Integrand size = 43, antiderivative size = 239 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 a (99 A+88 B+80 C) \tan (c+d x)}{495 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (99 A+88 B+80 C) \sec ^3(c+d x) \tan (c+d x)}{693 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a (11 B+C) \sec ^4(c+d x) \tan (c+d x)}{99 d \sqrt {a+a \sec (c+d x)}}-\frac {8 (99 A+88 B+80 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{3465 d}+\frac {2 C \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{11 d}+\frac {4 (99 A+88 B+80 C) (a+a \sec (c+d x))^{3/2} \tan (c+d x)}{1155 a d} \]
4/1155*(99*A+88*B+80*C)*(a+a*sec(d*x+c))^(3/2)*tan(d*x+c)/a/d+4/495*a*(99* A+88*B+80*C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/693*a*(99*A+88*B+80*C)* sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+2/99*a*(11*B+C)*sec(d*x+c )^4*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-8/3465*(99*A+88*B+80*C)*(a+a*sec(d *x+c))^(1/2)*tan(d*x+c)/d+2/11*C*sec(d*x+c)^4*(a+a*sec(d*x+c))^(1/2)*tan(d *x+c)/d
Time = 7.92 (sec) , antiderivative size = 400, normalized size of antiderivative = 1.67 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 A \left (35 \sqrt {1-\sec (c+d x)}-35 (1-\sec (c+d x))^{3/2}+21 (1-\sec (c+d x))^{5/2}-5 (1-\sec (c+d x))^{7/2}\right ) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{35 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))}+\frac {2 B \left (315 \sqrt {1-\sec (c+d x)}-420 (1-\sec (c+d x))^{3/2}+378 (1-\sec (c+d x))^{5/2}-180 (1-\sec (c+d x))^{7/2}+35 (1-\sec (c+d x))^{9/2}\right ) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{315 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))}+\frac {2 C \left (693 \sqrt {1-\sec (c+d x)}-1155 (1-\sec (c+d x))^{3/2}+1386 (1-\sec (c+d x))^{5/2}-990 (1-\sec (c+d x))^{7/2}+385 (1-\sec (c+d x))^{9/2}-63 (1-\sec (c+d x))^{11/2}\right ) \sqrt {a (1+\sec (c+d x))} \tan (c+d x)}{693 d \sqrt {1-\sec (c+d x)} (1+\sec (c+d x))} \]
(2*A*(35*Sqrt[1 - Sec[c + d*x]] - 35*(1 - Sec[c + d*x])^(3/2) + 21*(1 - Se c[c + d*x])^(5/2) - 5*(1 - Sec[c + d*x])^(7/2))*Sqrt[a*(1 + Sec[c + d*x])] *Tan[c + d*x])/(35*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])) + (2*B*(31 5*Sqrt[1 - Sec[c + d*x]] - 420*(1 - Sec[c + d*x])^(3/2) + 378*(1 - Sec[c + d*x])^(5/2) - 180*(1 - Sec[c + d*x])^(7/2) + 35*(1 - Sec[c + d*x])^(9/2)) *Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(315*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d*x])) + (2*C*(693*Sqrt[1 - Sec[c + d*x]] - 1155*(1 - Sec[c + d *x])^(3/2) + 1386*(1 - Sec[c + d*x])^(5/2) - 990*(1 - Sec[c + d*x])^(7/2) + 385*(1 - Sec[c + d*x])^(9/2) - 63*(1 - Sec[c + d*x])^(11/2))*Sqrt[a*(1 + Sec[c + d*x])]*Tan[c + d*x])/(693*d*Sqrt[1 - Sec[c + d*x]]*(1 + Sec[c + d *x]))
Time = 1.36 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4576, 27, 3042, 4504, 3042, 4290, 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.
| \(\displaystyle \int \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {a \csc \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 4576 |
| \(\displaystyle \frac {2 \int \frac {1}{2} \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a} (a (11 A+8 C)+a (11 B+C) \sec (c+d x))dx}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a} (a (11 A+8 C)+a (11 B+C) \sec (c+d x))dx}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (11 A+8 C)+a (11 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 4504 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \int \sec ^4(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 4290 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \int \sec ^3(c+d x) \sqrt {\sec (c+d x) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 4287 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \left (\frac {2 \int \frac {1}{2} \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \left (\frac {\int \sec (c+d x) (3 a-2 a \sec (c+d x)) \sqrt {\sec (c+d x) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \left (\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (3 a-2 a \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 4489 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \sec (c+d x) \sqrt {\sec (c+d x) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \left (\frac {\frac {7}{3} a \int \csc \left (c+d x+\frac {\pi }{2}\right ) \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
| \(\Big \downarrow \) 4279 |
| \(\displaystyle \frac {\frac {1}{9} a (99 A+88 B+80 C) \left (\frac {6}{7} \left (\frac {\frac {14 a^2 \tan (c+d x)}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {4 a \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d}}{5 a}+\frac {2 \tan (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 a d}\right )+\frac {2 a \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^2 (11 B+C) \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt {a \sec (c+d x)+a}}}{11 a}+\frac {2 C \tan (c+d x) \sec ^4(c+d x) \sqrt {a \sec (c+d x)+a}}{11 d}\) |
(2*C*Sec[c + d*x]^4*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(11*d) + ((2*a^ 2*(11*B + C)*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(99*A + 88*B + 80*C)*((2*a*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (6*((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)/(11*a)
3.5.82.3.1 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_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*b*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/( f*(2*n - 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[2*a*d*((n - 1)/(b*(2*n - 1))) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n - 1), x], x] /; Fre eQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && IntegerQ[2*n]
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 = 1.10 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.82
| method | result | size |
| default | \(\frac {2 \left (1584 A \cos \left (d x +c \right )^{5}+1408 B \cos \left (d x +c \right )^{5}+1280 C \cos \left (d x +c \right )^{5}+792 A \cos \left (d x +c \right )^{4}+704 B \cos \left (d x +c \right )^{4}+640 C \cos \left (d x +c \right )^{4}+594 A \cos \left (d x +c \right )^{3}+528 B \cos \left (d x +c \right )^{3}+480 C \cos \left (d x +c \right )^{3}+495 A \cos \left (d x +c \right )^{2}+440 B \cos \left (d x +c \right )^{2}+400 C \cos \left (d x +c \right )^{2}+385 B \cos \left (d x +c \right )+350 C \cos \left (d x +c \right )+315 C \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{3465 d \left (\cos \left (d x +c \right )+1\right )}\) | \(196\) |
| parts | \(\frac {2 A \left (16 \cos \left (d x +c \right )^{3}+8 \cos \left (d x +c \right )^{2}+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{2}}{35 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 B \left (128 \cos \left (d x +c \right )^{4}+64 \cos \left (d x +c \right )^{3}+48 \cos \left (d x +c \right )^{2}+40 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{3}}{315 d \left (\cos \left (d x +c \right )+1\right )}+\frac {2 C \left (256 \cos \left (d x +c \right )^{5}+128 \cos \left (d x +c \right )^{4}+96 \cos \left (d x +c \right )^{3}+80 \cos \left (d x +c \right )^{2}+70 \cos \left (d x +c \right )+63\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right ) \sec \left (d x +c \right )^{4}}{693 d \left (\cos \left (d x +c \right )+1\right )}\) | \(248\) |
int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1/2),x, method=_RETURNVERBOSE)
2/3465/d*(1584*A*cos(d*x+c)^5+1408*B*cos(d*x+c)^5+1280*C*cos(d*x+c)^5+792* A*cos(d*x+c)^4+704*B*cos(d*x+c)^4+640*C*cos(d*x+c)^4+594*A*cos(d*x+c)^3+52 8*B*cos(d*x+c)^3+480*C*cos(d*x+c)^3+495*A*cos(d*x+c)^2+440*B*cos(d*x+c)^2+ 400*C*cos(d*x+c)^2+385*B*cos(d*x+c)+350*C*cos(d*x+c)+315*C)*(a*(1+sec(d*x+ c)))^(1/2)/(cos(d*x+c)+1)*tan(d*x+c)*sec(d*x+c)^4
Time = 0.26 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.63 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (16 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{5} + 8 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{4} + 6 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{3} + 5 \, {\left (99 \, A + 88 \, B + 80 \, C\right )} \cos \left (d x + c\right )^{2} + 35 \, {\left (11 \, B + 10 \, C\right )} \cos \left (d x + c\right ) + 315 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3465 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="fricas")
2/3465*(16*(99*A + 88*B + 80*C)*cos(d*x + c)^5 + 8*(99*A + 88*B + 80*C)*co s(d*x + c)^4 + 6*(99*A + 88*B + 80*C)*cos(d*x + c)^3 + 5*(99*A + 88*B + 80 *C)*cos(d*x + c)^2 + 35*(11*B + 10*C)*cos(d*x + c) + 315*C)*sqrt((a*cos(d* x + c) + a)/cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^6 + d*cos(d*x + c)^ 5)
\[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \]
Integral(sqrt(a*(sec(c + d*x) + 1))*(A + B*sec(c + d*x) + C*sec(c + d*x)** 2)*sec(c + d*x)**4, x)
\[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4} \,d x } \]
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="maxima")
16/3465*(3465*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c ) + 1)^(3/4)*((A*d*cos(2*d*x + 2*c)^4 + A*d*sin(2*d*x + 2*c)^4 + 4*A*d*cos (2*d*x + 2*c)^3 + 6*A*d*cos(2*d*x + 2*c)^2 + 4*A*d*cos(2*d*x + 2*c) + 2*(A *d*cos(2*d*x + 2*c)^2 + 2*A*d*cos(2*d*x + 2*c) + A*d)*sin(2*d*x + 2*c)^2 + A*d)*integrate((((cos(14*d*x + 14*c)*cos(2*d*x + 2*c) + 6*cos(12*d*x + 12 *c)*cos(2*d*x + 2*c) + 15*cos(10*d*x + 10*c)*cos(2*d*x + 2*c) + 20*cos(8*d *x + 8*c)*cos(2*d*x + 2*c) + 15*cos(6*d*x + 6*c)*cos(2*d*x + 2*c) + 6*cos( 4*d*x + 4*c)*cos(2*d*x + 2*c) + cos(2*d*x + 2*c)^2 + sin(14*d*x + 14*c)*si n(2*d*x + 2*c) + 6*sin(12*d*x + 12*c)*sin(2*d*x + 2*c) + 15*sin(10*d*x + 1 0*c)*sin(2*d*x + 2*c) + 20*sin(8*d*x + 8*c)*sin(2*d*x + 2*c) + 15*sin(6*d* x + 6*c)*sin(2*d*x + 2*c) + 6*sin(4*d*x + 4*c)*sin(2*d*x + 2*c) + sin(2*d* x + 2*c)^2)*cos(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) + (cos(2* d*x + 2*c)*sin(14*d*x + 14*c) + 6*cos(2*d*x + 2*c)*sin(12*d*x + 12*c) + 15 *cos(2*d*x + 2*c)*sin(10*d*x + 10*c) + 20*cos(2*d*x + 2*c)*sin(8*d*x + 8*c ) + 15*cos(2*d*x + 2*c)*sin(6*d*x + 6*c) + 6*cos(2*d*x + 2*c)*sin(4*d*x + 4*c) - cos(14*d*x + 14*c)*sin(2*d*x + 2*c) - 6*cos(12*d*x + 12*c)*sin(2*d* x + 2*c) - 15*cos(10*d*x + 10*c)*sin(2*d*x + 2*c) - 20*cos(8*d*x + 8*c)*si n(2*d*x + 2*c) - 15*cos(6*d*x + 6*c)*sin(2*d*x + 2*c) - 6*cos(4*d*x + 4*c) *sin(2*d*x + 2*c))*sin(9/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*c os(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - ((cos(2*d*x +...
\[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {a \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4} \,d x } \]
integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)*(a+a*sec(d*x+c))^(1 /2),x, algorithm="giac")
Time = 27.24 (sec) , antiderivative size = 724, normalized size of antiderivative = 3.03 \[ \int \sec ^4(c+d x) \sqrt {a+a \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
((exp(c*1i + d*x*1i)*((A*16i)/(5*d) - ((352*B - 528*A + 320*C)*1i)/(1155*d )) + ((3696*A + 7392*B)*1i)/(1155*d))*(a + a/(exp(- c*1i - d*x*1i)/2 + exp (c*1i + d*x*1i)/2))^(1/2))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^2) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(( C*64i)/(9*d) + exp(c*1i + d*x*1i)*((C*256i)/(33*d) - (A*16i)/(9*d) + ((176 *A + 352*B + 704*C)*1i)/(99*d)) - ((176*A + 352*B)*1i)/(99*d) + ((176*A + 704*C)*1i)/(99*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^4) - ((a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(((1584*A + 3168*B)*1i)/(693*d) - exp(c*1i + d*x*1i)*(((352*B + 896*C)*1i)/(693*d) - (A*16i)/(7*d) + ((3168*B + 6336*C)*1i)/(693*d)) + ((3168*B - 6336*C)*1i) /(693*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 )*((A*16i)/(11*d) - ((32*A + 32*B + 64*C)*1i)/(11*d) + ((16*A + 32*B)*1i)/ (11*d)) + (A*16i)/(11*d) - ((32*A + 32*B + 64*C)*1i)/(11*d) + ((16*A + 32* B)*1i)/(11*d)))/((exp(c*1i + d*x*1i) + 1)*(exp(c*2i + d*x*2i) + 1)^5) - (e xp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^ (1/2)*(3168*A + 2816*B + 2560*C)*1i)/(3465*d*(exp(c*1i + d*x*1i) + 1)) - ( exp(c*1i + d*x*1i)*(a + a/(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)) ^(1/2)*(1584*A + 1408*B + 1280*C)*1i)/(3465*d*(exp(c*1i + d*x*1i) + 1)*(ex p(c*2i + d*x*2i) + 1))