Integrand size = 35, antiderivative size = 178 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {64 a^3 (5 A+7 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {16 a^2 (5 A+7 B) \sqrt {\cos (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 a (5 A+7 B) \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{7 d} \]
2/35*a*(5*A+7*B)*cos(d*x+c)^(3/2)*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/7* A*cos(d*x+c)^(5/2)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+64/105*a^3*(5*A+7*B )*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+16/105*a^2*(5*A+7*B )*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d
Time = 0.44 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.56 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 a^2 \sqrt {\cos (c+d x)} \left (230 A+301 B+(115 A+98 B) \cos (c+d x)+3 (20 A+7 B) \cos ^2(c+d x)+15 A \cos ^3(c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \sin (c+d x)}{105 d (1+\cos (c+d x))} \]
(2*a^2*Sqrt[Cos[c + d*x]]*(230*A + 301*B + (115*A + 98*B)*Cos[c + d*x] + 3 *(20*A + 7*B)*Cos[c + d*x]^2 + 15*A*Cos[c + d*x]^3)*Sqrt[a*(1 + Sec[c + d* x])]*Sin[c + d*x])/(105*d*(1 + Cos[c + d*x]))
Time = 0.98 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3434, 3042, 4501, 3042, 4296, 3042, 4296, 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.
| \(\displaystyle \int \cos ^{\frac {7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2} (A+B \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3434 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(\sec (c+d x) a+a)^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 4501 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} (5 A+7 B) \int \frac {(\sec (c+d x) a+a)^{5/2}}{\sec ^{\frac {5}{2}}(c+d x)}dx+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} (5 A+7 B) \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{5/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4296 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} (5 A+7 B) \left (\frac {8}{5} a \int \frac {(\sec (c+d x) a+a)^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} (5 A+7 B) \left (\frac {8}{5} a \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4296 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} (5 A+7 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \frac {\sqrt {\sec (c+d x) a+a}}{\sqrt {\sec (c+d x)}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} (5 A+7 B) \left (\frac {8}{5} a \left (\frac {4}{3} a \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 4291 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} (5 A+7 B) \left (\frac {8}{5} a \left (\frac {8 a^2 \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}+\frac {2 a \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{3 d \sqrt {\sec (c+d x)}}\right )+\frac {2 a \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{5/2}}{7 d \sec ^{\frac {5}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*A*(a + a*Sec[c + d*x])^(5/2)*Sin [c + d*x])/(7*d*Sec[c + d*x]^(5/2)) + ((5*A + 7*B)*((2*a*(a + a*Sec[c + d* x])^(3/2)*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a*((8*a^2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) + (2*a*Sqrt[a + a*Se c[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])))/5))/7)
3.6.34.3.1 Defintions of rubi rules used
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p Int[(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g , m, n, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && !(IntegerQ[m] && I ntegerQ[n])
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_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-a)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1) *((d*Csc[e + f*x])^n/(f*m)), x] + Simp[b*((2*m - 1)/(d*m)) Int[(a + b*Csc [e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f , m, n}, x] && EqQ[a^2 - b^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2] && Integer Q[2*m]
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]
Time = 3.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.56
\[-\frac {2 a^{2} \left (\left (15 \cos \left (d x +c \right )^{3}+60 \cos \left (d x +c \right )^{2}+115 \cos \left (d x +c \right )+230\right ) A +\left (21 \cos \left (d x +c \right )^{2}+98 \cos \left (d x +c \right )+301\right ) B \right ) \sqrt {\cos \left (d x +c \right )}\, \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right )}{105 d}\]
-2/105*a^2/d*((15*cos(d*x+c)^3+60*cos(d*x+c)^2+115*cos(d*x+c)+230)*A+(21*c os(d*x+c)^2+98*cos(d*x+c)+301)*B)*cos(d*x+c)^(1/2)*(a*(1+sec(d*x+c)))^(1/2 )*(cot(d*x+c)-csc(d*x+c))
Time = 0.26 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.63 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {2 \, {\left (15 \, A a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (20 \, A + 7 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (115 \, A + 98 \, B\right )} a^{2} \cos \left (d x + c\right ) + {\left (230 \, A + 301 \, B\right )} a^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \]
2/105*(15*A*a^2*cos(d*x + c)^3 + 3*(20*A + 7*B)*a^2*cos(d*x + c)^2 + (115* A + 98*B)*a^2*cos(d*x + c) + (230*A + 301*B)*a^2)*sqrt((a*cos(d*x + c) + a )/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(d*cos(d*x + c) + d)
Timed out. \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 482 vs. \(2 (154) = 308\).
Time = 0.46 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.71 \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\frac {5 \, \sqrt {2} {\left (315 \, a^{2} \cos \left (\frac {6}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 77 \, a^{2} \cos \left (\frac {4}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, a^{2} \cos \left (\frac {2}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) - 315 \, a^{2} \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {6}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 77 \, a^{2} \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {4}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) - 21 \, a^{2} \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) \sin \left (\frac {2}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 6 \, a^{2} \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 21 \, a^{2} \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 77 \, a^{2} \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 315 \, a^{2} \sin \left (\frac {1}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} A \sqrt {a} - 28 \, {\left (75 \, \sqrt {2} a^{2} \cos \left (\frac {5}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) \sin \left (2 \, d x + 2 \, c\right ) - 25 \, \sqrt {2} a^{2} \sin \left (\frac {3}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 75 \, \sqrt {2} a^{2} \sin \left (\frac {1}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right ) - 3 \, {\left (25 \, \sqrt {2} a^{2} \cos \left (2 \, d x + 2 \, c\right ) + \sqrt {2} a^{2}\right )} \sin \left (\frac {5}{4} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right )\right )\right )\right )} B \sqrt {a}}{840 \, d} \]
1/840*(5*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)))*sin(7/2*d*x + 7/2*c) + 21*a^2*cos(2/7*arctan 2(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*d*x + 7/2*c)*sin(4/7*arctan2(sin(7/2*d*x + 7/ 2*c), cos(7/2*d*x + 7/2*c))) - 21*a^2*cos(7/2*d*x + 7/2*c)*sin(2/7*arctan2 (sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 6*a^2*sin(7/2*d*x + 7/2*c) + 21*a^2*sin(5/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 7 7*a^2*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 315*a ^2*sin(1/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*A*sqrt(a) - 28*(75*sqrt(2)*a^2*cos(5/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) *sin(2*d*x + 2*c) - 25*sqrt(2)*a^2*sin(3/4*arctan2(sin(2*d*x + 2*c), cos(2 *d*x + 2*c))) - 75*sqrt(2)*a^2*sin(1/4*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))) - 3*(25*sqrt(2)*a^2*cos(2*d*x + 2*c) + sqrt(2)*a^2)*sin(5/4*arct an2(sin(2*d*x + 2*c), cos(2*d*x + 2*c))))*B*sqrt(a))/d
\[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int { {\left (B \sec \left (d x + c\right ) + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {7}{2}} \,d x } \]
Timed out. \[ \int \cos ^{\frac {7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx=\int {\cos \left (c+d\,x\right )}^{7/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]