Integrand size = 33, antiderivative size = 187 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a b (3 A+4 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \]
1/4*a*b*(3*A+4*C)*arctanh(sin(d*x+c))/d+1/15*(5*b^2*(2*A+3*C)+2*a^2*(4*A+5 *C))*tan(d*x+c)/d+1/4*a*b*(3*A+4*C)*sec(d*x+c)*tan(d*x+c)/d+1/15*(2*A*b^2+ a^2*(4*A+5*C))*sec(d*x+c)^2*tan(d*x+c)/d+1/10*a*A*b*sec(d*x+c)^3*tan(d*x+c )/d+1/5*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d
Time = 1.18 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.61 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 a b (3 A+4 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (60 \left (a^2+b^2\right ) (A+C)+15 a b (3 A+4 C) \sec (c+d x)+30 a A b \sec ^3(c+d x)+20 \left (A b^2+a^2 (2 A+C)\right ) \tan ^2(c+d x)+12 a^2 A \tan ^4(c+d x)\right )}{60 d} \]
(15*a*b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(60*(a^2 + b^2)*( A + C) + 15*a*b*(3*A + 4*C)*Sec[c + d*x] + 30*a*A*b*Sec[c + d*x]^3 + 20*(A *b^2 + a^2*(2*A + C))*Tan[c + d*x]^2 + 12*a^2*A*Tan[c + d*x]^4))/(60*d)
Time = 1.25 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3527, 3042, 3510, 27, 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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \frac {1}{5} \int (a+b \cos (c+d x)) \left (b (2 A+5 C) \cos ^2(c+d x)+a (4 A+5 C) \cos (c+d x)+2 A b\right ) \sec ^5(c+d x)dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}-\frac {1}{4} \int -2 \left (2 b^2 (2 A+5 C) \cos ^2(c+d x)+5 a b (3 A+4 C) \cos (c+d x)+2 \left ((4 A+5 C) a^2+2 A b^2\right )\right ) \sec ^4(c+d x)dx\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \int \left (2 b^2 (2 A+5 C) \cos ^2(c+d x)+5 a b (3 A+4 C) \cos (c+d x)+2 \left ((4 A+5 C) a^2+2 A b^2\right )\right ) \sec ^4(c+d x)dx+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \int \frac {2 b^2 (2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+5 a b (3 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 \left ((4 A+5 C) a^2+2 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \int \left (15 a b (3 A+4 C)+2 \left (2 (4 A+5 C) a^2+5 b^2 (2 A+3 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \int \frac {15 a b (3 A+4 C)+2 \left (2 (4 A+5 C) a^2+5 b^2 (2 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \left (2 \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \int \sec ^2(c+d x)dx+15 a b (3 A+4 C) \int \sec ^3(c+d x)dx\right )+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \left (2 \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+15 a b (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \left (15 a b (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {2 \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \left (15 a b (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {2 \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{d}\right )+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \left (15 a b (3 A+4 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{d}\right )+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \left (15 a b (3 A+4 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{d}\right )+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \left (\frac {2 \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{d}+15 a b (3 A+4 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )\right )+\frac {2 \left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{2 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
(A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a*A*b*Sec [c + d*x]^3*Tan[c + d*x])/(2*d) + ((2*(2*A*b^2 + a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((2*(5*b^2*(2*A + 3*C) + 2*a^2*(4*A + 5*C))*T an[c + d*x])/d + 15*a*b*(3*A + 4*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)))/3)/2)/5
3.6.39.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(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 = 10.86 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.95
| method | result | size |
| parts | \(-\frac {A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (A \,b^{2}+a^{2} C \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{2} C \tan \left (d x +c \right )}{d}+\frac {2 A a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {2 C a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(178\) |
| derivativedivides | \(\frac {-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 A a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 C a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right ) b^{2}}{d}\) | \(184\) |
| default | \(\frac {-A \,a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (d x +c \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (d x +c \right )\right )}{15}\right ) \tan \left (d x +c \right )-a^{2} C \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 A a b \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+2 C a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+C \tan \left (d x +c \right ) b^{2}}{d}\) | \(184\) |
| parallelrisch | \(\frac {-45 b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {4 C}{3}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 b \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A +\frac {4 C}{3}\right ) a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (160 A +200 C \right ) a^{2}+200 b^{2} \left (A +\frac {9 C}{10}\right )\right ) \sin \left (3 d x +3 c \right )+\left (\left (32 A +40 C \right ) a^{2}+40 b^{2} \left (A +\frac {3 C}{2}\right )\right ) \sin \left (5 d x +5 c \right )+420 b \left (A +\frac {4 C}{7}\right ) a \sin \left (2 d x +2 c \right )+90 b \left (A +\frac {4 C}{3}\right ) a \sin \left (4 d x +4 c \right )+320 \left (\frac {\left (A +\frac {3 C}{4}\right ) b^{2}}{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) \sin \left (d x +c \right )}{60 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\) | \(266\) |
| risch | \(-\frac {i \left (45 A a b \,{\mathrm e}^{9 i \left (d x +c \right )}+60 C a b \,{\mathrm e}^{9 i \left (d x +c \right )}-60 C \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+210 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+120 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-120 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-120 C \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-240 C \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-320 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-280 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-360 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-210 A a b \,{\mathrm e}^{3 i \left (d x +c \right )}-120 C a b \,{\mathrm e}^{3 i \left (d x +c \right )}-160 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-200 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-240 b^{2} C \,{\mathrm e}^{2 i \left (d x +c \right )}-45 A a b \,{\mathrm e}^{i \left (d x +c \right )}-60 C a b \,{\mathrm e}^{i \left (d x +c \right )}-32 A \,a^{2}-40 A \,b^{2}-40 a^{2} C -60 b^{2} C \right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 A a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}+\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {3 A a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}-\frac {a b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\) | \(421\) |
-A*a^2/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)-(A*b^2+C*a^ 2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b^2*C/d*tan(d*x+c)+2*A*a*b/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*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)))
Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, {\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, A a b \cos \left (d x + c\right ) + 12 \, A a^{2} + 4 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]
1/120*(15*(3*A + 4*C)*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*A + 4*C)*a*b*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(15*(3*A + 4*C)*a*b*co s(d*x + c)^3 + 4*(2*(4*A + 5*C)*a^2 + 5*(2*A + 3*C)*b^2)*cos(d*x + c)^4 + 30*A*a*b*cos(d*x + c) + 12*A*a^2 + 4*((4*A + 5*C)*a^2 + 5*A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 15 \, A a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C b^{2} \tan \left (d x + c\right )}{120 \, d} \]
1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d* x + c))*A*b^2 - 15*A*a*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*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)) + 120*C*b^2*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 532 vs. \(2 (175) = 350\).
Time = 0.34 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.84 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, A a b + 4 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, A a b + 4 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 80 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 240 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 232 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 360 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, A 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 ) + 75 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b^{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 )}^{5}}}{60 \, d} \]
1/60*(15*(3*A*a*b + 4*C*a*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*A* a*b + 4*C*a*b)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(60*A*a^2*tan(1/2*d* x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*A*a*b*tan(1/2*d*x + 1/ 2*c)^9 - 60*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*A*b^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*b^2*tan(1/2*d*x + 1/2*c)^9 - 80*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 160 *C*a^2*tan(1/2*d*x + 1/2*c)^7 + 30*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 120*C*a* b*tan(1/2*d*x + 1/2*c)^7 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 240*C*b^2*ta n(1/2*d*x + 1/2*c)^7 + 232*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*C*a^2*tan(1/ 2*d*x + 1/2*c)^5 + 200*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 360*C*b^2*tan(1/2*d* x + 1/2*c)^5 - 80*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 160*C*a^2*tan(1/2*d*x + 1 /2*c)^3 - 30*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 120*C*a*b*tan(1/2*d*x + 1/2*c) ^3 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 240*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 60*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1/2*d*x + 1/2*c) + 75*A*a*b* tan(1/2*d*x + 1/2*c) + 60*C*a*b*tan(1/2*d*x + 1/2*c) + 60*A*b^2*tan(1/2*d* x + 1/2*c) + 60*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5 )/d
Time = 5.86 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.72 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,C\right )}{2\,d}-\frac {\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2-\frac {5\,A\,a\,b}{2}-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (A\,a\,b-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-8\,C\,b^2-\frac {8\,A\,a^2}{3}+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^2}{15}+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+12\,C\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^2}{3}-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-8\,C\,b^2-A\,a\,b-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2+\frac {5\,A\,a\,b}{2}+2\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]
(a*b*atanh(tan(c/2 + (d*x)/2))*(3*A + 4*C))/(2*d) - (tan(c/2 + (d*x)/2)^9* (2*A*a^2 + 2*A*b^2 + 2*C*a^2 + 2*C*b^2 - (5*A*a*b)/2 - 2*C*a*b) - tan(c/2 + (d*x)/2)^3*((8*A*a^2)/3 + (16*A*b^2)/3 + (16*C*a^2)/3 + 8*C*b^2 + A*a*b + 4*C*a*b) - tan(c/2 + (d*x)/2)^7*((8*A*a^2)/3 + (16*A*b^2)/3 + (16*C*a^2) /3 + 8*C*b^2 - A*a*b - 4*C*a*b) + tan(c/2 + (d*x)/2)^5*((116*A*a^2)/15 + ( 20*A*b^2)/3 + (20*C*a^2)/3 + 12*C*b^2) + tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A *b^2 + 2*C*a^2 + 2*C*b^2 + (5*A*a*b)/2 + 2*C*a*b))/(d*(5*tan(c/2 + (d*x)/2 )^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x )/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))