Integrand size = 33, antiderivative size = 154 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {2 a b (2 A+3 C) \tan (c+d x)}{3 d}+\frac {\left (2 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a A b \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]
1/8*(4*b^2*(A+2*C)+a^2*(3*A+4*C))*arctanh(sin(d*x+c))/d+2/3*a*b*(2*A+3*C)* tan(d*x+c)/d+1/8*(2*A*b^2+a^2*(3*A+4*C))*sec(d*x+c)*tan(d*x+c)/d+1/6*a*A*b *sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c )/d
Time = 0.67 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \left (4 b^2 (A+2 C)+a^2 (3 A+4 C)\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 A b^2+a^2 (3 A+4 C)\right ) \sec (c+d x)+6 a^2 A \sec ^3(c+d x)+16 a b \left (3 (A+C)+A \tan ^2(c+d x)\right )\right )}{24 d} \]
(3*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x ]*(3*(4*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x] + 6*a^2*A*Sec[c + d*x]^3 + 1 6*a*b*(3*(A + C) + A*Tan[c + d*x]^2)))/(24*d)
Time = 1.10 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3527, 3042, 3510, 25, 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.
| \(\displaystyle \int \sec ^5(c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \frac {1}{4} \int (a+b \cos (c+d x)) \left (b (A+4 C) \cos ^2(c+d x)+a (3 A+4 C) \cos (c+d x)+2 A b\right ) \sec ^4(c+d x)dx+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (3 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{4} \left (\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}-\frac {1}{3} \int -\left (\left (3 b^2 (A+4 C) \cos ^2(c+d x)+8 a b (2 A+3 C) \cos (c+d x)+3 \left ((3 A+4 C) a^2+2 A b^2\right )\right ) \sec ^3(c+d x)\right )dx\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \left (3 b^2 (A+4 C) \cos ^2(c+d x)+8 a b (2 A+3 C) \cos (c+d x)+3 \left ((3 A+4 C) a^2+2 A b^2\right )\right ) \sec ^3(c+d x)dx+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int \frac {3 b^2 (A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+8 a b (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+3 \left ((3 A+4 C) a^2+2 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (16 a b (2 A+3 C)+3 \left ((3 A+4 C) a^2+4 b^2 (A+2 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x)dx+\frac {3 \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {16 a b (2 A+3 C)+3 \left ((3 A+4 C) a^2+4 b^2 (A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {3 \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \sec (c+d x)dx+16 a b (2 A+3 C) \int \sec ^2(c+d x)dx\right )+\frac {3 \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+16 a b (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {3 \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {16 a b (2 A+3 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {3 \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (3 \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {16 a b (2 A+3 C) \tan (c+d x)}{d}\right )+\frac {3 \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (\frac {3 \left (a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {16 a b (2 A+3 C) \tan (c+d x)}{d}\right )+\frac {3 \left (a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {2 a A b \tan (c+d x) \sec ^2(c+d x)}{3 d}\right )+\frac {A \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d}\) |
(A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((2*a*A*b*S ec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((3*(2*A*b^2 + a^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*(4*b^2*(A + 2*C) + a^2*(3*A + 4*C))*ArcTa nh[Sin[c + d*x]])/d + (16*a*b*(2*A + 3*C)*Tan[c + d*x])/d)/2)/3)/4
3.6.38.3.1 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_.)*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_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 9.68 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.05
| method | result | size |
| parts | \(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (A \,b^{2}+a^{2} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {2 A a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {2 C a b \tan \left (d x +c \right )}{d}\) | \(162\) |
| derivativedivides | \(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 A a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 C a b \tan \left (d x +c \right )+A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(180\) |
| default | \(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 A a b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 C a b \tan \left (d x +c \right )+A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(180\) |
| parallelrisch | \(\frac {-36 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+36 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (18 A +24 C \right ) a^{2}+24 A \,b^{2}\right ) \sin \left (3 d x +3 c \right )+128 b \left (A +\frac {3 C}{4}\right ) a \sin \left (2 d x +2 c \right )+32 b a \left (A +\frac {3 C}{2}\right ) \sin \left (4 d x +4 c \right )+66 \left (\left (A +\frac {4 C}{11}\right ) a^{2}+\frac {4 A \,b^{2}}{11}\right ) \sin \left (d x +c \right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(231\) |
| risch | \(-\frac {i \left (9 A \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-48 C a b \,{\mathrm e}^{6 i \left (d x +c \right )}+33 A \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 A \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-96 A a b \,{\mathrm e}^{4 i \left (d x +c \right )}-144 C a b \,{\mathrm e}^{4 i \left (d x +c \right )}-33 A \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-128 A a b \,{\mathrm e}^{2 i \left (d x +c \right )}-144 C a b \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-12 A \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-32 A a b -48 C a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a^{2} C}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2} C}{d}-\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a^{2} C}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2} C}{d}\) | \(457\) |
| norman | \(\frac {\frac {\left (5 A \,a^{2}-16 A a b +4 A \,b^{2}+4 a^{2} C -16 C a b \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (5 A \,a^{2}+16 A a b +4 A \,b^{2}+4 a^{2} C +16 C a b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (45 A \,a^{2}-16 A a b +4 A \,b^{2}+4 a^{2} C +48 C a b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (45 A \,a^{2}+16 A a b +4 A \,b^{2}+4 a^{2} C -48 C a b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (69 A \,a^{2}-112 A a b +36 A \,b^{2}+36 a^{2} C -48 C a b \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (69 A \,a^{2}+112 A a b +36 A \,b^{2}+36 a^{2} C +48 C a b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (165 A \,a^{2}-16 A a b -60 A \,b^{2}-60 a^{2} C -144 C a b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (165 A \,a^{2}+16 A a b -60 A \,b^{2}-60 a^{2} C +144 C a b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (3 A \,a^{2}+4 A \,b^{2}+4 a^{2} C +8 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 A \,a^{2}+4 A \,b^{2}+4 a^{2} C +8 b^{2} C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(476\) |
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)))+(A*b^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)))+b^2*C/d*ln(sec(d*x+c)+tan(d*x+c))-2*A*a*b/d*(-2/3-1/3*sec(d*x +c)^2)*tan(d*x+c)+2*C*a*b/d*tan(d*x+c)
Time = 0.30 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, {\left (A + 2 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, {\left (A + 2 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (2 \, A + 3 \, C\right )} a b \cos \left (d x + c\right )^{3} + 16 \, A a b \cos \left (d x + c\right ) + 6 \, A a^{2} + 3 \, {\left ({\left (3 \, A + 4 \, C\right )} a^{2} + 4 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]
1/48*(3*((3*A + 4*C)*a^2 + 4*(A + 2*C)*b^2)*cos(d*x + c)^4*log(sin(d*x + c ) + 1) - 3*((3*A + 4*C)*a^2 + 4*(A + 2*C)*b^2)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*(2*A + 3*C)*a*b*cos(d*x + c)^3 + 16*A*a*b*cos(d*x + c) + 6*A*a^2 + 3*((3*A + 4*C)*a^2 + 4*A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d *cos(d*x + c)^4)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.51 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b - 3 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, C a b \tan \left (d x + c\right )}{48 \, d} \]
1/48*(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(si n(d*x + c) + 1) + 3*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)) - 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)) + 24*C*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*C*a*b*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 426 vs. \(2 (144) = 288\).
Time = 0.34 (sec) , antiderivative size = 426, normalized size of antiderivative = 2.77 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 8 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a^{2} + 4 \, C a^{2} + 4 \, A b^{2} + 8 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 144 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 80 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 144 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, A 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 )}^{4}}}{24 \, d} \]
1/24*(3*(3*A*a^2 + 4*C*a^2 + 4*A*b^2 + 8*C*b^2)*log(abs(tan(1/2*d*x + 1/2* c) + 1)) - 3*(3*A*a^2 + 4*C*a^2 + 4*A*b^2 + 8*C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 48*A*a*b*tan(1/2*d*x + 1/2*c)^7 - 48*C*a*b*tan(1/2*d*x + 1/2* c)^7 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^7 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 80*A*a*b*tan(1/2*d*x + 1/2*c)^5 + 144*C* a*b*tan(1/2*d*x + 1/2*c)^5 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 9*A*a^2*tan (1/2*d*x + 1/2*c)^3 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*A*a*b*tan(1/2*d *x + 1/2*c)^3 - 144*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 15*A*a^2*tan(1/2*d*x + 1/2*c) + 12*C*a^2*tan(1/2*d*x + 1/2*c) + 48*A*a*b*tan(1/2*d*x + 1/2*c) + 48*C*a*b*tan(1/2*d*x + 1/2*c) + 12*A*b^2* tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 5.77 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.99 \[ \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {\left (\frac {5\,A\,a^2}{4}+A\,b^2+C\,a^2-4\,A\,a\,b-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a^2}{4}-A\,b^2-C\,a^2+\frac {20\,A\,a\,b}{3}+12\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,A\,a^2}{4}-A\,b^2-C\,a^2-\frac {20\,A\,a\,b}{3}-12\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a^2}{4}+A\,b^2+C\,a^2+4\,A\,a\,b+4\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,A\,a^2}{8}+\frac {A\,b^2}{2}+\frac {C\,a^2}{2}+C\,b^2\right )}{\frac {3\,A\,a^2}{2}+2\,A\,b^2+2\,C\,a^2+4\,C\,b^2}\right )\,\left (\frac {3\,A\,a^2}{4}+A\,b^2+C\,a^2+2\,C\,b^2\right )}{d} \]
(tan(c/2 + (d*x)/2)*((5*A*a^2)/4 + A*b^2 + C*a^2 + 4*A*a*b + 4*C*a*b) + ta n(c/2 + (d*x)/2)^7*((5*A*a^2)/4 + A*b^2 + C*a^2 - 4*A*a*b - 4*C*a*b) - tan (c/2 + (d*x)/2)^3*(A*b^2 - (3*A*a^2)/4 + C*a^2 + (20*A*a*b)/3 + 12*C*a*b) + tan(c/2 + (d*x)/2)^5*((3*A*a^2)/4 - A*b^2 - C*a^2 + (20*A*a*b)/3 + 12*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)) + (atanh((4*tan(c/2 + (d*x)/2)*((3*A *a^2)/8 + (A*b^2)/2 + (C*a^2)/2 + C*b^2))/((3*A*a^2)/2 + 2*A*b^2 + 2*C*a^2 + 4*C*b^2))*((3*A*a^2)/4 + A*b^2 + C*a^2 + 2*C*b^2))/d