Integrand size = 31, antiderivative size = 156 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\left (3 a^2 A+4 A b^2+8 a b B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (4 a A b+2 a^2 B+3 b^2 B\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^2 A+4 A b^2+8 a b B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (2 A b+a B) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a^2 A \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
1/8*(3*A*a^2+4*A*b^2+8*B*a*b)*arctanh(sin(d*x+c))/d+1/3*(4*A*a*b+2*B*a^2+3 *B*b^2)*tan(d*x+c)/d+1/8*(3*A*a^2+4*A*b^2+8*B*a*b)*sec(d*x+c)*tan(d*x+c)/d +1/3*a*(2*A*b+B*a)*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a^2*A*sec(d*x+c)^3*tan(d* x+c)/d
Time = 0.85 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.77 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \left (3 a^2 A+4 A b^2+8 a b B\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (24 \left (2 a A b+a^2 B+b^2 B\right )+3 \left (3 a^2 A+4 A b^2+8 a b B\right ) \sec (c+d x)+6 a^2 A \sec ^3(c+d x)+8 a (2 A b+a B) \tan ^2(c+d x)\right )}{24 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]
Output:
(3*(3*a^2*A + 4*A*b^2 + 8*a*b*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24* (2*a*A*b + a^2*B + b^2*B) + 3*(3*a^2*A + 4*A*b^2 + 8*a*b*B)*Sec[c + d*x] + 6*a^2*A*Sec[c + d*x]^3 + 8*a*(2*A*b + a*B)*Tan[c + d*x]^2))/(24*d)
Time = 0.94 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3467, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 3042, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3467 |
| \(\displaystyle \frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {1}{4} \int -\left (\left (4 b^2 B \cos ^2(c+d x)+\left (3 A a^2+8 b B a+4 A b^2\right ) \cos (c+d x)+4 a (2 A b+a B)\right ) \sec ^4(c+d x)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \int \left (4 b^2 B \cos ^2(c+d x)+\left (3 A a^2+8 b B a+4 A b^2\right ) \cos (c+d x)+4 a (2 A b+a B)\right ) \sec ^4(c+d x)dx+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {4 b^2 B \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 A a^2+8 b B a+4 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+4 a (2 A b+a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left (3 \left (3 A a^2+8 b B a+4 A b^2\right )+4 \left (2 B a^2+4 A b a+3 b^2 B\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {3 \left (3 A a^2+8 b B a+4 A b^2\right )+4 \left (2 B a^2+4 A b a+3 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \int \sec ^3(c+d x)dx+4 \left (2 a^2 B+4 a A b+3 b^2 B\right ) \int \sec ^2(c+d x)dx\right )+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (4 \left (2 a^2 B+4 a A b+3 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 \left (2 a^2 B+4 a A b+3 b^2 B\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {4 \left (2 a^2 B+4 a A b+3 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \left (2 a^2 B+4 a A b+3 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \left (2 a^2 B+4 a A b+3 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 A+8 a b B+4 A b^2\right ) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \left (2 a^2 B+4 a A b+3 b^2 B\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a B+2 A b) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 A \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]
Output:
(a^2*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*a*(2*A*b + a*B)*Sec[c + d* x]^2*Tan[c + d*x])/(3*d) + ((4*(4*a*A*b + 2*a^2*B + 3*b^2*B)*Tan[c + d*x]) /d + 3*(3*a^2*A + 4*A*b^2 + 8*a*b*B)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/3)/4
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_)])^2*((A_.) + (B_.)*sin[(e_.) + (f _.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ (B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 *(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2)) Int[(c + d* Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*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] && 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[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_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 8.53 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95
| method | result | size |
| parts | \(\frac {a^{2} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (A \,b^{2}+2 B a b \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (2 A a b +a^{2} B \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,b^{2} \tan \left (d x +c \right )}{d}\) | \(148\) |
| derivativedivides | \(\frac {a^{2} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-2 A a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right ) b^{2}}{d}\) | \(185\) |
| default | \(\frac {a^{2} A \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-2 A a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \tan \left (d x +c \right ) b^{2}}{d}\) | \(185\) |
| parallelrisch | \(\frac {-36 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2} A +\frac {4}{3} A \,b^{2}+\frac {8}{3} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+36 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a^{2} A +\frac {4}{3} A \,b^{2}+\frac {8}{3} B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (128 A a b +64 a^{2} B +48 B \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (18 a^{2} A +24 A \,b^{2}+48 B a b \right ) \sin \left (3 d x +3 c \right )+\left (32 A a b +16 a^{2} B +24 B \,b^{2}\right ) \sin \left (4 d x +4 c \right )+66 \sin \left (d x +c \right ) \left (a^{2} A +\frac {4}{11} A \,b^{2}+\frac {8}{11} B a b \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(246\) |
| norman | \(\frac {\frac {\left (7 a^{2} A -4 A \,b^{2}-8 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (27 a^{2} A -32 A a b +12 A \,b^{2}-16 a^{2} B +24 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {\left (27 a^{2} A +32 A a b +12 A \,b^{2}+16 a^{2} B +24 B a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}+\frac {\left (5 a^{2} A -16 A a b +4 A \,b^{2}-8 a^{2} B +8 B a b -8 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {\left (5 a^{2} A +16 A a b +4 A \,b^{2}+8 a^{2} B +8 B a b +8 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (81 a^{2} A -16 A a b -12 A \,b^{2}-8 a^{2} B -24 B a b -72 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {\left (81 a^{2} A +16 A a b -12 A \,b^{2}+8 a^{2} B -24 B a b +72 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (3 a^{2} A +4 A \,b^{2}+8 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a^{2} A +4 A \,b^{2}+8 B a b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(429\) |
| risch | \(-\frac {i \left (9 A \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+24 B a b \,{\mathrm e}^{7 i \left (d x +c \right )}-24 B \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+33 A \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+24 B a b \,{\mathrm e}^{5 i \left (d x +c \right )}-96 A a b \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-72 B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-33 A \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{3 i \left (d x +c \right )}-128 A a b \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-72 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{i \left (d x +c \right )}-32 A a b -16 a^{2} B -24 B \,b^{2}\right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}-\frac {3 a^{2} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}\) | \(447\) |
Input:
int((a+cos(d*x+c)*b)^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x,method=_RETURNVERBO SE)
Output:
a^2*A/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)))+(A*b^2+2*B*a*b)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c )+tan(d*x+c)))-(2*A*a*b+B*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*b^2/ d*tan(d*x+c)
Time = 0.11 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.15 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, B a^{2} + 4 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, A a^{2} + 3 \, {\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="f ricas")
Output:
1/48*(3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(2*B*a^2 + 4*A*a*b + 3*B*b^2)*cos(d*x + c)^3 + 6*A*a^2 + 3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*cos(d*x + c)^2 + 8*(B*a^2 + 2*A*a*b)*cos(d*x + c))*si n(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.46 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b - 3 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B b^{2} \tan \left (d x + c\right )}{48 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="m axima")
Output:
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 32*(tan(d*x + c)^3 + 3* tan(d*x + c))*A*a*b - 3*A*a^2*(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)) - 24*B*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d* x + c) + 1) + log(sin(d*x + c) - 1)) - 12*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)) + 48*B*b^2*tan (d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (146) = 292\).
Time = 0.18 (sec) , antiderivative size = 478, normalized size of antiderivative = 3.06 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="g iac")
Output:
1/24*(3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(1 5*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*A*a* b*tan(1/2*d*x + 1/2*c)^7 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^2*tan( 1/2*d*x + 1/2*c)^7 - 24*B*b^2*tan(1/2*d*x + 1/2*c)^7 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 40*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 80*A*a*b*tan(1/2*d*x + 1/2 *c)^5 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*B* a^2*tan(1/2*d*x + 1/2*c)^3 - 80*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*B*a*b*ta n(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*B*b^2*tan(1/2* d*x + 1/2*c)^3 + 15*A*a^2*tan(1/2*d*x + 1/2*c) + 24*B*a^2*tan(1/2*d*x + 1/ 2*c) + 48*A*a*b*tan(1/2*d*x + 1/2*c) + 24*B*a*b*tan(1/2*d*x + 1/2*c) + 12* A*b^2*tan(1/2*d*x + 1/2*c) + 24*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 45.42 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.01 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,a^2}{8}+B\,a\,b+\frac {A\,b^2}{2}\right )}{\frac {3\,A\,a^2}{2}+4\,B\,a\,b+2\,A\,b^2}\right )\,\left (\frac {3\,A\,a^2}{4}+2\,B\,a\,b+A\,b^2\right )}{d}+\frac {\left (\frac {5\,A\,a^2}{4}+A\,b^2-2\,B\,a^2-2\,B\,b^2-4\,A\,a\,b+2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a^2}{4}-A\,b^2+\frac {10\,B\,a^2}{3}+6\,B\,b^2+\frac {20\,A\,a\,b}{3}-2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,A\,a^2}{4}-A\,b^2-\frac {10\,B\,a^2}{3}-6\,B\,b^2-\frac {20\,A\,a\,b}{3}-2\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a^2}{4}+A\,b^2+2\,B\,a^2+2\,B\,b^2+4\,A\,a\,b+2\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^2)/cos(c + d*x)^5,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((3*A*a^2)/8 + (A*b^2)/2 + B*a*b))/((3*A*a^2) /2 + 2*A*b^2 + 4*B*a*b))*((3*A*a^2)/4 + A*b^2 + 2*B*a*b))/d + (tan(c/2 + ( d*x)/2)^7*((5*A*a^2)/4 + A*b^2 - 2*B*a^2 - 2*B*b^2 - 4*A*a*b + 2*B*a*b) - tan(c/2 + (d*x)/2)^3*(A*b^2 - (3*A*a^2)/4 + (10*B*a^2)/3 + 6*B*b^2 + (20*A *a*b)/3 + 2*B*a*b) + tan(c/2 + (d*x)/2)^5*((3*A*a^2)/4 - A*b^2 + (10*B*a^2 )/3 + 6*B*b^2 + (20*A*a*b)/3 - 2*B*a*b) + tan(c/2 + (d*x)/2)*((5*A*a^2)/4 + A*b^2 + 2*B*a^2 + 2*B*b^2 + 4*A*a*b + 2*B*a*b))/(d*(6*tan(c/2 + (d*x)/2) ^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^ 8 + 1))
Time = 0.24 (sec) , antiderivative size = 533, normalized size of antiderivative = 3.42 \[ \int (a+b \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x)
Output:
( - 3*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3 - 12*cos (c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**2 + 6*cos(c + d*x )*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3 + 24*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**2 - 3*cos(c + d*x)*log(tan((c + d* x)/2) - 1)*a**3 - 12*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**2 + 3*cos (c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**3 + 12*cos(c + d*x) *log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**2 - 6*cos(c + d*x)*log(tan ((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3 - 24*cos(c + d*x)*log(tan((c + d*x )/2) + 1)*sin(c + d*x)**2*a*b**2 + 3*cos(c + d*x)*log(tan((c + d*x)/2) + 1 )*a**3 + 12*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**2 - 3*cos(c + d*x) *sin(c + d*x)**3*a**3 - 12*cos(c + d*x)*sin(c + d*x)**3*a*b**2 + 5*cos(c + d*x)*sin(c + d*x)*a**3 + 12*cos(c + d*x)*sin(c + d*x)*a*b**2 + 16*sin(c + d*x)**5*a**2*b + 8*sin(c + d*x)**5*b**3 - 40*sin(c + d*x)**3*a**2*b - 16* sin(c + d*x)**3*b**3 + 24*sin(c + d*x)*a**2*b + 8*sin(c + d*x)*b**3)/(8*co s(c + d*x)*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))