Integrand size = 41, antiderivative size = 160 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^2 (7 A+8 B+12 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (4 A+5 B+6 C) \tan (c+d x)}{3 d}+\frac {a^2 (11 A+16 B+12 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(A+2 B) \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/8*a^2*(7*A+8*B+12*C)*arctanh(sin(d*x+c))/d+1/3*a^2*(4*A+5*B+6*C)*tan(d*x +c)/d+1/24*a^2*(11*A+16*B+12*C)*sec(d*x+c)*tan(d*x+c)/d+1/6*(A+2*B)*(a^2+a ^2*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a+a*cos(d*x+c))^2*sec(d*x+ c)^3*tan(d*x+c)/d
Time = 1.88 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.55 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^2 \left (3 (7 A+8 B+12 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (48 (A+B+C)+3 (7 A+8 B+4 C) \sec (c+d x)+6 A \sec ^3(c+d x)+8 (2 A+B) \tan ^2(c+d x)\right )\right )}{24 d} \]
(a^2*(3*(7*A + 8*B + 12*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(48*(A + B + C) + 3*(7*A + 8*B + 4*C)*Sec[c + d*x] + 6*A*Sec[c + d*x]^3 + 8*(2*A + B )*Tan[c + d*x]^2)))/(24*d)
Time = 1.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.09, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {3042, 3522, 3042, 3454, 3042, 3447, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (2 a (A+2 B)+a (A+4 C) \cos (c+d x)) \sec ^4(c+d x)dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a (A+2 B)+a (A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\frac {1}{3} \int (\cos (c+d x) a+a) \left ((11 A+16 B+12 C) a^2+(5 A+4 B+12 C) \cos (c+d x) a^2\right ) \sec ^3(c+d x)dx+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((11 A+16 B+12 C) a^2+(5 A+4 B+12 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{3} \int \left ((5 A+4 B+12 C) \cos ^2(c+d x) a^3+(11 A+16 B+12 C) a^3+\left ((5 A+4 B+12 C) a^3+(11 A+16 B+12 C) a^3\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {(5 A+4 B+12 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+(11 A+16 B+12 C) a^3+\left ((5 A+4 B+12 C) a^3+(11 A+16 B+12 C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \left (8 (4 A+5 B+6 C) a^3+3 (7 A+8 B+12 C) \cos (c+d x) a^3\right ) \sec ^2(c+d x)dx+\frac {a^3 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \int \frac {8 (4 A+5 B+6 C) a^3+3 (7 A+8 B+12 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (8 a^3 (4 A+5 B+6 C) \int \sec ^2(c+d x)dx+3 a^3 (7 A+8 B+12 C) \int \sec (c+d x)dx\right )+\frac {a^3 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 a^3 (7 A+8 B+12 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+8 a^3 (4 A+5 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^3 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 a^3 (7 A+8 B+12 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {8 a^3 (4 A+5 B+6 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a^3 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (3 a^3 (7 A+8 B+12 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {8 a^3 (4 A+5 B+6 C) \tan (c+d x)}{d}\right )+\frac {a^3 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{3} \left (\frac {1}{2} \left (\frac {3 a^3 (7 A+8 B+12 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {8 a^3 (4 A+5 B+6 C) \tan (c+d x)}{d}\right )+\frac {a^3 (11 A+16 B+12 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 (A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{3 d}}{4 a}+\frac {A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d}\) |
(A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*(A + 2* B)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^3*(11 *A + 16*B + 12*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*a^3*(7*A + 8*B + 12*C)*ArcTanh[Sin[c + d*x]])/d + (8*a^3*(4*A + 5*B + 6*C)*Tan[c + d*x])/d) /2)/3)/(4*a)
3.4.16.3.1 Defintions of rubi rules used
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_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(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)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
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_)])^(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/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 9.24 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.16
| method | result | size |
| parts | \(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (B \,a^{2}+2 a^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}+a^{2} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{2}}{d}\) | \(185\) |
| parallelrisch | \(\frac {15 a^{2} \left (-\frac {14 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {8 B}{7}+\frac {12 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{15}+\frac {14 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {8 B}{7}+\frac {12 C}{7}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{15}+\frac {8 \left (\frac {7 B}{3}+2 C +\frac {8 A}{3}\right ) \sin \left (2 d x +2 c \right )}{15}+\frac {\left (7 A +8 B +4 C \right ) \sin \left (3 d x +3 c \right )}{15}+\frac {4 \left (\frac {B}{3}+\frac {4 A}{15}+\frac {2 C}{5}\right ) \sin \left (4 d x +4 c \right )}{3}+\sin \left (d x +c \right ) \left (\frac {4 C}{15}+\frac {8 B}{15}+A \right )\right )}{4 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(200\) |
| derivativedivides | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-2 A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} C \tan \left (d x +c \right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(254\) |
| default | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-2 A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} C \tan \left (d x +c \right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(254\) |
| norman | \(\frac {-\frac {a^{2} \left (7 A +8 B +12 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {a^{2} \left (7 A +8 B +12 C \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{2} \left (25 A +24 B +20 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{2} \left (33 A +24 B +52 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{2} \left (65 A -8 B -44 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{2} \left (89 A -200 B -204 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{2} \left (121 A -40 B +84 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{2} \left (217 A +152 B +84 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a^{2} \left (7 A +8 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{2} \left (7 A +8 B +12 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(324\) |
| risch | \(-\frac {i a^{2} \left (21 A \,{\mathrm e}^{7 i \left (d x +c \right )}+24 B \,{\mathrm e}^{7 i \left (d x +c \right )}+12 C \,{\mathrm e}^{7 i \left (d x +c \right )}-24 B \,{\mathrm e}^{6 i \left (d x +c \right )}-48 C \,{\mathrm e}^{6 i \left (d x +c \right )}+45 A \,{\mathrm e}^{5 i \left (d x +c \right )}+24 B \,{\mathrm e}^{5 i \left (d x +c \right )}+12 C \,{\mathrm e}^{5 i \left (d x +c \right )}-96 A \,{\mathrm e}^{4 i \left (d x +c \right )}-120 B \,{\mathrm e}^{4 i \left (d x +c \right )}-144 C \,{\mathrm e}^{4 i \left (d x +c \right )}-45 A \,{\mathrm e}^{3 i \left (d x +c \right )}-24 B \,{\mathrm e}^{3 i \left (d x +c \right )}-12 C \,{\mathrm e}^{3 i \left (d x +c \right )}-128 A \,{\mathrm e}^{2 i \left (d x +c \right )}-136 B \,{\mathrm e}^{2 i \left (d x +c \right )}-144 C \,{\mathrm e}^{2 i \left (d x +c \right )}-21 A \,{\mathrm e}^{i \left (d x +c \right )}-24 B \,{\mathrm e}^{i \left (d x +c \right )}-12 C \,{\mathrm e}^{i \left (d x +c \right )}-32 A -40 B -48 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {7 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {7 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(405\) |
A*a^2/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)))-(2*A*a^2+B*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(B*a^2+2 *C*a^2)/d*tan(d*x+c)+(A*a^2+2*B*a^2+C*a^2)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/ 2*ln(sec(d*x+c)+tan(d*x+c)))+1/d*C*ln(sec(d*x+c)+tan(d*x+c))*a^2
Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (7 \, A + 8 \, B + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (7 \, A + 8 \, B + 12 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (4 \, A + 5 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 8 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5, x, algorithm="fricas")
1/48*(3*(7*A + 8*B + 12*C)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(7 *A + 8*B + 12*C)*a^2*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(4*A + 5 *B + 6*C)*a^2*cos(d*x + c)^3 + 3*(7*A + 8*B + 4*C)*a^2*cos(d*x + c)^2 + 8* (2*A + B)*a^2*cos(d*x + c) + 6*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (150) = 300\).
Time = 0.21 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.98 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 3 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{2} \tan \left (d x + c\right ) + 96 \, C a^{2} \tan \left (d x + c\right )}{48 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5, x, algorithm="maxima")
1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 16*(tan(d*x + c)^3 + 3* tan(d*x + c))*B*a^2 - 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)) - 12*A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d* x + c) + 1) + log(sin(d*x + c) - 1)) - 24*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*C*a^2*(2* sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 24*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*B *a^2*tan(d*x + c) + 96*C*a^2*tan(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.81 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (7 \, A a^{2} + 8 \, B a^{2} + 12 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (7 \, A a^{2} + 8 \, B a^{2} + 12 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 77 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 88 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 132 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 136 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 156 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5, x, algorithm="giac")
1/24*(3*(7*A*a^2 + 8*B*a^2 + 12*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(7*A*a^2 + 8*B*a^2 + 12*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2* (21*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 + 36*C* a^2*tan(1/2*d*x + 1/2*c)^7 - 77*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 88*B*a^2*ta n(1/2*d*x + 1/2*c)^5 - 132*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2 *d*x + 1/2*c)^3 + 136*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 156*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 75*A*a^2*tan(1/2*d*x + 1/2*c) - 72*B*a^2*tan(1/2*d*x + 1/2*c ) - 60*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 5.01 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.53 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\frac {4\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {7\,A}{8}+B+\frac {3\,C}{2}\right )}{\frac {7\,A\,a^2}{2}+4\,B\,a^2+6\,C\,a^2}\right )\,\left (\frac {7\,A}{8}+B+\frac {3\,C}{2}\right )}{d}-\frac {\left (\frac {7\,A\,a^2}{4}+2\,B\,a^2+3\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (-\frac {77\,A\,a^2}{12}-\frac {22\,B\,a^2}{3}-11\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {83\,A\,a^2}{12}+\frac {34\,B\,a^2}{3}+13\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {25\,A\,a^2}{4}-6\,B\,a^2-5\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(2*a^2*atanh((4*a^2*tan(c/2 + (d*x)/2)*((7*A)/8 + B + (3*C)/2))/((7*A*a^2) /2 + 4*B*a^2 + 6*C*a^2))*((7*A)/8 + B + (3*C)/2))/d - (tan(c/2 + (d*x)/2)^ 7*((7*A*a^2)/4 + 2*B*a^2 + 3*C*a^2) - tan(c/2 + (d*x)/2)^5*((77*A*a^2)/12 + (22*B*a^2)/3 + 11*C*a^2) + tan(c/2 + (d*x)/2)^3*((83*A*a^2)/12 + (34*B*a ^2)/3 + 13*C*a^2) - tan(c/2 + (d*x)/2)*((25*A*a^2)/4 + 6*B*a^2 + 5*C*a^2)) /(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))