Integrand size = 35, antiderivative size = 181 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a^2 (52 A+63 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {4 a^2 (52 A+63 B) \sin (c+d x)}{105 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 a A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]
2/35*a^2*(8*A+7*B)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+2/ 105*a^2*(52*A+63*B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2)+4 /105*a^2*(52*A+63*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2)+ 2/7*a*A*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(7/2)
Time = 0.31 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.56 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {a \sqrt {a (1+\cos (c+d x))} (82 A+63 B+3 (78 A+77 B) \cos (c+d x)+(52 A+63 B) \cos (2 (c+d x))+52 A \cos (3 (c+d x))+63 B \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{105 d \cos ^{\frac {7}{2}}(c+d x)} \]
(a*Sqrt[a*(1 + Cos[c + d*x])]*(82*A + 63*B + 3*(78*A + 77*B)*Cos[c + d*x] + (52*A + 63*B)*Cos[2*(c + d*x)] + 52*A*Cos[3*(c + d*x)] + 63*B*Cos[3*(c + d*x)])*Tan[(c + d*x)/2])/(105*d*Cos[c + d*x]^(7/2))
Time = 0.89 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3454, 27, 3042, 3459, 3042, 3251, 3042, 3250}
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.
| \(\displaystyle \int \frac {(a \cos (c+d x)+a)^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {2}{7} \int \frac {\sqrt {\cos (c+d x) a+a} (a (8 A+7 B)+a (4 A+7 B) \cos (c+d x))}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} \int \frac {\sqrt {\cos (c+d x) a+a} (a (8 A+7 B)+a (4 A+7 B) \cos (c+d x))}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (8 A+7 B)+a (4 A+7 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} \left (\frac {1}{5} a (52 A+63 B) \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3250 |
| \(\displaystyle \frac {1}{7} \left (\frac {2 a^2 (8 A+7 B) \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {1}{5} a (52 A+63 B) \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )+\frac {2 a A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{7 d \cos ^{\frac {7}{2}}(c+d x)}\) |
(2*a*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + ( (2*a^2*(8*A + 7*B)*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)*Sqrt[a + a*Cos[c + d*x]]) + (a*(52*A + 63*B)*((2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)*Sq rt[a + a*Cos[c + d*x]]) + (4*a*Sin[c + d*x])/(3*d*Sqrt[Cos[c + d*x]]*Sqrt[ a + a*Cos[c + d*x]])))/5)/7
3.2.80.3.1 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[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(3/2), x_Symbol] :> Simp[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*Sq rt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] + Sim p[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1) *(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b* c - 2*a*d*(n + 1)))/(2*d*(n + 1)*(b*c + a*d)) Int[Sqrt[a + b*Sin[e + f*x] ]*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x ] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[n, -1]
Time = 7.66 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.60
| method | result | size |
| default | \(\frac {2 a \sin \left (d x +c \right ) \left (104 A \left (\cos ^{3}\left (d x +c \right )\right )+126 B \left (\cos ^{3}\left (d x +c \right )\right )+52 A \left (\cos ^{2}\left (d x +c \right )\right )+63 B \left (\cos ^{2}\left (d x +c \right )\right )+39 A \cos \left (d x +c \right )+21 B \cos \left (d x +c \right )+15 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{105 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {7}{2}}}\) | \(109\) |
| parts | \(\frac {2 A \sin \left (d x +c \right ) \left (104 \left (\cos ^{3}\left (d x +c \right )\right )+52 \left (\cos ^{2}\left (d x +c \right )\right )+39 \cos \left (d x +c \right )+15\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a}{105 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {2 B \sin \left (d x +c \right ) \left (6 \left (\cos ^{2}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}\, a}{5 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(138\) |
2/105*a/d*sin(d*x+c)*(104*A*cos(d*x+c)^3+126*B*cos(d*x+c)^3+52*A*cos(d*x+c )^2+63*B*cos(d*x+c)^2+39*A*cos(d*x+c)+21*B*cos(d*x+c)+15*A)*(a*(1+cos(d*x+ c)))^(1/2)/(1+cos(d*x+c))/cos(d*x+c)^(7/2)
Time = 0.29 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.59 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2 \, {\left (2 \, {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{3} + {\left (52 \, A + 63 \, B\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (13 \, A + 7 \, B\right )} a \cos \left (d x + c\right ) + 15 \, A a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{105 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4}\right )}} \]
2/105*(2*(52*A + 63*B)*a*cos(d*x + c)^3 + (52*A + 63*B)*a*cos(d*x + c)^2 + 3*(13*A + 7*B)*a*cos(d*x + c) + 15*A*a)*sqrt(a*cos(d*x + c) + a)*sqrt(cos (d*x + c))*sin(d*x + c)/(d*cos(d*x + c)^5 + d*cos(d*x + c)^4)
Timed out. \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (157) = 314\).
Time = 0.34 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.66 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {4 \, {\left (\frac {21 \, {\left (\frac {5 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {7 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {2 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )} B {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{2}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 1\right )}} + \frac {{\left (\frac {105 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {245 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {273 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {171 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {38 \, \sqrt {2} a^{\frac {3}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}\right )} A {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{{\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {9}{2}} {\left (\frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {\sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1\right )}}\right )}}{105 \, d} \]
4/105*(21*(5*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sqrt(2)* a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 7*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 2*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*B*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^2/((sin(d*x + c)/(cos (d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(2* sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 1)) + (105*sqrt(2)*a^(3/2)*sin(d*x + c)/(cos(d*x + c) + 1) - 245*sqrt(2) *a^(3/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 273*sqrt(2)*a^(3/2)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 171*sqrt(2)*a^(3/2)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 38*sqrt(2)*a^(3/2)*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*A*( sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(3*sin(d*x + c) ^2/(cos(d*x + c) + 1)^2 + 3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1)))/d
Result contains complex when optimal does not.
Time = 260.78 (sec) , antiderivative size = 147871, normalized size of antiderivative = 816.97 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Too large to display} \]
67108864/105*sqrt(2)*sqrt(-tan(1/4*d*x + c)^4*tan(1/2*c)^8 + 14*tan(1/4*d* x + c)^4*tan(1/2*c)^6 - 24*tan(1/4*d*x + c)^3*tan(1/2*c)^7 + 6*tan(1/4*d*x + c)^2*tan(1/2*c)^8 + 56*tan(1/4*d*x + c)^3*tan(1/2*c)^5 - 84*tan(1/4*d*x + c)^2*tan(1/2*c)^6 + 24*tan(1/4*d*x + c)*tan(1/2*c)^7 - tan(1/2*c)^8 - 1 4*tan(1/4*d*x + c)^4*tan(1/2*c)^2 + 56*tan(1/4*d*x + c)^3*tan(1/2*c)^3 - 5 6*tan(1/4*d*x + c)*tan(1/2*c)^5 + 14*tan(1/2*c)^6 + tan(1/4*d*x + c)^4 - 2 4*tan(1/4*d*x + c)^3*tan(1/2*c) + 84*tan(1/4*d*x + c)^2*tan(1/2*c)^2 - 56* tan(1/4*d*x + c)*tan(1/2*c)^3 - 6*tan(1/4*d*x + c)^2 + 24*tan(1/4*d*x + c) *tan(1/2*c) - 14*tan(1/2*c)^2 + 1)*(((((((((((((((-52*I*A*a*e^(1055/2*I*c) - 63*I*B*a*e^(1055/2*I*c) - 21008*I*A*a*e^(1053/2*I*c) - 25452*I*B*a*e^(1 053/2*I*c) - 4233112*I*A*a*e^(1051/2*I*c) - 5128578*I*B*a*e^(1051/2*I*c) - 567237008*I*A*a*e^(1049/2*I*c) - 687229452*I*B*a*e^(1049/2*I*c) - 5686551 0052*I*A*a*e^(1047/2*I*c) - 68894752563*I*B*a*e^(1047/2*I*c) - 45492408041 60*I*A*a*e^(1045/2*I*c) - 5511580205040*I*B*a*e^(1045/2*I*c) - 30252451347 6978*I*A*a*e^(1043/2*I*c) - 366520083635517*I*B*a*e^(1043/2*I*c) - 1720067 9480665512*I*A*a*e^(1041/2*I*c) - 20839284755400468*I*B*a*e^(1041/2*I*c) - 853583719248764868*I*A*a*e^(1039/2*I*c) - 1034149506008652852*I*B*a*e^(10 39/2*I*c) - 37557683649422024642*I*A*a*e^(1037/2*I*c) - 455025782669963001 63*I*B*a*e^(1037/2*I*c) - 1483528504376157661378*I*A*a*e^(1035/2*I*c) - 17 97351841782932557017*I*B*a*e^(1035/2*I*c) - 53137293718531292117900*I*A...
Time = 7.07 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.30 \[ \int \frac {(a+a \cos (c+d x))^{3/2} (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {\sqrt {a+a\,\cos \left (c+d\,x\right )}\,\left (-\frac {8\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+3\,B\right )}{3\,d}+\frac {16\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )\,\left (13\,A+12\,B\right )}{15\,d}+\frac {8\,a\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\sin \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )\,\left (52\,A+63\,B\right )}{105\,d}\right )}{6\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )+6\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {3\,c}{2}+\frac {3\,d\,x}{2}\right )+2\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {5\,c}{2}+\frac {5\,d\,x}{2}\right )+2\,\sqrt {\cos \left (c+d\,x\right )}\,{\mathrm {e}}^{\frac {c\,7{}\mathrm {i}}{2}+\frac {d\,x\,7{}\mathrm {i}}{2}}\,\cos \left (\frac {7\,c}{2}+\frac {7\,d\,x}{2}\right )} \]
((a + a*cos(c + d*x))^(1/2)*((16*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((3*c)/2 + (3*d*x)/2)*(13*A + 12*B))/(15*d) - (8*a*exp((c*7i)/2 + (d*x*7i)/2)*sin(c /2 + (d*x)/2)*(2*A + 3*B))/(3*d) + (8*a*exp((c*7i)/2 + (d*x*7i)/2)*sin((7* c)/2 + (7*d*x)/2)*(52*A + 63*B))/(105*d)))/(6*cos(c + d*x)^(1/2)*exp((c*7i )/2 + (d*x*7i)/2)*cos(c/2 + (d*x)/2) + 6*cos(c + d*x)^(1/2)*exp((c*7i)/2 + (d*x*7i)/2)*cos((3*c)/2 + (3*d*x)/2) + 2*cos(c + d*x)^(1/2)*exp((c*7i)/2 + (d*x*7i)/2)*cos((5*c)/2 + (5*d*x)/2) + 2*cos(c + d*x)^(1/2)*exp((c*7i)/2 + (d*x*7i)/2)*cos((7*c)/2 + (7*d*x)/2))