Integrand size = 41, antiderivative size = 149 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^2 (7 A+8 B+12 C) x+\frac {a^2 (7 A+8 B+12 C) \sin (c+d x)}{6 d}+\frac {a^2 (7 A+8 B+12 C) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(A+2 B) \cos ^2(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{4 d} \] Output:
1/8*a^2*(7*A+8*B+12*C)*x+1/6*a^2*(7*A+8*B+12*C)*sin(d*x+c)/d+1/24*a^2*(7*A +8*B+12*C)*cos(d*x+c)*sin(d*x+c)/d+1/6*(A+2*B)*cos(d*x+c)^2*(a+a*sec(d*x+c ))^2*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d*x+c))^2*sin(d*x+c)/d
Time = 0.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.77 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \sin (c+d x) \left (8 (4 A+5 B+6 C)+3 (7 A+8 B+4 C) \cos (c+d x)+8 (2 A+B) \cos ^2(c+d x)+6 A \cos ^3(c+d x)+\frac {6 (7 A+8 B+12 C) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )}{\sqrt {\sin ^2(c+d x)}}\right )}{24 d} \] Input:
Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Se c[c + d*x]^2),x]
Output:
(a^2*Sin[c + d*x]*(8*(4*A + 5*B + 6*C) + 3*(7*A + 8*B + 4*C)*Cos[c + d*x] + 8*(2*A + B)*Cos[c + d*x]^2 + 6*A*Cos[c + d*x]^3 + (6*(7*A + 8*B + 12*C)* ArcSin[Sqrt[Sin[(c + d*x)/2]^2]])/Sqrt[Sin[c + d*x]^2]))/(24*d)
Time = 0.88 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.95, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 4574, 3042, 4501, 3042, 4275, 3042, 3117, 4533, 24}
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.
| \(\displaystyle \int \cos ^4(c+d x) (a \sec (c+d x)+a)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int \cos ^3(c+d x) (\sec (c+d x) a+a)^2 (2 a (A+2 B)+a (A+4 C) \sec (c+d x))dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a (A+2 B)+a (A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \frac {\frac {1}{3} a (7 A+8 B+12 C) \int \cos ^2(c+d x) (\sec (c+d x) a+a)^2dx+\frac {2 a (A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} a (7 A+8 B+12 C) \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {2 a (A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 4275 |
| \(\displaystyle \frac {\frac {1}{3} a (7 A+8 B+12 C) \left (2 a^2 \int \cos (c+d x)dx+\int \cos ^2(c+d x) \left (\sec ^2(c+d x) a^2+a^2\right )dx\right )+\frac {2 a (A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} a (7 A+8 B+12 C) \left (2 a^2 \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx\right )+\frac {2 a (A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {\frac {1}{3} a (7 A+8 B+12 C) \left (\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {2 a^2 \sin (c+d x)}{d}\right )+\frac {2 a (A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {\frac {1}{3} a (7 A+8 B+12 C) \left (\frac {3 a^2 \int 1dx}{2}+\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}\right )+\frac {2 a (A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{3} a (7 A+8 B+12 C) \left (\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 x}{2}\right )+\frac {2 a (A+2 B) \sin (c+d x) \cos ^2(c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^2}{4 d}\) |
Input:
Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(4*d) + ((2*a*(A + 2*B)*Cos[c + d*x]^2*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + (a*(7*A + 8*B + 12*C)*((3*a^2*x)/2 + (2*a^2*Sin[c + d*x])/d + (a^2*Cos[c + d*x]*Sin [c + d*x])/(2*d)))/3)/(4*a)
Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[2*a*(b/d) Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), 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[(a*A*m - b*B*n)/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x] , x] /; FreeQ[{a, b, d, e, f, A, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a ^2 - b^2, 0] && EqQ[m + n + 1, 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
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/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 0.66 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.56
| method | result | size |
| parallelrisch | \(\frac {\left (3 \left (A +B +\frac {C}{2}\right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {B}{2}\right ) \sin \left (3 d x +3 c \right )+\frac {3 A \sin \left (4 d x +4 c \right )}{16}+3 \left (3 A +\frac {7 B}{2}+4 C \right ) \sin \left (d x +c \right )+\frac {21 x d \left (\frac {8 B}{7}+A +\frac {12 C}{7}\right )}{4}\right ) a^{2}}{6 d}\) | \(84\) |
| risch | \(\frac {7 a^{2} A x}{8}+a^{2} B x +\frac {3 a^{2} x C}{2}+\frac {3 \sin \left (d x +c \right ) a^{2} A}{2 d}+\frac {7 a^{2} B \sin \left (d x +c \right )}{4 d}+\frac {2 \sin \left (d x +c \right ) C \,a^{2}}{d}+\frac {a^{2} A \sin \left (4 d x +4 c \right )}{32 d}+\frac {a^{2} A \sin \left (3 d x +3 c \right )}{6 d}+\frac {B \,a^{2} \sin \left (3 d x +3 c \right )}{12 d}+\frac {a^{2} A \sin \left (2 d x +2 c \right )}{2 d}+\frac {B \,a^{2} \sin \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{4 d}\) | \(175\) |
| derivativedivides | \(\frac {a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )+\frac {2 a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C \,a^{2} \sin \left (d x +c \right )+a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(203\) |
| default | \(\frac {a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,a^{2} \sin \left (d x +c \right )+C \,a^{2} \left (d x +c \right )+\frac {2 a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C \,a^{2} \sin \left (d x +c \right )+a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(203\) |
| norman | \(\frac {\frac {a^{2} \left (3 A -8 B -4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}-\frac {a^{2} \left (7 A +8 B +12 C \right ) x}{8}+\frac {a^{2} \left (7 A +8 B +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{6 d}+\frac {a^{2} \left (7 A +8 B +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}-\frac {a^{2} \left (7 A +8 B +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {3 a^{2} \left (7 A +8 B +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {3 a^{2} \left (7 A +8 B +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}-\frac {3 a^{2} \left (7 A +8 B +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}-\frac {3 a^{2} \left (7 A +8 B +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {a^{2} \left (7 A +8 B +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a^{2} \left (7 A +8 B +12 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}-\frac {a^{2} \left (25 A +24 B +20 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a^{2} \left (53 A -104 B -156 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {a^{2} \left (71 A +40 B +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {a^{2} \left (85 A +56 B +132 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\) | \(438\) |
Input:
int(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,meth od=_RETURNVERBOSE)
Output:
1/6*(3*(A+B+1/2*C)*sin(2*d*x+2*c)+(A+1/2*B)*sin(3*d*x+3*c)+3/16*A*sin(4*d* x+4*c)+3*(3*A+7/2*B+4*C)*sin(d*x+c)+21/4*x*d*(8/7*B+A+12/7*C))*a^2/d
Time = 0.08 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.66 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (7 \, A + 8 \, B + 12 \, C\right )} a^{2} d x + {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, A + 8 \, B + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (4 \, A + 5 \, B + 6 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
Output:
1/24*(3*(7*A + 8*B + 12*C)*a^2*d*x + (6*A*a^2*cos(d*x + c)^3 + 8*(2*A + B) *a^2*cos(d*x + c)^2 + 3*(7*A + 8*B + 4*C)*a^2*cos(d*x + c) + 8*(4*A + 5*B + 6*C)*a^2)*sin(d*x + c))/d
Timed out. \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)** 2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.28 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 96 \, {\left (d x + c\right )} C a^{2} - 96 \, B a^{2} \sin \left (d x + c\right ) - 192 \, C a^{2} \sin \left (d x + c\right )}{96 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
Output:
-1/96*(64*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 3*(12*d*x + 12*c + sin (4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 - 9 6*(d*x + c)*C*a^2 - 96*B*a^2*sin(d*x + c) - 192*C*a^2*sin(d*x + c))/d
Time = 0.24 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.66 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (7 \, A a^{2} + 8 \, B a^{2} + 12 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (21 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 77 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 88 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 132 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 136 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 156 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 75 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \] Input:
integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
Output:
1/24*(3*(7*A*a^2 + 8*B*a^2 + 12*C*a^2)*(d*x + c) + 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 36*C*a^2*tan(1/2*d*x + 1/2 *c)^7 + 77*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 88*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 132*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 136 *B*a^2*tan(1/2*d*x + 1/2*c)^3 + 156*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 75*A*a^ 2*tan(1/2*d*x + 1/2*c) + 72*B*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1/2* d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d
Time = 12.22 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.17 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {7\,A\,a^2\,x}{8}+B\,a^2\,x+\frac {3\,C\,a^2\,x}{2}+\frac {3\,A\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {7\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {A\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \] Input:
int(cos(c + d*x)^4*(a + a/cos(c + d*x))^2*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
Output:
(7*A*a^2*x)/8 + B*a^2*x + (3*C*a^2*x)/2 + (3*A*a^2*sin(c + d*x))/(2*d) + ( 7*B*a^2*sin(c + d*x))/(4*d) + (2*C*a^2*sin(c + d*x))/d + (A*a^2*sin(2*c + 2*d*x))/(2*d) + (A*a^2*sin(3*c + 3*d*x))/(6*d) + (A*a^2*sin(4*c + 4*d*x))/ (32*d) + (B*a^2*sin(2*c + 2*d*x))/(2*d) + (B*a^2*sin(3*c + 3*d*x))/(12*d) + (C*a^2*sin(2*c + 2*d*x))/(4*d)
Time = 0.16 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.91 \[ \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{2} \left (-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +27 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -16 \sin \left (d x +c \right )^{3} a -8 \sin \left (d x +c \right )^{3} b +48 \sin \left (d x +c \right ) a +48 \sin \left (d x +c \right ) b +48 \sin \left (d x +c \right ) c +21 a d x +24 b d x +36 c d x \right )}{24 d} \] Input:
int(cos(d*x+c)^4*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
Output:
(a**2*( - 6*cos(c + d*x)*sin(c + d*x)**3*a + 27*cos(c + d*x)*sin(c + d*x)* a + 24*cos(c + d*x)*sin(c + d*x)*b + 12*cos(c + d*x)*sin(c + d*x)*c - 16*s in(c + d*x)**3*a - 8*sin(c + d*x)**3*b + 48*sin(c + d*x)*a + 48*sin(c + d* x)*b + 48*sin(c + d*x)*c + 21*a*d*x + 24*b*d*x + 36*c*d*x))/(24*d)