Integrand size = 35, antiderivative size = 228 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^3 (124 A+135 B) \sin (c+d x)}{315 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {2 a^3 (292 A+345 B) \sin (c+d x)}{315 d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {4 a^3 (292 A+345 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{315 d \sqrt {a+a \sec (c+d x)}}+\frac {2 a^2 (4 A+3 B) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{21 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {2 a A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{9 d \sec ^{\frac {7}{2}}(c+d x)} \]
2/9*a*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d/sec(d*x+c)^(7/2)+2/315*a^3*(12 4*A+135*B)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)+2/315*a^3* (292*A+345*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+4/315*a ^3*(292*A+345*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+2/21 *a^2*(4*A+3*B)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(5/2)
Time = 0.64 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.47 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^3 \left (35 A+5 (26 A+9 B) \sec (c+d x)+3 (73 A+60 B) \sec ^2(c+d x)+(292 A+345 B) \sec ^3(c+d x)+(584 A+690 B) \sec ^4(c+d x)\right ) \sin (c+d x)}{315 d \sec ^{\frac {7}{2}}(c+d x) \sqrt {a (1+\sec (c+d x))}} \]
(2*a^3*(35*A + 5*(26*A + 9*B)*Sec[c + d*x] + 3*(73*A + 60*B)*Sec[c + d*x]^ 2 + (292*A + 345*B)*Sec[c + d*x]^3 + (584*A + 690*B)*Sec[c + d*x]^4)*Sin[c + d*x])/(315*d*Sec[c + d*x]^(7/2)*Sqrt[a*(1 + Sec[c + d*x])])
Time = 1.37 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4505, 27, 3042, 4505, 27, 3042, 4503, 3042, 4292, 3042, 4291}
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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {2}{9} \int \frac {(\sec (c+d x) a+a)^{3/2} (3 a (4 A+3 B)+a (4 A+9 B) \sec (c+d x))}{2 \sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \int \frac {(\sec (c+d x) a+a)^{3/2} (3 a (4 A+3 B)+a (4 A+9 B) \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2} \left (3 a (4 A+3 B)+a (4 A+9 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{9} \left (\frac {2}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((124 A+135 B) a^2+(76 A+99 B) \sec (c+d x) a^2\right )}{2 \sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\sec (c+d x) a+a} \left ((124 A+135 B) a^2+(76 A+99 B) \sec (c+d x) a^2\right )}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{7} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a} \left ((124 A+135 B) a^2+(76 A+99 B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4503 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^2 (292 A+345 B) \int \frac {\sqrt {\sec (c+d x) a+a}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a^3 (124 A+135 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^2 (292 A+345 B) \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a^3 (124 A+135 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4292 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^2 (292 A+345 B) \left (\frac {2}{3} \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (124 A+135 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{9} \left (\frac {1}{7} \left (\frac {3}{5} a^2 (292 A+345 B) \left (\frac {2}{3} \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )+\frac {2 a^3 (124 A+135 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}\right )+\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \frac {1}{9} \left (\frac {6 a^2 (4 A+3 B) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{7 d \sec ^{\frac {5}{2}}(c+d x)}+\frac {1}{7} \left (\frac {2 a^3 (124 A+135 B) \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}+\frac {3}{5} a^2 (292 A+345 B) \left (\frac {4 a \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}\right )\right )\right )+\frac {2 a A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{9 d \sec ^{\frac {7}{2}}(c+d x)}\) |
(2*a*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(9*d*Sec[c + d*x]^(7/2)) + ((6*a^2*(4*A + 3*B)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(7*d*Sec[c + d *x]^(5/2)) + ((2*a^3*(124*A + 135*B)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2) *Sqrt[a + a*Sec[c + d*x]]) + (3*a^2*(292*A + 345*B)*((2*a*Sin[c + d*x])/(3 *d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + (4*a*Sqrt[Sec[c + d*x]]* Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])))/5)/7)/9
3.3.46.3.1 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[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)] *(d_.)], x_Symbol] :> Simp[-2*a*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*S qrt[d*Csc[e + f*x]])), x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp[a*((2*n + 1)/(2*b*d*n)) Int[Sqrt[a + b*Csc [e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[n, -2^(-1)] && IntegerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Co t[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] + Simp [(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n) Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[ e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Time = 5.43 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.55
| method | result | size |
| default | \(\frac {2 a^{2} \left (35 A \cos \left (d x +c \right )^{4}+130 A \cos \left (d x +c \right )^{3}+45 B \cos \left (d x +c \right )^{3}+219 A \cos \left (d x +c \right )^{2}+180 B \cos \left (d x +c \right )^{2}+292 A \cos \left (d x +c \right )+345 B \cos \left (d x +c \right )+584 A +690 B \right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(125\) |
| parts | \(\frac {2 A \,a^{2} \left (35 \cos \left (d x +c \right )^{4}+130 \cos \left (d x +c \right )^{3}+219 \cos \left (d x +c \right )^{2}+292 \cos \left (d x +c \right )+584\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{315 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 B \,a^{2} \left (3 \cos \left (d x +c \right )^{3}+12 \cos \left (d x +c \right )^{2}+23 \cos \left (d x +c \right )+46\right ) \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \tan \left (d x +c \right )}{21 d \left (\cos \left (d x +c \right )+1\right ) \sec \left (d x +c \right )^{\frac {3}{2}}}\) | \(162\) |
2/315*a^2/d*(35*A*cos(d*x+c)^4+130*A*cos(d*x+c)^3+45*B*cos(d*x+c)^3+219*A* cos(d*x+c)^2+180*B*cos(d*x+c)^2+292*A*cos(d*x+c)+345*B*cos(d*x+c)+584*A+69 0*B)*(a*(1+sec(d*x+c)))^(1/2)/(cos(d*x+c)+1)/sec(d*x+c)^(3/2)*tan(d*x+c)
Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.62 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (35 \, A a^{2} \cos \left (d x + c\right )^{5} + 5 \, {\left (26 \, A + 9 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} + 3 \, {\left (73 \, A + 60 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + {\left (292 \, A + 345 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (292 \, A + 345 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right ) + d\right )} \sqrt {\cos \left (d x + c\right )}} \]
2/315*(35*A*a^2*cos(d*x + c)^5 + 5*(26*A + 9*B)*a^2*cos(d*x + c)^4 + 3*(73 *A + 60*B)*a^2*cos(d*x + c)^3 + (292*A + 345*B)*a^2*cos(d*x + c)^2 + 2*(29 2*A + 345*B)*a^2*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin (d*x + c)/((d*cos(d*x + c) + d)*sqrt(cos(d*x + c)))
Timed out. \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 746 vs. \(2 (198) = 396\).
Time = 0.56 (sec) , antiderivative size = 746, normalized size of antiderivative = 3.27 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/5040*(sqrt(2)*(8190*a^2*cos(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d* x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 2100*a^2*cos(2/3*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 756*a^2*cos(4/9*arc tan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/2*d*x + 9/2*c) + 2 25*a^2*cos(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c)))*sin(9/ 2*d*x + 9/2*c) - 8190*a^2*cos(9/2*d*x + 9/2*c)*sin(8/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 2100*a^2*cos(9/2*d*x + 9/2*c)*sin(2/3* arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 756*a^2*cos(9/2*d*x + 9/2*c)*sin(4/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) - 2 25*a^2*cos(9/2*d*x + 9/2*c)*sin(2/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2* d*x + 9/2*c))) + 70*a^2*sin(9/2*d*x + 9/2*c) + 225*a^2*sin(7/9*arctan2(sin (9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 756*a^2*sin(5/9*arctan2(sin(9/ 2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 2100*a^2*sin(1/3*arctan2(sin(9/2* d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 8190*a^2*sin(1/9*arctan2(sin(9/2*d* x + 9/2*c), cos(9/2*d*x + 9/2*c))))*A*sqrt(a) + 30*sqrt(2)*(315*a^2*cos(6/ 7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c ) + 77*a^2*cos(4/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c)))*si n(7/2*d*x + 7/2*c) + 21*a^2*cos(2/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2* d*x + 7/2*c)))*sin(7/2*d*x + 7/2*c) - 315*a^2*cos(7/2*d*x + 7/2*c)*sin(6/7 *arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) - 77*a^2*cos(7/2*...
\[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\sec \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]
Time = 17.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.69 \[ \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {9}{2}}(c+d x)} \, dx=\frac {a^2\,\cos \left (c+d\,x\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (c+d\,x\right )+1\right )}{\cos \left (c+d\,x\right )}}\,\left (10290\,A\,\sin \left (c+d\,x\right )+11760\,B\,\sin \left (c+d\,x\right )+2856\,A\,\sin \left (2\,c+2\,d\,x\right )+981\,A\,\sin \left (3\,c+3\,d\,x\right )+260\,A\,\sin \left (4\,c+4\,d\,x\right )+35\,A\,\sin \left (5\,c+5\,d\,x\right )+2940\,B\,\sin \left (2\,c+2\,d\,x\right )+720\,B\,\sin \left (3\,c+3\,d\,x\right )+90\,B\,\sin \left (4\,c+4\,d\,x\right )\right )}{2520\,d\,\left (\cos \left (c+d\,x\right )+1\right )} \]
(a^2*cos(c + d*x)*(1/cos(c + d*x))^(1/2)*((a*(cos(c + d*x) + 1))/cos(c + d *x))^(1/2)*(10290*A*sin(c + d*x) + 11760*B*sin(c + d*x) + 2856*A*sin(2*c + 2*d*x) + 981*A*sin(3*c + 3*d*x) + 260*A*sin(4*c + 4*d*x) + 35*A*sin(5*c + 5*d*x) + 2940*B*sin(2*c + 2*d*x) + 720*B*sin(3*c + 3*d*x) + 90*B*sin(4*c + 4*d*x)))/(2520*d*(cos(c + d*x) + 1))