Integrand size = 31, antiderivative size = 167 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{8} b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) x+\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sin (c+d x)}{2 d}+\frac {b \left (2 a^2 C+b^2 (4 A+3 C)\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a C (a+b \cos (c+d x))^2 \sin (c+d x)}{4 d}+\frac {C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d} \]
1/8*b*(12*a^2*(2*A+C)+b^2*(4*A+3*C))*x+a^3*A*arctanh(sin(d*x+c))/d+1/2*a*( 6*A*b^2+(a^2+4*b^2)*C)*sin(d*x+c)/d+1/8*b*(2*a^2*C+b^2*(4*A+3*C))*cos(d*x+ c)*sin(d*x+c)/d+1/4*a*C*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/4*C*(a+b*cos(d*x +c))^3*sin(d*x+c)/d
Time = 2.29 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.08 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {4 b \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) (c+d x)-32 a^3 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 a^3 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 a \left (12 A b^2+4 a^2 C+9 b^2 C\right ) \sin (c+d x)+8 b \left (A b^2+\left (3 a^2+b^2\right ) C\right ) \sin (2 (c+d x))+8 a b^2 C \sin (3 (c+d x))+b^3 C \sin (4 (c+d x))}{32 d} \]
(4*b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*(c + d*x) - 32*a^3*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*a^3*A*Log[Cos[(c + d*x)/2] + Sin[(c + d *x)/2]] + 8*a*(12*A*b^2 + 4*a^2*C + 9*b^2*C)*Sin[c + d*x] + 8*b*(A*b^2 + ( 3*a^2 + b^2)*C)*Sin[2*(c + d*x)] + 8*a*b^2*C*Sin[3*(c + d*x)] + b^3*C*Sin[ 4*(c + d*x)])/(32*d)
Time = 1.24 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 3529, 3042, 3528, 27, 3042, 3512, 3042, 3502, 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.
| \(\displaystyle \int \sec (c+d x) (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {1}{4} \int (a+b \cos (c+d x))^2 \left (3 a C \cos ^2(c+d x)+b (4 A+3 C) \cos (c+d x)+4 a A\right ) \sec (c+d x)dx+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (3 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (4 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )+4 a A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3528 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{3} \int 3 (a+b \cos (c+d x)) \left (4 A a^2+b (8 A+5 C) \cos (c+d x) a+\left (2 C a^2+b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\int (a+b \cos (c+d x)) \left (4 A a^2+b (8 A+5 C) \cos (c+d x) a+\left (2 C a^2+b^2 (4 A+3 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x)dx+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (4 A a^2+b (8 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+\left (2 C a^2+b^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \left (8 A a^3+4 \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \cos ^2(c+d x) a+b \left (12 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {b \left (2 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {8 A a^3+4 \left (6 A b^2+\left (a^2+4 b^2\right ) C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a+b \left (12 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {b \left (2 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\int \left (8 A a^3+b \left (12 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {4 a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \sin (c+d x)}{d}\right )+\frac {b \left (2 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (\int \frac {8 A a^3+b \left (12 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {4 a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \sin (c+d x)}{d}\right )+\frac {b \left (2 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (8 a^3 A \int \sec (c+d x)dx+\frac {4 a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \sin (c+d x)}{d}+b x \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right )+\frac {b \left (2 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \left (8 a^3 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {4 a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \sin (c+d x)}{d}+b x \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right )+\frac {b \left (2 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{4} \left (\frac {b \left (2 a^2 C+4 A b^2+3 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {1}{2} \left (\frac {8 a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {4 a \left (C \left (a^2+4 b^2\right )+6 A b^2\right ) \sin (c+d x)}{d}+b x \left (12 a^2 (2 A+C)+b^2 (4 A+3 C)\right )\right )+\frac {a C \sin (c+d x) (a+b \cos (c+d x))^2}{d}\right )+\frac {C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}\) |
(C*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + ((b*(4*A*b^2 + 2*a^2*C + 3 *b^2*C)*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (a*C*(a + b*Cos[c + d*x])^2*Sin [c + d*x])/d + (b*(12*a^2*(2*A + C) + b^2*(4*A + 3*C))*x + (8*a^3*A*ArcTan h[Sin[c + d*x]])/d + (4*a*(6*A*b^2 + (a^2 + 4*b^2)*C)*Sin[c + d*x])/d)/2)/ 4
3.6.42.3.1 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_.) + (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[((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)*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)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* (a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , A, 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 = 4.51 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.90
| method | result | size |
| parallelrisch | \(\frac {-32 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+32 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 \left (b^{2} \left (A +C \right )+3 a^{2} C \right ) b \sin \left (2 d x +2 c \right )+8 C \sin \left (3 d x +3 c \right ) a \,b^{2}+C \sin \left (4 d x +4 c \right ) b^{3}+96 \left (\left (A +\frac {3 C}{4}\right ) b^{2}+\frac {a^{2} C}{3}\right ) a \sin \left (d x +c \right )+96 x b d \left (\left (\frac {A}{6}+\frac {C}{8}\right ) b^{2}+a^{2} \left (A +\frac {C}{2}\right )\right )}{32 d}\) | \(150\) |
| parts | \(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,b^{3}+3 C \,a^{2} b \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A a \,b^{2}+C \,a^{3}\right ) \sin \left (d x +c \right )}{d}+\frac {C \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {3 A \,a^{2} b \left (d x +c \right )}{d}+\frac {C a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d}\) | \(167\) |
| derivativedivides | \(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \sin \left (d x +c \right )+3 A \,a^{2} b \left (d x +c \right )+3 C \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \sin \left (d x +c \right ) a \,b^{2}+C a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(177\) |
| default | \(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{3} \sin \left (d x +c \right )+3 A \,a^{2} b \left (d x +c \right )+3 C \,a^{2} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 A \sin \left (d x +c \right ) a \,b^{2}+C a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,b^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(177\) |
| risch | \(3 A \,a^{2} b x +\frac {A \,b^{3} x}{2}+\frac {3 C \,a^{2} b x}{2}+\frac {3 b^{3} C x}{8}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A a \,b^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} C \,a^{3}}{2 d}-\frac {9 i {\mathrm e}^{i \left (d x +c \right )} C a \,b^{2}}{8 d}+\frac {9 i {\mathrm e}^{-i \left (d x +c \right )} C a \,b^{2}}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{3}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A a \,b^{2}}{2 d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (4 d x +4 c \right ) C \,b^{3}}{32 d}+\frac {\sin \left (3 d x +3 c \right ) C a \,b^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) C \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C \,b^{3}}{4 d}\) | \(285\) |
| norman | \(\frac {\left (\frac {3}{8} C \,b^{3}+3 A \,a^{2} b +\frac {1}{2} A \,b^{3}+\frac {3}{2} C \,a^{2} b \right ) x +\left (\frac {3}{8} C \,b^{3}+3 A \,a^{2} b +\frac {1}{2} A \,b^{3}+\frac {3}{2} C \,a^{2} b \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{4} C \,b^{3}+30 A \,a^{2} b +5 A \,b^{3}+15 C \,a^{2} b \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{4} C \,b^{3}+30 A \,a^{2} b +5 A \,b^{3}+15 C \,a^{2} b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{8} C \,b^{3}+15 A \,a^{2} b +\frac {5}{2} A \,b^{3}+\frac {15}{2} C \,a^{2} b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{8} C \,b^{3}+15 A \,a^{2} b +\frac {5}{2} A \,b^{3}+\frac {15}{2} C \,a^{2} b \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (24 A a \,b^{2}-4 A \,b^{3}+8 C \,a^{3}-12 C \,a^{2} b +24 C a \,b^{2}-5 C \,b^{3}\right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (24 A a \,b^{2}+4 A \,b^{3}+8 C \,a^{3}+12 C \,a^{2} b +24 C a \,b^{2}+5 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (48 A a \,b^{2}-4 A \,b^{3}+16 C \,a^{3}-12 C \,a^{2} b +32 C a \,b^{2}-C \,b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (48 A a \,b^{2}+4 A \,b^{3}+16 C \,a^{3}+12 C \,a^{2} b +32 C a \,b^{2}+C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {4 a \left (9 A \,b^{2}+3 a^{2} C +5 b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}+\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(546\) |
1/32*(-32*A*a^3*ln(tan(1/2*d*x+1/2*c)-1)+32*A*a^3*ln(tan(1/2*d*x+1/2*c)+1) +8*(b^2*(A+C)+3*a^2*C)*b*sin(2*d*x+2*c)+8*C*sin(3*d*x+3*c)*a*b^2+C*sin(4*d *x+4*c)*b^3+96*((A+3/4*C)*b^2+1/3*a^2*C)*a*sin(d*x+c)+96*x*b*d*((1/6*A+1/8 *C)*b^2+a^2*(A+1/2*C)))/d
Time = 0.29 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.87 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {4 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (12 \, {\left (2 \, A + C\right )} a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} d x + {\left (2 \, C b^{3} \cos \left (d x + c\right )^{3} + 8 \, C a b^{2} \cos \left (d x + c\right )^{2} + 8 \, C a^{3} + 8 \, {\left (3 \, A + 2 \, C\right )} a b^{2} + {\left (12 \, C a^{2} b + {\left (4 \, A + 3 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/8*(4*A*a^3*log(sin(d*x + c) + 1) - 4*A*a^3*log(-sin(d*x + c) + 1) + (12* (2*A + C)*a^2*b + (4*A + 3*C)*b^3)*d*x + (2*C*b^3*cos(d*x + c)^3 + 8*C*a*b ^2*cos(d*x + c)^2 + 8*C*a^3 + 8*(3*A + 2*C)*a*b^2 + (12*C*a^2*b + (4*A + 3 *C)*b^3)*cos(d*x + c))*sin(d*x + c))/d
\[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int \left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, dx \]
Time = 0.20 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {96 \, {\left (d x + c\right )} A a^{2} b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{2} + 8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} + 32 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 32 \, C a^{3} \sin \left (d x + c\right ) + 96 \, A a b^{2} \sin \left (d x + c\right )}{32 \, d} \]
1/32*(96*(d*x + c)*A*a^2*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2*b - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a*b^2 + 8*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^3 + (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C *b^3 + 32*A*a^3*log(sec(d*x + c) + tan(d*x + c)) + 32*C*a^3*sin(d*x + c) + 96*A*a*b^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 503 vs. \(2 (157) = 314\).
Time = 0.32 (sec) , antiderivative size = 503, normalized size of antiderivative = 3.01 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {8 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (24 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 3 \, C b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 72 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 8 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, 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 )}^{4}}}{8 \, d} \]
1/8*(8*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*A*a^3*log(abs(tan(1/2* d*x + 1/2*c) - 1)) + (24*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 3*C*b^3)*(d*x + c) + 2*(8*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 4*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 5*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 24*C*a^ 3*tan(1/2*d*x + 1/2*c)^5 - 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 72*A*a*b^2* tan(1/2*d*x + 1/2*c)^5 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 4*A*b^3*tan(1 /2*d*x + 1/2*c)^5 + 3*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 24*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 72*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 40*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*A*b^3*tan(1/2*d*x + 1/2*c )^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 8*C*a^3*tan(1/2*d*x + 1/2*c) + 12*C *a^2*b*tan(1/2*d*x + 1/2*c) + 24*A*a*b^2*tan(1/2*d*x + 1/2*c) + 24*C*a*b^2 *tan(1/2*d*x + 1/2*c) + 4*A*b^3*tan(1/2*d*x + 1/2*c) + 5*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 3.59 (sec) , antiderivative size = 2008, normalized size of antiderivative = 12.02 \[ \int (a+b \cos (c+d x))^3 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\text {Too large to display} \]
(tan(c/2 + (d*x)/2)*(A*b^3 + 2*C*a^3 + (5*C*b^3)/4 + 6*A*a*b^2 + 6*C*a*b^2 + 3*C*a^2*b) - tan(c/2 + (d*x)/2)^7*(A*b^3 - 2*C*a^3 + (5*C*b^3)/4 - 6*A* a*b^2 - 6*C*a*b^2 + 3*C*a^2*b) + tan(c/2 + (d*x)/2)^3*(A*b^3 + 6*C*a^3 - ( 3*C*b^3)/4 + 18*A*a*b^2 + 10*C*a*b^2 + 3*C*a^2*b) + tan(c/2 + (d*x)/2)^5*( 6*C*a^3 - A*b^3 + (3*C*b^3)/4 + 18*A*a*b^2 + 10*C*a*b^2 - 3*C*a^2*b))/(d*( 4*tan(c/2 + (d*x)/2)^2 + 6*tan(c/2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) - (b*atan(((b*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^ 2*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*b^2) - (b*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*(32*A*a^3 + 16*A*b^3 + 12*C* b^3 + 96*A*a^2*b + 48*C*a^2*b)*1i)/8)*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C *b^2))/8 + (b*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12* A*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*b^2) + (b*(24*A*a^2 + 4*A*b^2 + 12 *C*a^2 + 3*C*b^2)*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 + 96*A*a^2*b + 48*C*a^2* b)*1i)/8)*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2))/8)/((b*(tan(c/2 + (d* x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + (9*C^2*b^6)/2 + 96*A^2*a^2*b^4 + 288*A^2*a ^4*b^2 + 36*C^2*a^2*b^4 + 72*C^2*a^4*b^2 + 12*A*C*b^6 + 120*A*C*a^2*b^4 + 288*A*C*a^4*b^2) + (b*(24*A*a^2 + 4*A*b^2 + 12*C*a^2 + 3*C*b^2)*(32*A*a^3 + 16*A*b^3 + 12*C*b^3 + 96*A*a^2*b + 48*C*a^2*b)*1i)/8)*(24*A*a^2 + 4*A...