Integrand size = 43, antiderivative size = 193 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 a (21 A+18 B+16 C) \tan (c+d x)}{45 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (9 B+C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt {a+a \sec (c+d x)}}-\frac {4 (21 A+18 B+16 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{315 d}+\frac {2 C \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{9 d}+\frac {2 (21 A+18 B+16 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{105 a d} \]
2/105*(21*A+18*B+16*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+2/45*a*(21*A+ 18*B+16*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/63*a*(9*B+C)*sec(d*x+c)^3 *tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-4/315*(21*A+18*B+16*C)*(a+a*sec(d*x+c ))^(1/2)*tan(d*x+c)/d+2/9*C*sec(d*x+c)^3*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c) /d
Time = 1.75 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.78 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (189 A+162 B+214 C+2 (63 A+99 B+88 C) \cos (c+d x)+11 (21 A+18 B+16 C) \cos (2 (c+d x))+42 A \cos (3 (c+d x))+36 B \cos (3 (c+d x))+32 C \cos (3 (c+d x))+42 A \cos (4 (c+d x))+36 B \cos (4 (c+d x))+32 C \cos (4 (c+d x))) \sec ^4(c+d x) \tan (c+d x)}{315 d \sqrt {a (1+\sec (c+d x))}} \]
(a*(189*A + 162*B + 214*C + 2*(63*A + 99*B + 88*C)*Cos[c + d*x] + 11*(21*A + 18*B + 16*C)*Cos[2*(c + d*x)] + 42*A*Cos[3*(c + d*x)] + 36*B*Cos[3*(c + d*x)] + 32*C*Cos[3*(c + d*x)] + 42*A*Cos[4*(c + d*x)] + 36*B*Cos[4*(c + d *x)] + 32*C*Cos[4*(c + d*x)])*Sec[c + d*x]^4*Tan[c + d*x])/(315*d*Sqrt[a*( 1 + Sec[c + d*x])])
Time = 1.09 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4576, 27, 3042, 4504, 3042, 4287, 27, 3042, 4489, 3042, 4279}
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.
| \(\displaystyle \int \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (3 A+2 C)+a (9 B+C) \sec (c+d x))dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a} (3 a (3 A+2 C)+a (9 B+C) \sec (c+d x))dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (3 a (3 A+2 C)+a (9 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {3}{7} a (21 A+18 B+16 C) \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a^2 (9 B+C) \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}}{9 a}+\frac {2 C \tan (c+d x) \sec ^3(c+d x) \sqrt {a \sec (c+d x)+a}}{9 d}\) |
(2*C*Sec[c + d*x]^3*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(9*d) + ((2*a^2 *(9*B + C)*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + ( 3*a*(21*A + 18*B + 16*C)*((2*(a + a*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*a *d) + ((14*a^2*Tan[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (4*a*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(3*d))/(5*a)))/7)/(9*a)
3.5.83.3.1 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[csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2*b*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]])), x] /; Free Q[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-Cot[e + f*x])*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b*( m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B , e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b *(m + 1), 0] && !LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[-2*b*B*C ot[e + f*x]*((d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[(A*b*(2*n + 1) + 2*a*B*n)/(b*(2*n + 1)) Int[Sqrt[a + b*Csc[e + f* x]]*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ [A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] && !LtQ[n, 0]
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[(-C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Cs c[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Cs c[e + f*x])^m*(d*Csc[e + f*x])^n*Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b* B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m , n}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 0]
Time = 0.85 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.84
| method | result | size |
| default | \(\frac {2 \left (168 A \cos \left (d x +c \right )^{4}+144 B \cos \left (d x +c \right )^{4}+128 C \cos \left (d x +c \right )^{4}+84 A \cos \left (d x +c \right )^{3}+72 B \cos \left (d x +c \right )^{3}+64 C \cos \left (d x +c \right )^{3}+63 A \cos \left (d x +c \right )^{2}+54 B \cos \left (d x +c \right )^{2}+48 C \cos \left (d x +c \right )^{2}+45 B \cos \left (d x +c \right )+40 C \cos \left (d x +c \right )+35 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(163\) |
| parts | \(\frac {2 A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (8 \sin \left (d x +c \right )+4 \tan \left (d x +c \right )+3 \sec \left (d x +c \right ) \tan \left (d x +c \right )\right )}{15 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \left (16 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{35 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \left (128 \cos \left (d x +c \right )^{4}+64 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+40 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}\) | \(215\) |
int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, method=_RETURNVERBOSE)
2/315/d*(168*A*cos(d*x+c)^4+144*B*cos(d*x+c)^4+128*C*cos(d*x+c)^4+84*A*cos (d*x+c)^3+72*B*cos(d*x+c)^3+64*C*cos(d*x+c)^3+63*A*cos(d*x+c)^2+54*B*cos(d *x+c)^2+48*C*cos(d*x+c)^2+45*B*cos(d*x+c)+40*C*cos(d*x+c)+35*C)*(a*(1+sec( d*x+c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^3
Time = 0.26 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.68 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (8 \, {\left (21 \, A + 18 \, B + 16 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (21 \, A + 18 \, B + 16 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (21 \, A + 18 \, B + 16 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (9 \, B + 8 \, C\right )} \cos \left (d x + c\right ) + 35 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="fricas")
2/315*(8*(21*A + 18*B + 16*C)*cos(d*x + c)^4 + 4*(21*A + 18*B + 16*C)*cos( d*x + c)^3 + 3*(21*A + 18*B + 16*C)*cos(d*x + c)^2 + 5*(9*B + 8*C)*cos(d*x + c) + 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos( d*x + c)^5 + d*cos(d*x + c)^4)
\[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}\, dx \]
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x) + C*sec(c + d*x)** 2)*sec(c + d*x)**3, x)
\[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="maxima")
8/315*(315*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(((A + 2*B)*d*cos(2*d*x + 2*c)^4 + (A + 2*B)*d*sin(2*d*x + 2*c)^ 4 + 4*(A + 2*B)*d*cos(2*d*x + 2*c)^3 + 6*(A + 2*B)*d*cos(2*d*x + 2*c)^2 + 4*(A + 2*B)*d*cos(2*d*x + 2*c) + 2*((A + 2*B)*d*cos(2*d*x + 2*c)^2 + 2*(A + 2*B)*d*cos(2*d*x + 2*c) + (A + 2*B)*d)*sin(2*d*x + 2*c)^2 + (A + 2*B)*d) *integrate((((cos(12*d*x + 12*c)*cos(2*d*x + 2*c) + 5*cos(10*d*x + 10*c)*c os(2*d*x + 2*c) + 10*cos(8*d*x + 8*c)*cos(2*d*x + 2*c) + 10*cos(6*d*x + 6* c)*cos(2*d*x + 2*c) + 5*cos(4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2* c)^2 + sin(12*d*x + 12*c)*sin(2*d*x + 2*c) + 5*sin(10*d*x + 10*c)*sin(2*d* x + 2*c) + 10*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 10*sin(6*d*x + 6*c)*sin( 2*d*x + 2*c) + 5*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d*x + 2*c)^2)*c os(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2*d*x + 2*c)*si n(12*d*x + 12*c) + 5*cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 10*cos(2*d*x + 2*c)*sin(8*d*x + 8*c) + 10*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 5*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(12*d*x + 12*c)*sin(2*d*x + 2*c) - 5*cos(10* d*x + 10*c)*sin(2*d*x + 2*c) - 10*cos(8*d*x + 8*c)*sin(2*d*x + 2*c) - 10*c os(6*d*x + 6*c)*sin(2*d*x + 2*c) - 5*cos(4*d*x + 4*c)*sin(2*d*x + 2*c))*si n(7/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*cos(1/2*arctan2(sin(2* d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x + 2*c)*sin(12*d*x + 12*c) + 5*cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 10*cos(2*d*x + 2*c)*sin(8*d*...
\[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3} \,d x } \]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="giac")
Time = 29.31 (sec) , antiderivative size = 600, normalized size of antiderivative = 3.11 \[ \int \sec ^3(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {A\,8{}\mathrm {i}}{3\,d}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (168\,A+144\,B+128\,C\right )\,1{}\mathrm {i}}{315\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {A\,8{}\mathrm {i}}{9\,d}-\frac {\left (16\,A+16\,B+32\,C\right )\,1{}\mathrm {i}}{9\,d}+\frac {\left (8\,A+16\,B\right )\,1{}\mathrm {i}}{9\,d}\right )-\frac {A\,8{}\mathrm {i}}{9\,d}+\frac {\left (16\,A+16\,B+32\,C\right )\,1{}\mathrm {i}}{9\,d}-\frac {\left (8\,A+16\,B\right )\,1{}\mathrm {i}}{9\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^4}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (\frac {A\,8{}\mathrm {i}}{7\,d}-\frac {C\,32{}\mathrm {i}}{7\,d}-\frac {\left (72\,A+144\,B+288\,C\right )\,1{}\mathrm {i}}{63\,d}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {C\,32{}\mathrm {i}}{63\,d}-\frac {\left (72\,A+144\,B\right )\,1{}\mathrm {i}}{63\,d}+\frac {\left (72\,A+288\,C\right )\,1{}\mathrm {i}}{63\,d}\right )\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\left (\frac {\left (168\,A+336\,B\right )\,1{}\mathrm {i}}{105\,d}+\frac {\left (48\,B-32\,C\right )\,1{}\mathrm {i}}{105\,d}\right )-\frac {A\,8{}\mathrm {i}}{5\,d}+\frac {\left (336\,B+672\,C\right )\,1{}\mathrm {i}}{105\,d}\right )}{\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {a+\frac {a}{\frac {{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}}{2}}}\,\left (336\,A+288\,B+256\,C\right )\,1{}\mathrm {i}}{315\,d\,\left ({\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}+1\right )} \]
((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*((A*8i)/(3* d) - (exp(c*1i + d*x*1i)*(168*A + 144*B + 128*C)*1i)/(315*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*((A*8i)/(9*d) - ((16*A + 16*B + 32*C)*1i)/(9*d) + ((8*A + 16*B)*1i)/(9*d)) - (A*8i)/(9*d) + ((16 *A + 16*B + 32*C)*1i)/(9*d) - ((8*A + 16*B)*1i)/(9*d)))/((exp(c*1i + d*x*1 i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) + ((a + a/(exp(- c*1i - d*x*1i)/2 + ex p(c*1i + d*x*1i)/2))^(1/2)*((A*8i)/(7*d) - (C*32i)/(7*d) - ((72*A + 144*B + 288*C)*1i)/(63*d) + exp(c*1i + d*x*1i)*((C*32i)/(63*d) - ((72*A + 144*B) *1i)/(63*d) + ((72*A + 288*C)*1i)/(63*d))))/((exp(c*1i + d*x*1i) + 1)*(exp (c*2i + d*x*2i) + 1)^3) + ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x *1i)/2))^(1/2)*(exp(c*1i + d*x*1i)*(((168*A + 336*B)*1i)/(105*d) + ((48*B - 32*C)*1i)/(105*d)) - (A*8i)/(5*d) + ((336*B + 672*C)*1i)/(105*d)))/((exp (c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - (exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(336*A + 288*B + 256*C)*1i)/(315*d*(exp(c*1i + d*x*1i) + 1))