Integrand size = 33, antiderivative size = 167 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} a \left (6 A b^2+2 a^2 C+3 b^2 C\right ) x+\frac {3 a^2 A b \text {arctanh}(\sin (c+d x))}{d}-\frac {b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac {a b^2 (6 A-5 C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d} \]
1/2*a*(6*A*b^2+2*C*a^2+3*C*b^2)*x+3*a^2*A*b*arctanh(sin(d*x+c))/d-1/3*b*(a ^2*(6*A-8*C)-b^2*(3*A+2*C))*sin(d*x+c)/d-1/6*a*b^2*(6*A-5*C)*cos(d*x+c)*si n(d*x+c)/d-1/3*b*(3*A-C)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+A*(a+b*cos(d*x+c) )^3*tan(d*x+c)/d
Time = 2.86 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.11 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {36 a A b^2 c+12 a^3 c C+18 a b^2 c C+36 a A b^2 d x+12 a^3 C d x+18 a b^2 C d x-36 a^2 A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+36 a^2 A b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 b \left (4 A b^2+3 \left (4 a^2+b^2\right ) C\right ) \sin (c+d x)+9 a b^2 C \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))+12 a^3 A \tan (c+d x)}{12 d} \]
(36*a*A*b^2*c + 12*a^3*c*C + 18*a*b^2*c*C + 36*a*A*b^2*d*x + 12*a^3*C*d*x + 18*a*b^2*C*d*x - 36*a^2*A*b*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3 6*a^2*A*b*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*b*(4*A*b^2 + 3*(4*a ^2 + b^2)*C)*Sin[c + d*x] + 9*a*b^2*C*Sin[2*(c + d*x)] + b^3*C*Sin[3*(c + d*x)] + 12*a^3*A*Tan[c + d*x])/(12*d)
Time = 1.24 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.05, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.394, Rules used = {3042, 3527, 3042, 3528, 3042, 3512, 3042, 3502, 27, 3042, 3214, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \int (a+b \cos (c+d x))^2 \left (-b (3 A-C) \cos ^2(c+d x)+a C \cos (c+d x)+3 A b\right ) \sec (c+d x)dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (-b (3 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+a C \sin \left (c+d x+\frac {\pi }{2}\right )+3 A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {1}{3} \int (a+b \cos (c+d x)) \left (-a b (6 A-5 C) \cos ^2(c+d x)+\left (3 C a^2+3 A b^2+2 b^2 C\right ) \cos (c+d x)+9 a A b\right ) \sec (c+d x)dx-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-a b (6 A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 C a^2+3 A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+9 a A b\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (18 A b a^2+3 \left (2 C a^2+6 A b^2+3 b^2 C\right ) \cos (c+d x) a-2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {18 A b a^2+3 \left (2 C a^2+6 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a-2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int 3 \left (6 A b a^2+\left (2 C a^2+6 A b^2+3 b^2 C\right ) \cos (c+d x) a\right ) \sec (c+d x)dx-\frac {2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \left (6 A b a^2+\left (2 C a^2+6 A b^2+3 b^2 C\right ) \cos (c+d x) a\right ) \sec (c+d x)dx-\frac {2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \int \frac {6 A b a^2+\left (2 C a^2+6 A b^2+3 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) a}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (6 a^2 A b \int \sec (c+d x)dx+a x \left (2 a^2 C+6 A b^2+3 b^2 C\right )\right )-\frac {2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (6 a^2 A b \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+a x \left (2 a^2 C+6 A b^2+3 b^2 C\right )\right )-\frac {2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 \left (\frac {6 a^2 A b \text {arctanh}(\sin (c+d x))}{d}+a x \left (2 a^2 C+6 A b^2+3 b^2 C\right )\right )-\frac {2 b \left (a^2 (6 A-8 C)-b^2 (3 A+2 C)\right ) \sin (c+d x)}{d}\right )-\frac {a b^2 (6 A-5 C) \sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d}\) |
-1/3*(b*(3*A - C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/d + (-1/2*(a*b^2*(6 *A - 5*C)*Cos[c + d*x]*Sin[c + d*x])/d + (3*(a*(6*A*b^2 + 2*a^2*C + 3*b^2* C)*x + (6*a^2*A*b*ArcTanh[Sin[c + d*x]])/d) - (2*b*(a^2*(6*A - 8*C) - b^2* (3*A + 2*C))*Sin[c + d*x])/d)/2)/3 + (A*(a + b*Cos[c + d*x])^3*Tan[c + d*x ])/d
3.6.43.3.1 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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, 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[((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)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A* d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2) - C*(a *c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[ m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 5.84 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {A \,a^{3} \tan \left (d x +c \right )+C \,a^{3} \left (d x +c \right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a^{2} b +3 A a \,b^{2} \left (d x +c \right )+3 C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{3}+\frac {C \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(135\) |
| default | \(\frac {A \,a^{3} \tan \left (d x +c \right )+C \,a^{3} \left (d x +c \right )+3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C \sin \left (d x +c \right ) a^{2} b +3 A a \,b^{2} \left (d x +c \right )+3 C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \sin \left (d x +c \right ) b^{3}+\frac {C \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(135\) |
| parts | \(\frac {A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {\left (A \,b^{3}+3 C \,a^{2} b \right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 A a \,b^{2}+C \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {C \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(142\) |
| parallelrisch | \(\frac {-72 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b \cos \left (d x +c \right )+72 A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b \cos \left (d x +c \right )+12 \left (\left (A +\frac {5 C}{6}\right ) b^{2}+3 a^{2} C \right ) b \sin \left (2 d x +2 c \right )+9 C \sin \left (3 d x +3 c \right ) a \,b^{2}+C \sin \left (4 d x +4 c \right ) b^{3}+24 a \left (3 x \left (\left (A +\frac {C}{2}\right ) b^{2}+\frac {a^{2} C}{3}\right ) d \cos \left (d x +c \right )+\sin \left (d x +c \right ) \left (A \,a^{2}+\frac {3 b^{2} C}{8}\right )\right )}{24 d \cos \left (d x +c \right )}\) | \(172\) |
| risch | \(3 A a \,b^{2} x +a^{3} C x +\frac {3 a \,b^{2} C x}{2}-\frac {3 i C a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,b^{3}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{2} b}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} C \,b^{3}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,b^{3}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{2} b}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} C \,b^{3}}{8 d}+\frac {3 i C a \,b^{2} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i A \,a^{3}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 A \,a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {3 A \,a^{2} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) C \,b^{3}}{12 d}\) | \(267\) |
| norman | \(\frac {\left (-3 A a \,b^{2}-C \,a^{3}-\frac {3}{2} C a \,b^{2}\right ) x +\left (-15 A a \,b^{2}-5 C \,a^{3}-\frac {15}{2} C a \,b^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A a \,b^{2}+C \,a^{3}+\frac {3}{2} C a \,b^{2}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (15 A a \,b^{2}+5 C \,a^{3}+\frac {15}{2} C a \,b^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-12 A a \,b^{2}-4 C \,a^{3}-6 C a \,b^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A a \,b^{2}+4 C \,a^{3}+6 C a \,b^{2}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {\left (2 A \,a^{3}-2 A \,b^{3}-6 C \,a^{2} b +3 C a \,b^{2}-2 C \,b^{3}\right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {\left (2 A \,a^{3}+2 A \,b^{3}+6 C \,a^{2} b +3 C a \,b^{2}+2 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (30 A \,a^{3}-18 A \,b^{3}-54 C \,a^{2} b +9 C a \,b^{2}-10 C \,b^{3}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (30 A \,a^{3}-6 A \,b^{3}-18 C \,a^{2} b -9 C a \,b^{2}-2 C \,b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 \left (30 A \,a^{3}+6 A \,b^{3}+18 C \,a^{2} b -9 C a \,b^{2}+2 C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {\left (30 A \,a^{3}+18 A \,b^{3}+54 C \,a^{2} b +9 C a \,b^{2}+10 C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 A \,a^{2} b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {3 A \,a^{2} b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(562\) |
1/d*(A*a^3*tan(d*x+c)+C*a^3*(d*x+c)+3*A*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3* C*sin(d*x+c)*a^2*b+3*A*a*b^2*(d*x+c)+3*C*a*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+ 1/2*d*x+1/2*c)+A*sin(d*x+c)*b^3+1/3*C*b^3*(2+cos(d*x+c)^2)*sin(d*x+c))
Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.95 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {9 \, A a^{2} b \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, A a^{2} b \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, C a^{3} + 3 \, {\left (2 \, A + C\right )} a b^{2}\right )} d x \cos \left (d x + c\right ) + {\left (2 \, C b^{3} \cos \left (d x + c\right )^{3} + 9 \, C a b^{2} \cos \left (d x + c\right )^{2} + 6 \, A a^{3} + 2 \, {\left (9 \, C a^{2} b + {\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]
1/6*(9*A*a^2*b*cos(d*x + c)*log(sin(d*x + c) + 1) - 9*A*a^2*b*cos(d*x + c) *log(-sin(d*x + c) + 1) + 3*(2*C*a^3 + 3*(2*A + C)*a*b^2)*d*x*cos(d*x + c) + (2*C*b^3*cos(d*x + c)^3 + 9*C*a*b^2*cos(d*x + c)^2 + 6*A*a^3 + 2*(9*C*a ^2*b + (3*A + 2*C)*b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.84 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} C a^{3} + 36 \, {\left (d x + c\right )} A a b^{2} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 18 \, A a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, C a^{2} b \sin \left (d x + c\right ) + 12 \, A b^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \]
1/12*(12*(d*x + c)*C*a^3 + 36*(d*x + c)*A*a*b^2 + 9*(2*d*x + 2*c + sin(2*d *x + 2*c))*C*a*b^2 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*b^3 + 18*A*a^2* b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*C*a^2*b*sin(d*x + c ) + 12*A*b^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 306, normalized size of antiderivative = 1.83 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {18 \, A a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 18 \, A a^{2} b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + 3 \, {\left (2 \, C a^{3} + 6 \, A a b^{2} + 3 \, C a b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
1/6*(18*A*a^2*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 18*A*a^2*b*log(abs(ta n(1/2*d*x + 1/2*c) - 1)) - 12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/ 2*c)^2 - 1) + 3*(2*C*a^3 + 6*A*a*b^2 + 3*C*a*b^2)*(d*x + c) + 2*(18*C*a^2* b*tan(1/2*d*x + 1/2*c)^5 - 9*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^3*tan( 1/2*d*x + 1/2*c)^5 + 6*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 36*C*a^2*b*tan(1/2*d *x + 1/2*c)^3 + 12*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^3*tan(1/2*d*x + 1/ 2*c)^3 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c) + 9*C*a*b^2*tan(1/2*d*x + 1/2*c) + 6*A*b^3*tan(1/2*d*x + 1/2*c) + 6*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d* x + 1/2*c)^2 + 1)^3)/d
Time = 2.66 (sec) , antiderivative size = 238, normalized size of antiderivative = 1.43 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+6\,A\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+3\,C\,a\,b^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,a^2\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,6{}\mathrm {i}}{d}+\frac {\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {5\,C\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{12}+\frac {C\,b^3\,\sin \left (4\,c+4\,d\,x\right )}{24}+A\,a^3\,\sin \left (c+d\,x\right )+\frac {3\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{8}+\frac {3\,C\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {3\,C\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{8}}{d\,\cos \left (c+d\,x\right )} \]
(2*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 6*A*a*b^2*atan(sin( c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*a^2*b*atan((sin(c/2 + (d*x)/2)*1i)/ cos(c/2 + (d*x)/2))*6i + 3*C*a*b^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x) /2)))/d + ((A*b^3*sin(2*c + 2*d*x))/2 + (5*C*b^3*sin(2*c + 2*d*x))/12 + (C *b^3*sin(4*c + 4*d*x))/24 + A*a^3*sin(c + d*x) + (3*C*a*b^2*sin(c + d*x))/ 8 + (3*C*a^2*b*sin(2*c + 2*d*x))/2 + (3*C*a*b^2*sin(3*c + 3*d*x))/8)/(d*co s(c + d*x))