Integrand size = 40, antiderivative size = 156 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\left (3 a^2 B+4 b^2 B+8 a b C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (4 a b B+2 a^2 C+3 b^2 C\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a (2 b B+a C) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {a^2 B \sec ^3(c+d x) \tan (c+d x)}{4 d} \] Output:
1/8*(3*B*a^2+4*B*b^2+8*C*a*b)*arctanh(sin(d*x+c))/d+1/3*(4*B*a*b+2*C*a^2+3 *C*b^2)*tan(d*x+c)/d+1/8*(3*B*a^2+4*B*b^2+8*C*a*b)*sec(d*x+c)*tan(d*x+c)/d +1/3*a*(2*B*b+C*a)*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a^2*B*sec(d*x+c)^3*tan(d* x+c)/d
Time = 0.76 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.77 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 \left (3 a^2 B+4 b^2 B+8 a b C\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (24 \left (2 a b B+a^2 C+b^2 C\right )+3 \left (3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x)+6 a^2 B \sec ^3(c+d x)+8 a (2 b B+a C) \tan ^2(c+d x)\right )}{24 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
Output:
(3*(3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24* (2*a*b*B + a^2*C + b^2*C) + 3*(3*a^2*B + 4*b^2*B + 8*a*b*C)*Sec[c + d*x] + 6*a^2*B*Sec[c + d*x]^3 + 8*a*(2*b*B + a*C)*Tan[c + d*x]^2))/(24*d)
Time = 1.08 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3508, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^2 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3467 |
| \(\displaystyle \frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {1}{4} \int -\left (\left (4 b^2 C \cos ^2(c+d x)+\left (3 B a^2+8 b C a+4 b^2 B\right ) \cos (c+d x)+4 a (2 b B+a C)\right ) \sec ^4(c+d x)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \int \left (4 b^2 C \cos ^2(c+d x)+\left (3 B a^2+8 b C a+4 b^2 B\right ) \cos (c+d x)+4 a (2 b B+a C)\right ) \sec ^4(c+d x)dx+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {4 b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B a^2+8 b C a+4 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+4 a (2 b B+a C)}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left (3 \left (3 B a^2+8 b C a+4 b^2 B\right )+4 \left (2 C a^2+4 b B a+3 b^2 C\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {3 \left (3 B a^2+8 b C a+4 b^2 B\right )+4 \left (2 C a^2+4 b B a+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \int \sec ^3(c+d x)dx+4 \left (2 a^2 C+4 a b B+3 b^2 C\right ) \int \sec ^2(c+d x)dx\right )+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (4 \left (2 a^2 C+4 a b B+3 b^2 C\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 \left (2 a^2 C+4 a b B+3 b^2 C\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {4 \left (2 a^2 C+4 a b B+3 b^2 C\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \left (2 a^2 C+4 a b B+3 b^2 C\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \left (2 a^2 C+4 a b B+3 b^2 C\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (3 \left (3 a^2 B+8 a b C+4 b^2 B\right ) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \left (2 a^2 C+4 a b B+3 b^2 C\right ) \tan (c+d x)}{d}\right )+\frac {4 a (a C+2 b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a^2 B \tan (c+d x) \sec ^3(c+d x)}{4 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x ]^6,x]
Output:
(a^2*B*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*a*(2*b*B + a*C)*Sec[c + d* x]^2*Tan[c + d*x])/(3*d) + ((4*(4*a*b*B + 2*a^2*C + 3*b^2*C)*Tan[c + d*x]) /d + 3*(3*a^2*B + 4*b^2*B + 8*a*b*C)*(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[((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[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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 = 0.95 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95
| method | result | size |
| parts | \(\frac {\left (B \,b^{2}+2 a b C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (2 B a b +a^{2} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {b^{2} C \tan \left (d x +c \right )}{d}\) | \(148\) |
| derivativedivides | \(\frac {-a^{2} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right ) b^{2}+B \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(185\) |
| default | \(\frac {-a^{2} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 a b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right ) b^{2}+B \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(185\) |
| parallelrisch | \(\frac {-36 \left (B \,a^{2}+\frac {4}{3} B \,b^{2}+\frac {8}{3} a b C \right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+36 \left (B \,a^{2}+\frac {4}{3} B \,b^{2}+\frac {8}{3} a b C \right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (128 B a b +64 a^{2} C +48 b^{2} C \right ) \sin \left (2 d x +2 c \right )+\left (18 B \,a^{2}+24 B \,b^{2}+48 a b C \right ) \sin \left (3 d x +3 c \right )+\left (32 B a b +16 a^{2} C +24 b^{2} C \right ) \sin \left (4 d x +4 c \right )+66 \sin \left (d x +c \right ) \left (B \,a^{2}+\frac {4}{11} B \,b^{2}+\frac {8}{11} a b C \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(246\) |
| risch | \(-\frac {i \left (9 B \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+24 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-24 C \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+33 B \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 B \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+24 C a b \,{\mathrm e}^{5 i \left (d x +c \right )}-96 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-48 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-72 b^{2} C \,{\mathrm e}^{4 i \left (d x +c \right )}-33 B \,{\mathrm e}^{3 i \left (d x +c \right )} a^{2}-12 B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-24 a b C \,{\mathrm e}^{3 i \left (d x +c \right )}-128 B \,{\mathrm e}^{2 i \left (d x +c \right )} a b -64 C \,{\mathrm e}^{2 i \left (d x +c \right )} a^{2}-72 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 B \,{\mathrm e}^{i \left (d x +c \right )} a^{2}-12 B \,{\mathrm e}^{i \left (d x +c \right )} b^{2}-24 C \,{\mathrm e}^{i \left (d x +c \right )} a b -32 B a b -16 a^{2} C -24 b^{2} C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 B \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b C}{d}+\frac {3 B \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b C}{d}\) | \(447\) |
Input:
int((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,method =_RETURNVERBOSE)
Output:
(B*b^2+2*C*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*B*a*b+C*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*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)))+b^2*C/ d*tan(d*x+c)
Time = 0.11 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.15 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {3 \, {\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, C a^{2} + 4 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, B a^{2} + 3 \, {\left (3 \, B a^{2} + 8 \, C a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")
Output:
1/48*(3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(2*C*a^2 + 4*B*a*b + 3*C*b^2)*cos(d*x + c)^3 + 6*B*a^2 + 3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*cos(d*x + c)^2 + 8*(C*a^2 + 2*B*a*b)*cos(d*x + c))*si n(d*x + c))/(d*cos(d*x + c)^4)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**6 ,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.46 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b - 3 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, C a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, C b^{2} \tan \left (d x + c\right )}{48 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")
Output:
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 + 32*(tan(d*x + c)^3 + 3* tan(d*x + c))*B*a*b - 3*B*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*C*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*B*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*C*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 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")
Output:
1/24*(3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*B*a^2 + 8*C*a*b + 4*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(1 5*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*B*a* b*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*B*b^2*tan( 1/2*d*x + 1/2*c)^7 - 24*C*b^2*tan(1/2*d*x + 1/2*c)^7 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 40*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 80*B*a*b*tan(1/2*d*x + 1/2 *c)^5 - 24*C*a*b*tan(1/2*d*x + 1/2*c)^5 - 12*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*C* a^2*tan(1/2*d*x + 1/2*c)^3 - 80*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 24*C*a*b*ta n(1/2*d*x + 1/2*c)^3 - 12*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*C*b^2*tan(1/2* d*x + 1/2*c)^3 + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 24*C*a^2*tan(1/2*d*x + 1/ 2*c) + 48*B*a*b*tan(1/2*d*x + 1/2*c) + 24*C*a*b*tan(1/2*d*x + 1/2*c) + 12* B*b^2*tan(1/2*d*x + 1/2*c) + 24*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 2.71 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.01 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^2}{8}+C\,a\,b+\frac {B\,b^2}{2}\right )}{\frac {3\,B\,a^2}{2}+4\,C\,a\,b+2\,B\,b^2}\right )\,\left (\frac {3\,B\,a^2}{4}+2\,C\,a\,b+B\,b^2\right )}{d}+\frac {\left (\frac {5\,B\,a^2}{4}+B\,b^2-2\,C\,a^2-2\,C\,b^2-4\,B\,a\,b+2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,B\,a^2}{4}-B\,b^2+\frac {10\,C\,a^2}{3}+6\,C\,b^2+\frac {20\,B\,a\,b}{3}-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,B\,a^2}{4}-B\,b^2-\frac {10\,C\,a^2}{3}-6\,C\,b^2-\frac {20\,B\,a\,b}{3}-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,B\,a^2}{4}+B\,b^2+2\,C\,a^2+2\,C\,b^2+4\,B\,a\,b+2\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \] Input:
int(((B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d *x)^6,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((3*B*a^2)/8 + (B*b^2)/2 + C*a*b))/((3*B*a^2) /2 + 2*B*b^2 + 4*C*a*b))*((3*B*a^2)/4 + B*b^2 + 2*C*a*b))/d + (tan(c/2 + ( d*x)/2)^7*((5*B*a^2)/4 + B*b^2 - 2*C*a^2 - 2*C*b^2 - 4*B*a*b + 2*C*a*b) - tan(c/2 + (d*x)/2)^3*(B*b^2 - (3*B*a^2)/4 + (10*C*a^2)/3 + 6*C*b^2 + (20*B *a*b)/3 + 2*C*a*b) + tan(c/2 + (d*x)/2)^5*((3*B*a^2)/4 - B*b^2 + (10*C*a^2 )/3 + 6*C*b^2 + (20*B*a*b)/3 - 2*C*a*b) + tan(c/2 + (d*x)/2)*((5*B*a^2)/4 + B*b^2 + 2*C*a^2 + 2*C*b^2 + 4*B*a*b + 2*C*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 = 0.16 (sec) , antiderivative size = 624, normalized size of antiderivative = 4.00 \[ \int (a+b \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
Output:
( - 16*cos(c + d*x)*sin(c + d*x)**3*a**2*c - 32*cos(c + d*x)*sin(c + d*x)* *3*a*b**2 - 24*cos(c + d*x)*sin(c + d*x)**3*b**2*c + 24*cos(c + d*x)*sin(c + d*x)*a**2*c + 48*cos(c + d*x)*sin(c + d*x)*a*b**2 + 24*cos(c + d*x)*sin (c + d*x)*b**2*c - 9*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b - 24 *log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b*c - 12*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**3 + 18*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2* a**2*b + 48*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b*c + 24*log(tan(( c + d*x)/2) - 1)*sin(c + d*x)**2*b**3 - 9*log(tan((c + d*x)/2) - 1)*a**2*b - 24*log(tan((c + d*x)/2) - 1)*a*b*c - 12*log(tan((c + d*x)/2) - 1)*b**3 + 9*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b + 24*log(tan((c + d*x )/2) + 1)*sin(c + d*x)**4*a*b*c + 12*log(tan((c + d*x)/2) + 1)*sin(c + d*x )**4*b**3 - 18*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b - 48*log(t an((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b*c - 24*log(tan((c + d*x)/2) + 1)* sin(c + d*x)**2*b**3 + 9*log(tan((c + d*x)/2) + 1)*a**2*b + 24*log(tan((c + d*x)/2) + 1)*a*b*c + 12*log(tan((c + d*x)/2) + 1)*b**3 - 9*sin(c + d*x)* *3*a**2*b - 24*sin(c + d*x)**3*a*b*c - 12*sin(c + d*x)**3*b**3 + 15*sin(c + d*x)*a**2*b + 24*sin(c + d*x)*a*b*c + 12*sin(c + d*x)*b**3)/(24*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))