Integrand size = 41, antiderivative size = 303 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (12 A b^2+15 a b B+a^2 (4 A+6 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(4 A b+3 a B) \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b \left (39 a^2 b B-6 b^3 B+4 a b^2 (11 A-6 C)+4 a^3 (2 A+3 C)\right ) \tan (c+d x)}{6 d}-\frac {b^2 \left (18 a b B+3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d} \] Output:
1/2*a*(8*A*b^3+B*a^3+12*B*a*b^2+4*a^2*b*(A+2*C))*x+1/2*b^2*(2*A*b^2+8*B*a* b+12*C*a^2+C*b^2)*arctanh(sin(d*x+c))/d+1/6*(12*A*b^2+15*B*a*b+a^2*(4*A+6* C))*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/6*(4*A*b+3*B*a)*cos(d*x+c)*(a+b*sec( d*x+c))^3*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*sec(d*x+c))^4*sin(d*x+c)/d- 1/6*b*(39*B*a^2*b-6*B*b^3+4*a*b^2*(11*A-6*C)+4*a^3*(2*A+3*C))*tan(d*x+c)/d -1/6*b^2*(18*B*a*b+3*b^2*(6*A-C)+a^2*(4*A+6*C))*sec(d*x+c)*tan(d*x+c)/d
Time = 6.03 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.22 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 a \left (8 A b^3+a^3 B+12 a b^2 B+4 a^2 b (A+2 C)\right ) (c+d x)-6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 b^2 \left (2 A b^2+8 a b B+12 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 b^3 (b B+4 a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {12 b^3 (b B+4 a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 a^2 \left (24 A b^2+16 a b B+a^2 (3 A+4 C)\right ) \sin (c+d x)+3 a^3 (4 A b+a B) \sin (2 (c+d x))+a^4 A \sin (3 (c+d x))}{12 d} \] Input:
Integrate[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
(6*a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*(c + d*x) - 6*b^2* (2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x )/2]] + 6*b^2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (3*b^4*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (12*b^3*(b*B + 4*a*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/ 2]) - (3*b^4*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (12*b^3*(b*B + 4 *a*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3*a^2*(24* A*b^2 + 16*a*b*B + a^2*(3*A + 4*C))*Sin[c + d*x] + 3*a^3*(4*A*b + a*B)*Sin [2*(c + d*x)] + a^4*A*Sin[3*(c + d*x)])/(12*d)
Time = 1.52 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {3042, 4582, 3042, 4582, 3042, 4582, 3042, 4536, 2009}
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.
| \(\displaystyle \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^4 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (-b (2 A-3 C) \sec ^2(c+d x)+(2 a A+3 b B+3 a C) \sec (c+d x)+4 A b+3 a B\right )dx+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (-b (2 A-3 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(2 a A+3 b B+3 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 A b+3 a B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left ((4 A+6 C) a^2+15 b B a+12 A b^2-6 b (2 A b-C b+a B) \sec ^2(c+d x)+\left (3 B a^2+4 b (A+3 C) a+6 b^2 B\right ) \sec (c+d x)\right )dx+\frac {(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left ((4 A+6 C) a^2+15 b B a+12 A b^2-6 b (2 A b-C b+a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (3 B a^2+4 b (A+3 C) a+6 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int (a+b \sec (c+d x)) \left (-2 b \left ((4 A+6 C) a^2+18 b B a+3 b^2 (6 A-C)\right ) \sec ^2(c+d x)-b \left (3 B a^2+2 b (4 A-9 C) a-6 b^2 B\right ) \sec (c+d x)+3 \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right )dx+\frac {\sin (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \sec (c+d x))^2}{d}\right )+\frac {(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-2 b \left ((4 A+6 C) a^2+18 b B a+3 b^2 (6 A-C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (3 B a^2+2 b (4 A-9 C) a-6 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right )dx+\frac {\sin (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \sec (c+d x))^2}{d}\right )+\frac {(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 4536 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {1}{2} \int \left (6 \left (12 C a^2+8 b B a+2 A b^2+b^2 C\right ) \sec (c+d x) b^2-2 \left (4 (2 A+3 C) a^3+39 b B a^2+4 b^2 (11 A-6 C) a-6 b^3 B\right ) \sec ^2(c+d x) b+6 a \left (B a^3+4 b (A+2 C) a^2+12 b^2 B a+8 A b^3\right )\right )dx+\frac {\sin (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \sec (c+d x))^2}{d}-\frac {b^2 \tan (c+d x) \sec (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}\right )+\frac {(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {\sin (c+d x) \left (a^2 (4 A+6 C)+15 a b B+12 A b^2\right ) (a+b \sec (c+d x))^2}{d}-\frac {b^2 \tan (c+d x) \sec (c+d x) \left (a^2 (4 A+6 C)+18 a b B+3 b^2 (6 A-C)\right )}{d}+\frac {1}{2} \left (\frac {6 b^2 \left (12 a^2 C+8 a b B+2 A b^2+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 b \tan (c+d x) \left (4 a^3 (2 A+3 C)+39 a^2 b B+4 a b^2 (11 A-6 C)-6 b^3 B\right )}{d}+6 a x \left (a^3 B+4 a^2 b (A+2 C)+12 a b^2 B+8 A b^3\right )\right )\right )+\frac {(3 a B+4 A b) \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}\) |
Input:
Int[Cos[c + d*x]^3*(a + b*Sec[c + d*x])^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(3*d) + (((4*A*b + 3*a*B)*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) + (((12*A*b ^2 + 15*a*b*B + a^2*(4*A + 6*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/d - (b^2*(18*a*b*B + 3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Sec[c + d*x]*Tan[c + d *x])/d + (6*a*(8*A*b^3 + a^3*B + 12*a*b^2*B + 4*a^2*b*(A + 2*C))*x + (6*b^ 2*(2*A*b^2 + 8*a*b*B + 12*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/d - (2*b*( 39*a^2*b*B - 6*b^3*B + 4*a*b^2*(11*A - 6*C) + 4*a^3*(2*A + 3*C))*Tan[c + d *x])/d)/2)/2)/3
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Simp[1/2 Int[Simp[2*A*a + (2*B*a + b*(2* A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B, C}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Time = 2.10 (sec) , antiderivative size = 284, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \sin \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{3} b \sin \left (d x +c \right )+4 a^{3} b C \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 B \,a^{2} b^{2} \left (d x +c \right )+6 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a A \,b^{3} \left (d x +c \right )+4 B a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 C a \,b^{3} \tan \left (d x +c \right )+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{4}+C \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(284\) |
| default | \(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} C \sin \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 B \,a^{3} b \sin \left (d x +c \right )+4 a^{3} b C \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 B \,a^{2} b^{2} \left (d x +c \right )+6 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a A \,b^{3} \left (d x +c \right )+4 B a \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 C a \,b^{3} \tan \left (d x +c \right )+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{4}+C \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(284\) |
| parallelrisch | \(\frac {-24 b^{2} \left (\left (A +\frac {C}{2}\right ) b^{2}+4 B a b +6 C \,a^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+24 b^{2} \left (\left (A +\frac {C}{2}\right ) b^{2}+4 B a b +6 C \,a^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+48 x \left (2 A \,b^{3}+3 B a \,b^{2}+a^{2} b \left (A +2 C \right )+\frac {B \,a^{3}}{4}\right ) a d \cos \left (2 d x +2 c \right )+6 \left (4 A \,a^{3} b +B \,a^{4}+4 B \,b^{4}+16 C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+\left (72 A \,a^{2} b^{2}+48 B \,a^{3} b +11 \left (A +\frac {12 C}{11}\right ) a^{4}\right ) \sin \left (3 d x +3 c \right )+3 \left (4 A \,a^{3} b +B \,a^{4}\right ) \sin \left (4 d x +4 c \right )+a^{4} A \sin \left (5 d x +5 c \right )+2 \left (12 C \,b^{4}+36 A \,a^{2} b^{2}+24 B \,a^{3} b +5 \left (A +\frac {6 C}{5}\right ) a^{4}\right ) \sin \left (d x +c \right )+48 x \left (2 A \,b^{3}+3 B a \,b^{2}+a^{2} b \left (A +2 C \right )+\frac {B \,a^{3}}{4}\right ) a d}{24 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(350\) |
| risch | \(6 B \,a^{2} b^{2} x +2 a^{3} A b x +4 A a \,b^{3} x +4 C \,a^{3} b x -\frac {2 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{3} b}{d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} A \,a^{3} b}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{4}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{4}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{4}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{4}}{2 d}-\frac {i a^{4} A \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {i a^{4} A \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {i b^{3} \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-8 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 B b -8 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} B \,a^{4}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{2 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} B \,a^{4}}{8 d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,a^{2} b^{2}}{d}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a \,b^{3}}{d}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,a^{2} b^{2}}{d}-\frac {3 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {3 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a \,b^{3}}{d}+\frac {B \,a^{4} x}{2}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} A \,a^{3} b}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2} b^{2}}{d}+\frac {2 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3} b}{d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2} b^{2}}{d}\) | \(584\) |
| norman | \(\text {Expression too large to display}\) | \(1117\) |
Input:
int(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth od=_RETURNVERBOSE)
Output:
1/d*(1/3*a^4*A*(2+cos(d*x+c)^2)*sin(d*x+c)+B*a^4*(1/2*cos(d*x+c)*sin(d*x+c )+1/2*d*x+1/2*c)+a^4*C*sin(d*x+c)+4*A*a^3*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2 *d*x+1/2*c)+4*B*a^3*b*sin(d*x+c)+4*a^3*b*C*(d*x+c)+6*A*a^2*b^2*sin(d*x+c)+ 6*B*a^2*b^2*(d*x+c)+6*C*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+4*a*A*b^3*(d*x+c )+4*B*a*b^3*ln(sec(d*x+c)+tan(d*x+c))+4*C*a*b^3*tan(d*x+c)+A*b^4*ln(sec(d* x+c)+tan(d*x+c))+B*tan(d*x+c)*b^4+C*b^4*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln( sec(d*x+c)+tan(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 262, normalized size of antiderivative = 0.86 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 \, {\left (B a^{4} + 4 \, {\left (A + 2 \, C\right )} a^{3} b + 12 \, B a^{2} b^{2} + 8 \, A a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, C a^{2} b^{2} + 8 \, B a b^{3} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 3 \, C b^{4} + 3 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 12 \, B a^{3} b + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} + 6 \, {\left (4 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
Output:
1/12*(6*(B*a^4 + 4*(A + 2*C)*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*d*x*cos(d*x + c)^2 + 3*(12*C*a^2*b^2 + 8*B*a*b^3 + (2*A + C)*b^4)*cos(d*x + c)^2*log( sin(d*x + c) + 1) - 3*(12*C*a^2*b^2 + 8*B*a*b^3 + (2*A + C)*b^4)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^4*cos(d*x + c)^4 + 3*C*b^4 + 3*(B* a^4 + 4*A*a^3*b)*cos(d*x + c)^3 + 2*((2*A + 3*C)*a^4 + 12*B*a^3*b + 18*A*a ^2*b^2)*cos(d*x + c)^2 + 6*(4*C*a*b^3 + B*b^4)*cos(d*x + c))*sin(d*x + c)) /(d*cos(d*x + c)^2)
Timed out. \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*(a+b*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)** 2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.03 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 48 \, {\left (d x + c\right )} C a^{3} b - 72 \, {\left (d x + c\right )} B a^{2} b^{2} - 48 \, {\left (d x + c\right )} A a b^{3} + 3 \, C b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{4} \sin \left (d x + c\right ) - 48 \, B a^{3} b \sin \left (d x + c\right ) - 72 \, A a^{2} b^{2} \sin \left (d x + c\right ) - 48 \, C a b^{3} \tan \left (d x + c\right ) - 12 \, B b^{4} \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
Output:
-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 3*(2*d*x + 2*c + sin(2* d*x + 2*c))*B*a^4 - 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3*b - 48*(d*x + c)*C*a^3*b - 72*(d*x + c)*B*a^2*b^2 - 48*(d*x + c)*A*a*b^3 + 3*C*b^4*(2* sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 36*C*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 24*B*a*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*A*b^4*(log( sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 12*C*a^4*sin(d*x + c) - 48*B* a^3*b*sin(d*x + c) - 72*A*a^2*b^2*sin(d*x + c) - 48*C*a*b^3*tan(d*x + c) - 12*B*b^4*tan(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 543, normalized size of antiderivative = 1.79 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
Output:
1/6*(3*(B*a^4 + 4*A*a^3*b + 8*C*a^3*b + 12*B*a^2*b^2 + 8*A*a*b^3)*(d*x + c ) + 3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*A*b^4 + C*b^4)*log(abs(tan(1/2*d*x + 1 /2*c) + 1)) - 3*(12*C*a^2*b^2 + 8*B*a*b^3 + 2*A*b^4 + C*b^4)*log(abs(tan(1 /2*d*x + 1/2*c) - 1)) - 6*(8*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^4*tan( 1/2*d*x + 1/2*c)^3 - C*b^4*tan(1/2*d*x + 1/2*c)^3 - 8*C*a*b^3*tan(1/2*d*x + 1/2*c) - 2*B*b^4*tan(1/2*d*x + 1/2*c) - C*b^4*tan(1/2*d*x + 1/2*c))/(tan (1/2*d*x + 1/2*c)^2 - 1)^2 + 2*(6*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 3*B*a^4*t an(1/2*d*x + 1/2*c)^5 + 6*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 12*A*a^3*b*tan(1/ 2*d*x + 1/2*c)^5 + 24*B*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 36*A*a^2*b^2*tan(1/ 2*d*x + 1/2*c)^5 + 4*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 48*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 72*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 3*B*a^4*tan(1/2*d*x + 1/2*c) + 6*C*a^4*tan(1/2*d*x + 1/2*c) + 12*A*a^3*b*tan(1/2*d*x + 1/2*c) + 24*B*a^3* b*tan(1/2*d*x + 1/2*c) + 36*A*a^2*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 16.64 (sec) , antiderivative size = 4839, normalized size of antiderivative = 15.97 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
int(cos(c + d*x)^3*(a + b/cos(c + d*x))^4*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
- (tan(c/2 + (d*x)/2)^3*((8*A*a^4)/3 + 2*B*a^4 - 4*B*b^4 - 4*C*b^4 + 8*A*a ^3*b - 16*C*a*b^3) + tan(c/2 + (d*x)/2)^7*((8*A*a^4)/3 - 2*B*a^4 + 4*B*b^4 - 4*C*b^4 - 8*A*a^3*b + 16*C*a*b^3) - tan(c/2 + (d*x)/2)^9*(2*A*a^4 - B*a ^4 - 2*B*b^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 - 4*A*a^3*b + 8*B*a^3*b - 8* C*a*b^3) - tan(c/2 + (d*x)/2)*(2*A*a^4 + B*a^4 + 2*B*b^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 + 4*A*a^3*b + 8*B*a^3*b + 8*C*a*b^3) + tan(c/2 + (d*x)/2)^ 5*(4*C*a^4 - (4*A*a^4)/3 - 6*C*b^4 + 24*A*a^2*b^2 + 16*B*a^3*b))/(d*(tan(c /2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2)^4 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/ 2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (atan((((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2 + 4*B*a*b^3)*(32*A*b^4 + 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*B*a*b^3 + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*B^2*a^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 512*B^2*a^2*b^6 + 1152*B^2*a^4*b^4 + 192*B^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 256*A*B*a*b^7 + 64*A*B*a^7*b + 128*B*C*a*b^7 + 128*B*C*a^7*b + 1536*A*B*a^3*b^5 + 896*A*B*a^5*b^3 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2 + 1536*B*C*a^3*b^5 + 1536*B*C*a^5*b^3))*(A*b^4 + (C*b^4) /2 + 6*C*a^2*b^2 + 4*B*a*b^3)*1i - ((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2 + 4*B *a*b^3)*(32*A*b^4 + 16*B*a^4 + 16*C*b^4 + 192*B*a^2*b^2 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*B*a*b^3 + 128*C*a^3*b) - tan(c/2 + (d*x)...
Time = 0.16 (sec) , antiderivative size = 625, normalized size of antiderivative = 2.06 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
int(cos(d*x+c)^3*(a+b*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(15*cos(c + d*x)*sin(c + d*x)**3*a**4*b - 15*cos(c + d*x)*sin(c + d*x)*a** 4*b - 24*cos(c + d*x)*sin(c + d*x)*a*b**3*c - 6*cos(c + d*x)*sin(c + d*x)* b**5 - 36*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**2*c - 30*log(t an((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**4 - 3*log(tan((c + d*x)/2) - 1)* sin(c + d*x)**2*b**4*c + 36*log(tan((c + d*x)/2) - 1)*a**2*b**2*c + 30*log (tan((c + d*x)/2) - 1)*a*b**4 + 3*log(tan((c + d*x)/2) - 1)*b**4*c + 36*lo g(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2*c + 30*log(tan((c + d*x) /2) + 1)*sin(c + d*x)**2*a*b**4 + 3*log(tan((c + d*x)/2) + 1)*sin(c + d*x) **2*b**4*c - 36*log(tan((c + d*x)/2) + 1)*a**2*b**2*c - 30*log(tan((c + d* x)/2) + 1)*a*b**4 - 3*log(tan((c + d*x)/2) + 1)*b**4*c - 2*sin(c + d*x)**5 *a**5 + 8*sin(c + d*x)**3*a**5 + 6*sin(c + d*x)**3*a**4*c + 60*sin(c + d*x )**3*a**3*b**2 + 15*sin(c + d*x)**2*a**4*b*c + 15*sin(c + d*x)**2*a**4*b*d *x + 24*sin(c + d*x)**2*a**3*b*c**2 + 24*sin(c + d*x)**2*a**3*b*c*d*x + 60 *sin(c + d*x)**2*a**2*b**3*c + 60*sin(c + d*x)**2*a**2*b**3*d*x - 6*sin(c + d*x)*a**5 - 6*sin(c + d*x)*a**4*c - 60*sin(c + d*x)*a**3*b**2 - 3*sin(c + d*x)*b**4*c - 15*a**4*b*c - 15*a**4*b*d*x - 24*a**3*b*c**2 - 24*a**3*b*c *d*x - 60*a**2*b**3*c - 60*a**2*b**3*d*x)/(6*d*(sin(c + d*x)**2 - 1))