 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.33 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,b^3\,x}{2}+\frac {3\,B\,a^3\,x}{8}+C\,b^3\,x+\frac {9\,A\,a^2\,b\,x}{8}+\frac {3\,B\,a\,b^2\,x}{2}+\frac {3\,C\,a^2\,b\,x}{2}+\frac {5\,A\,a^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {B\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {5\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {A\,a^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,A\,a^2\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,C\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {9\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{d} \] Input:
int(cos(c + d*x)^5*(a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
(A*b^3*x)/2 + (3*B*a^3*x)/8 + C*b^3*x + (9*A*a^2*b*x)/8 + (3*B*a*b^2*x)/2 + (3*C*a^2*b*x)/2 + (5*A*a^3*sin(c + d*x))/(8*d) + (B*b^3*sin(c + d*x))/d + (3*C*a^3*sin(c + d*x))/(4*d) + (5*A*a^3*sin(3*c + 3*d*x))/(48*d) + (A*a^ 3*sin(5*c + 5*d*x))/(80*d) + (A*b^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(2 *c + 2*d*x))/(4*d) + (B*a^3*sin(4*c + 4*d*x))/(32*d) + (C*a^3*sin(3*c + 3* d*x))/(12*d) + (3*A*a^2*b*sin(2*c + 2*d*x))/(4*d) + (A*a*b^2*sin(3*c + 3*d *x))/(4*d) + (3*A*a^2*b*sin(4*c + 4*d*x))/(32*d) + (3*B*a*b^2*sin(2*c + 2* d*x))/(4*d) + (B*a^2*b*sin(3*c + 3*d*x))/(4*d) + (3*C*a^2*b*sin(2*c + 2*d* x))/(4*d) + (9*A*a*b^2*sin(c + d*x))/(4*d) + (9*B*a^2*b*sin(c + d*x))/(4*d ) + (3*C*a*b^2*sin(c + d*x))/d
Time = 0.16 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.86 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{3} b +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{3} b +45 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{2} b c +60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{3}+6 \sin \left (d x +c \right )^{5} a^{4}-20 \sin \left (d x +c \right )^{3} a^{4}-10 \sin \left (d x +c \right )^{3} a^{3} c -60 \sin \left (d x +c \right )^{3} a^{2} b^{2}+30 \sin \left (d x +c \right ) a^{4}+30 \sin \left (d x +c \right ) a^{3} c +180 \sin \left (d x +c \right ) a^{2} b^{2}+90 \sin \left (d x +c \right ) a \,b^{2} c +30 \sin \left (d x +c \right ) b^{4}+45 a^{3} b d x +45 a^{2} b c d x +60 a \,b^{3} d x +30 b^{3} c d x}{30 d} \] Input:
int(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
( - 30*cos(c + d*x)*sin(c + d*x)**3*a**3*b + 75*cos(c + d*x)*sin(c + d*x)* a**3*b + 45*cos(c + d*x)*sin(c + d*x)*a**2*b*c + 60*cos(c + d*x)*sin(c + d *x)*a*b**3 + 6*sin(c + d*x)**5*a**4 - 20*sin(c + d*x)**3*a**4 - 10*sin(c + d*x)**3*a**3*c - 60*sin(c + d*x)**3*a**2*b**2 + 30*sin(c + d*x)*a**4 + 30 *sin(c + d*x)*a**3*c + 180*sin(c + d*x)*a**2*b**2 + 90*sin(c + d*x)*a*b**2 *c + 30*sin(c + d*x)*b**4 + 45*a**3*b*d*x + 45*a**2*b*c*d*x + 60*a*b**3*d* x + 30*b**3*c*d*x)/(30*d)