Integrand size = 41, antiderivative size = 204 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{2} a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) x+\frac {b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {(3 A b+2 a B) (a+b \sec (c+d x))^2 \sin (c+d x)}{2 d}+\frac {A \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{2 d}-\frac {b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \tan (c+d x)}{2 d}-\frac {b^2 (4 A b+2 a B-b C) \sec (c+d x) \tan (c+d x)}{2 d} \] Output:
1/2*a*(6*A*b^2+6*B*a*b+a^2*(A+2*C))*x+1/2*b*(2*A*b^2+6*B*a*b+6*C*a^2+C*b^2 )*arctanh(sin(d*x+c))/d+1/2*(3*A*b+2*B*a)*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+ 1/2*A*cos(d*x+c)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d-1/2*b*(9*A*a*b+4*B*a^2-2* B*b^2-6*C*a*b)*tan(d*x+c)/d-1/2*b^2*(4*A*b+2*B*a-C*b)*sec(d*x+c)*tan(d*x+c )/d
Time = 4.86 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.57 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) (c+d x)-2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {b^3 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b^2 (b B+3 a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {b^3 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 b^2 (b B+3 a C) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 a^2 (3 A b+a B) \sin (c+d x)+a^3 A \sin (2 (c+d x))}{4 d} \] Input:
Integrate[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
(2*a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*(c + d*x) - 2*b*(2*A*b^2 + 6*a*b* B + 6*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*b*(2*A*b ^2 + 6*a*b*B + 6*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^3*C)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (4*b^2*(b*B + 3*a*C)*Si n[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) - (b^3*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*b^2*(b*B + 3*a*C)*Sin[(c + d*x)/2])/(Co s[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*a^2*(3*A*b + a*B)*Sin[c + d*x] + a^ 3*A*Sin[2*(c + d*x)])/(4*d)
Time = 0.90 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {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.
| \(\displaystyle \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (-2 b (A-C) \sec ^2(c+d x)+(2 b B+a (A+2 C)) \sec (c+d x)+3 A b+2 a B\right )dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (-2 b (A-C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(2 b B+a (A+2 C)) \csc \left (c+d x+\frac {\pi }{2}\right )+3 A b+2 a B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{2} \left (\int (a+b \sec (c+d x)) \left ((A+2 C) a^2+6 b B a+6 A b^2-2 b (4 A b-C b+2 a B) \sec ^2(c+d x)+b (2 b B-a (A-4 C)) \sec (c+d x)\right )dx+\frac {(2 a B+3 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{2} \left (\int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left ((A+2 C) a^2+6 b B a+6 A b^2-2 b (4 A b-C b+2 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b (2 b B-a (A-4 C)) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {(2 a B+3 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 4536 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \left (-2 b \left (4 B a^2+9 A b a-6 b C a-2 b^2 B\right ) \sec ^2(c+d x)+2 b \left (6 C a^2+6 b B a+2 A b^2+b^2 C\right ) \sec (c+d x)+2 a \left ((A+2 C) a^2+6 b B a+6 A b^2\right )\right )dx-\frac {b^2 \tan (c+d x) \sec (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{2} \left (\frac {2 b \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 b \tan (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{d}+2 a x \left (a^2 (A+2 C)+6 a b B+6 A b^2\right )\right )-\frac {b^2 \tan (c+d x) \sec (c+d x) (2 a B+4 A b-b C)}{d}+\frac {(2 a B+3 A b) \sin (c+d x) (a+b \sec (c+d x))^2}{d}\right )+\frac {A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{2 d}\) |
Input:
Int[Cos[c + d*x]^2*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(2*d) + (((3*A*b + 2* a*B)*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/d - (b^2*(4*A*b + 2*a*B - b*C)*S ec[c + d*x]*Tan[c + d*x])/d + (2*a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*x + (2*b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/d - (2* b*(9*a*A*b + 4*a^2*B - 2*b^2*B - 6*a*b*C)*Tan[c + d*x])/d)/2)/2
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 = 1.56 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )+3 A \,a^{2} b \sin \left (d x +c \right )+3 B \,a^{2} b \left (d x +c \right )+3 a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a A \,b^{2} \left (d x +c \right )+3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C a \,b^{2} \tan \left (d x +c \right )+A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{3} \tan \left (d x +c \right )+C \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(212\) |
| default | \(\frac {a^{3} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{3} \sin \left (d x +c \right )+a^{3} C \left (d x +c \right )+3 A \,a^{2} b \sin \left (d x +c \right )+3 B \,a^{2} b \left (d x +c \right )+3 a^{2} b C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a A \,b^{2} \left (d x +c \right )+3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 C a \,b^{2} \tan \left (d x +c \right )+A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,b^{3} \tan \left (d x +c \right )+C \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(212\) |
| parallelrisch | \(\frac {-8 b \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b^{2}+3 B a b +3 C \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+8 b \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b^{2}+3 B a b +3 C \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 x a \left (6 A \,b^{2}+6 B a b +a^{2} \left (A +2 C \right )\right ) d \cos \left (2 d x +2 c \right )+2 \left (a^{3} A +4 B \,b^{3}+12 C a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+4 \left (3 A \,a^{2} b +B \,a^{3}\right ) \sin \left (3 d x +3 c \right )+a^{3} A \sin \left (4 d x +4 c \right )+4 \left (3 A \,a^{2} b +B \,a^{3}+2 C \,b^{3}\right ) \sin \left (d x +c \right )+4 x a \left (6 A \,b^{2}+6 B a b +a^{2} \left (A +2 C \right )\right ) d}{8 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(270\) |
| risch | \(\frac {a^{3} A x}{2}+3 A a \,b^{2} x +3 B \,a^{2} b x +a^{3} x C -\frac {i a^{3} A \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {i a^{3} A \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {3 i A \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{2 d}-\frac {i b^{2} \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 B b -6 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {3 i A \,a^{2} b \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{3}}{2 d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a \,b^{2}}{d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,a^{2}}{d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a \,b^{2}}{d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,a^{2}}{d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(408\) |
| norman | \(\frac {\left (\frac {1}{2} a^{3} A +3 a A \,b^{2}+3 B \,a^{2} b +a^{3} C \right ) x +\left (-\frac {1}{2} a^{3} A -3 a A \,b^{2}-3 B \,a^{2} b -a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-\frac {1}{2} a^{3} A -3 a A \,b^{2}-3 B \,a^{2} b -a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (\frac {1}{2} a^{3} A +3 a A \,b^{2}+3 B \,a^{2} b +a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-a^{3} A -6 a A \,b^{2}-6 B \,a^{2} b -2 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-a^{3} A -6 a A \,b^{2}-6 B \,a^{2} b -2 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (2 a^{3} A +12 a A \,b^{2}+12 B \,a^{2} b +4 a^{3} C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (a^{3} A +6 A \,a^{2} b +2 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (5 a^{3} A -18 A \,a^{2} b -6 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {\left (a^{3} A -6 A \,a^{2} b -2 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}-\frac {2 \left (5 a^{3} A -6 A \,a^{2} b -2 B \,a^{3}-2 B \,b^{3}-6 C a \,b^{2}+C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {2 \left (5 a^{3} A +6 A \,a^{2} b +2 B \,a^{3}-2 B \,b^{3}-6 C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {\left (5 a^{3} A +18 A \,a^{2} b +6 B \,a^{3}+2 B \,b^{3}+6 C a \,b^{2}-C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {b \left (2 A \,b^{2}+6 B a b +6 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b \left (2 A \,b^{2}+6 B a b +6 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(704\) |
Input:
int(cos(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth od=_RETURNVERBOSE)
Output:
1/d*(a^3*A*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+B*a^3*sin(d*x+c)+a^3* C*(d*x+c)+3*A*a^2*b*sin(d*x+c)+3*B*a^2*b*(d*x+c)+3*a^2*b*C*ln(sec(d*x+c)+t an(d*x+c))+3*a*A*b^2*(d*x+c)+3*B*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+3*C*a*b^2 *tan(d*x+c)+A*b^3*ln(sec(d*x+c)+tan(d*x+c))+B*b^3*tan(d*x+c)+C*b^3*(1/2*se c(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.02 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} d x \cos \left (d x + c\right )^{2} + {\left (6 \, C a^{2} b + 6 \, B a b^{2} + {\left (2 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (6 \, C a^{2} b + 6 \, B a b^{2} + {\left (2 \, A + C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (A a^{3} \cos \left (d x + c\right )^{3} + C b^{3} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \] Input:
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
Output:
1/4*(2*((A + 2*C)*a^3 + 6*B*a^2*b + 6*A*a*b^2)*d*x*cos(d*x + c)^2 + (6*C*a ^2*b + 6*B*a*b^2 + (2*A + C)*b^3)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - ( 6*C*a^2*b + 6*B*a*b^2 + (2*A + C)*b^3)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(A*a^3*cos(d*x + c)^3 + C*b^3 + 2*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^ 2 + 2*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2)
Timed out. \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**2*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)** 2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 4 \, {\left (d x + c\right )} C a^{3} + 12 \, {\left (d x + c\right )} B a^{2} b + 12 \, {\left (d x + c\right )} A a b^{2} - C b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{3} \sin \left (d x + c\right ) + 12 \, A a^{2} b \sin \left (d x + c\right ) + 12 \, C a b^{2} \tan \left (d x + c\right ) + 4 \, B b^{3} \tan \left (d x + c\right )}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
Output:
1/4*((2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 + 4*(d*x + c)*C*a^3 + 12*(d*x + c)*B*a^2*b + 12*(d*x + c)*A*a*b^2 - C*b^3*(2*sin(d*x + c)/(sin(d*x + c)^ 2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*C*a^2*b*(log(s in(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*A*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B*a^3*sin(d*x + c) + 12*A*a^2*b*sin(d*x + c) + 12*C*a*b^2* tan(d*x + c) + 4*B*b^3*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 540 vs. \(2 (192) = 384\).
Time = 0.29 (sec) , antiderivative size = 540, normalized size of antiderivative = 2.65 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:
integrate(cos(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
Output:
1/2*((A*a^3 + 2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2)*(d*x + c) + (6*C*a^2*b + 6* B*a*b^2 + 2*A*b^3 + C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3 + C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(A* a^3*tan(1/2*d*x + 1/2*c)^7 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 6*A*a^2*b*ta n(1/2*d*x + 1/2*c)^7 + 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 2*B*b^3*tan(1/2* d*x + 1/2*c)^7 - C*b^3*tan(1/2*d*x + 1/2*c)^7 - 3*A*a^3*tan(1/2*d*x + 1/2* c)^5 + 2*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 2*B*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*C*b ^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3*tan(1 /2*d*x + 1/2*c)^3 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6*C*a*b^2*tan(1/2*d *x + 1/2*c)^3 - 2*B*b^3*tan(1/2*d*x + 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2 *c)^3 - A*a^3*tan(1/2*d*x + 1/2*c) - 2*B*a^3*tan(1/2*d*x + 1/2*c) - 6*A*a^ 2*b*tan(1/2*d*x + 1/2*c) - 6*C*a*b^2*tan(1/2*d*x + 1/2*c) - 2*B*b^3*tan(1/ 2*d*x + 1/2*c) - C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^ 2)/d
Time = 16.13 (sec) , antiderivative size = 3879, normalized size of antiderivative = 19.01 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:
int(cos(c + d*x)^2*(a + b/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
(a*atan(((a*(tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C ^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4* b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B *a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 1 92*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3) - (a*(A*a^2 + 6*A*b^2 + 2*C*a^2 + 6*B*a*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b ^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b)*1i)/2)*(A*a^2 + 6*A*b^2 + 2*C*a ^2 + 6*B*a*b))/2 + (a*(tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2 *a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 28 8*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a ^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3) + (a*(A*a^2 + 6*A*b^2 + 2*C*a^2 + 6*B*a*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b)*1i)/2)*(A*a^2 + 6*A*b ^2 + 2*C*a^2 + 6*B*a*b))/2)/(192*A^3*a*b^8 - 192*C^3*a^8*b + (a*(tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b ^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A *C*a^4*b^2 + 576*B*C*a^3*b^3) - (a*(A*a^2 + 6*A*b^2 + 2*C*a^2 + 6*B*a*b...
Time = 0.15 (sec) , antiderivative size = 535, normalized size of antiderivative = 2.62 \[ \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {-6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2} c -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a^{2} b c +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a^{2} b c +2 \sin \left (d x +c \right )^{2} a^{3} c^{2}-12 a^{2} b^{2} c +\sin \left (d x +c \right )^{2} a^{4} c +12 \sin \left (d x +c \right )^{2} a^{2} b^{2} c -2 a^{3} c d x -12 a^{2} b^{2} d x +\sin \left (d x +c \right )^{2} a^{4} d x +2 \sin \left (d x +c \right )^{2} a^{3} c d x -\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3} c +8 \sin \left (d x +c \right )^{3} a^{3} b -8 \sin \left (d x +c \right ) a^{3} b -\sin \left (d x +c \right ) b^{3} c +\cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a^{4}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a^{4}+8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3} c -8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{3}-8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a \,b^{3}-\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b^{3} c +6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a^{2} b c +8 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a \,b^{3}+\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b^{3} c -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a^{2} b c +12 \sin \left (d x +c \right )^{2} a^{2} b^{2} d x -a^{4} d x -a^{4} c -2 a^{3} c^{2}-2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b^{4}}{2 d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:
int(cos(d*x+c)^2*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(cos(c + d*x)*sin(c + d*x)**3*a**4 - cos(c + d*x)*sin(c + d*x)*a**4 - 6*co s(c + d*x)*sin(c + d*x)*a*b**2*c - 2*cos(c + d*x)*sin(c + d*x)*b**4 - 6*lo g(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b*c - 8*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b**3 - log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b* *3*c + 6*log(tan((c + d*x)/2) - 1)*a**2*b*c + 8*log(tan((c + d*x)/2) - 1)* a*b**3 + log(tan((c + d*x)/2) - 1)*b**3*c + 6*log(tan((c + d*x)/2) + 1)*si n(c + d*x)**2*a**2*b*c + 8*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b** 3 + log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3*c - 6*log(tan((c + d*x) /2) + 1)*a**2*b*c - 8*log(tan((c + d*x)/2) + 1)*a*b**3 - log(tan((c + d*x) /2) + 1)*b**3*c + 8*sin(c + d*x)**3*a**3*b + sin(c + d*x)**2*a**4*c + sin( c + d*x)**2*a**4*d*x + 2*sin(c + d*x)**2*a**3*c**2 + 2*sin(c + d*x)**2*a** 3*c*d*x + 12*sin(c + d*x)**2*a**2*b**2*c + 12*sin(c + d*x)**2*a**2*b**2*d* x - 8*sin(c + d*x)*a**3*b - sin(c + d*x)*b**3*c - a**4*c - a**4*d*x - 2*a* *3*c**2 - 2*a**3*c*d*x - 12*a**2*b**2*c - 12*a**2*b**2*d*x)/(2*d*(sin(c + d*x)**2 - 1))