Integrand size = 45, antiderivative size = 226 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 a (A+9 B) \sin (c+d x)}{63 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {2 a (16 A+18 B+21 C) \sin (c+d x)}{105 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {8 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+a \cos (c+d x)}}+\frac {16 a (16 A+18 B+21 C) \sin (c+d x)}{315 d \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}+\frac {2 A \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{9 d \cos ^{\frac {9}{2}}(c+d x)} \]
2/63*a*(A+9*B)*sin(d*x+c)/d/cos(d*x+c)^(7/2)/(a+a*cos(d*x+c))^(1/2)+2/105* a*(16*A+18*B+21*C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos(d*x+c))^(1/2)+8/ 315*a*(16*A+18*B+21*C)*sin(d*x+c)/d/cos(d*x+c)^(3/2)/(a+a*cos(d*x+c))^(1/2 )+16/315*a*(16*A+18*B+21*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c)) ^(1/2)+2/9*A*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(9/2)
Time = 0.91 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {a (1+\cos (c+d x))} (214 A+162 B+189 C+2 (88 A+99 B+63 C) \cos (c+d x)+11 (16 A+18 B+21 C) \cos (2 (c+d x))+32 A \cos (3 (c+d x))+36 B \cos (3 (c+d x))+42 C \cos (3 (c+d x))+32 A \cos (4 (c+d x))+36 B \cos (4 (c+d x))+42 C \cos (4 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )}{315 d \cos ^{\frac {9}{2}}(c+d x)} \]
Integrate[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2 ))/Cos[c + d*x]^(11/2),x]
(Sqrt[a*(1 + Cos[c + d*x])]*(214*A + 162*B + 189*C + 2*(88*A + 99*B + 63*C )*Cos[c + d*x] + 11*(16*A + 18*B + 21*C)*Cos[2*(c + d*x)] + 32*A*Cos[3*(c + d*x)] + 36*B*Cos[3*(c + d*x)] + 42*C*Cos[3*(c + d*x)] + 32*A*Cos[4*(c + d*x)] + 36*B*Cos[4*(c + d*x)] + 42*C*Cos[4*(c + d*x)])*Tan[(c + d*x)/2])/( 315*d*Cos[c + d*x]^(9/2))
Time = 1.10 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.244, Rules used = {3042, 3522, 27, 3042, 3459, 3042, 3251, 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.
| \(\displaystyle \int \frac {\sqrt {a \cos (c+d x)+a} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{11/2}}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {2 \int \frac {\sqrt {\cos (c+d x) a+a} (a (A+9 B)+3 a (2 A+3 C) \cos (c+d x))}{2 \cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {\cos (c+d x) a+a} (a (A+9 B)+3 a (2 A+3 C) \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)}dx}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a} \left (a (A+9 B)+3 a (2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3459 |
| \(\displaystyle \frac {\frac {3}{7} a (16 A+18 B+21 C) \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a^2 (A+9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (16 A+18 B+21 C) \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a^2 (A+9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {3}{7} a (16 A+18 B+21 C) \left (\frac {4}{5} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (A+9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (16 A+18 B+21 C) \left (\frac {4}{5} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (A+9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3251 |
| \(\displaystyle \frac {\frac {3}{7} a (16 A+18 B+21 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\cos (c+d x) a+a}}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (A+9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{7} a (16 A+18 B+21 C) \left (\frac {4}{5} \left (\frac {2}{3} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}\right )+\frac {2 a^2 (A+9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3250 |
| \(\displaystyle \frac {\frac {2 a^2 (A+9 B) \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {3}{7} a (16 A+18 B+21 C) \left (\frac {2 a \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4}{5} \left (\frac {2 a \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) \sqrt {a \cos (c+d x)+a}}+\frac {4 a \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )\right )}{9 a}+\frac {2 A \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{9 d \cos ^{\frac {9}{2}}(c+d x)}\) |
Int[(Sqrt[a + a*Cos[c + d*x]]*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos [c + d*x]^(11/2),x]
(2*A*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(9*d*Cos[c + d*x]^(9/2)) + ((2 *a^2*(A + 9*B)*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)*Sqrt[a + a*Cos[c + d* x]]) + (3*a*(16*A + 18*B + 21*C)*((2*a*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/ 2)*Sqrt[a + a*Cos[c + d*x]]) + (4*((2*a*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3 /2)*Sqrt[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)/(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[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[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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Time = 13.43 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.72
| method | result | size |
| default | \(\frac {2 \sin \left (d x +c \right ) \left (128 A \left (\cos ^{4}\left (d x +c \right )\right )+144 B \left (\cos ^{4}\left (d x +c \right )\right )+168 C \left (\cos ^{4}\left (d x +c \right )\right )+64 A \left (\cos ^{3}\left (d x +c \right )\right )+72 B \left (\cos ^{3}\left (d x +c \right )\right )+84 C \left (\cos ^{3}\left (d x +c \right )\right )+48 A \left (\cos ^{2}\left (d x +c \right )\right )+54 B \left (\cos ^{2}\left (d x +c \right )\right )+63 C \left (\cos ^{2}\left (d x +c \right )\right )+40 A \cos \left (d x +c \right )+45 B \cos \left (d x +c \right )+35 A \right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{315 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {9}{2}}}\) | \(163\) |
| parts | \(\frac {2 A \sin \left (d x +c \right ) \left (128 \left (\cos ^{4}\left (d x +c \right )\right )+64 \left (\cos ^{3}\left (d x +c \right )\right )+48 \left (\cos ^{2}\left (d x +c \right )\right )+40 \cos \left (d x +c \right )+35\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{315 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {9}{2}}}+\frac {2 B \sin \left (d x +c \right ) \left (16 \left (\cos ^{3}\left (d x +c \right )\right )+8 \left (\cos ^{2}\left (d x +c \right )\right )+6 \cos \left (d x +c \right )+5\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{35 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {7}{2}}}+\frac {2 C \sin \left (d x +c \right ) \left (8 \left (\cos ^{2}\left (d x +c \right )\right )+4 \cos \left (d x +c \right )+3\right ) \sqrt {a \left (1+\cos \left (d x +c \right )\right )}}{15 d \left (1+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right )^{\frac {5}{2}}}\) | \(218\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+cos(d*x+c)*a)^(1/2)/cos(d*x+c)^(11/ 2),x,method=_RETURNVERBOSE)
2/315/d*sin(d*x+c)*(128*A*cos(d*x+c)^4+144*B*cos(d*x+c)^4+168*C*cos(d*x+c) ^4+64*A*cos(d*x+c)^3+72*B*cos(d*x+c)^3+84*C*cos(d*x+c)^3+48*A*cos(d*x+c)^2 +54*B*cos(d*x+c)^2+63*C*cos(d*x+c)^2+40*A*cos(d*x+c)+45*B*cos(d*x+c)+35*A) *(a*(1+cos(d*x+c)))^(1/2)/(1+cos(d*x+c))/cos(d*x+c)^(9/2)
Time = 0.29 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.58 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {2 \, {\left (8 \, {\left (16 \, A + 18 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{4} + 4 \, {\left (16 \, A + 18 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (16 \, A + 18 \, B + 21 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (8 \, A + 9 \, B\right )} \cos \left (d x + c\right ) + 35 \, A\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{315 \, {\left (d \cos \left (d x + c\right )^{6} + d \cos \left (d x + c\right )^{5}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c )^(11/2),x, algorithm="fricas")
2/315*(8*(16*A + 18*B + 21*C)*cos(d*x + c)^4 + 4*(16*A + 18*B + 21*C)*cos( d*x + c)^3 + 3*(16*A + 18*B + 21*C)*cos(d*x + c)^2 + 5*(8*A + 9*B)*cos(d*x + c) + 35*A)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))*sin(d*x + c)/(d* cos(d*x + c)^6 + d*cos(d*x + c)^5)
Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 848 vs. \(2 (196) = 392\).
Time = 0.40 (sec) , antiderivative size = 848, normalized size of antiderivative = 3.75 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Too large to display} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c )^(11/2),x, algorithm="maxima")
2/315*(21*C*(15*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt( 2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 17*sqrt(2)*sqrt(a)*sin(d* x + c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^3/((sin(d*x + c)/(c os(d*x + c) + 1) + 1)^(7/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*( 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)) + 9*B*(35*sqrt(2)*sqrt(a)* sin(d*x + c)/(cos(d*x + c) + 1) - 70*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d *x + c) + 1)^3 + 84*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 58*sqrt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9*sqrt(2)*sqrt(a) *sin(d*x + c)^9/(cos(d*x + c) + 1)^9)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^4/((sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(9/2)*(-sin(d*x + c)/(cos(d *x + c) + 1) + 1)^(9/2)*(4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + sin (d*x + c)^8/(cos(d*x + c) + 1)^8 + 1)) + A*(315*sqrt(2)*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) - 735*sqrt(2)*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 1302*sqrt(2)*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 1206*sq rt(2)*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 431*sqrt(2)*sqrt(a)*si n(d*x + c)^9/(cos(d*x + c) + 1)^9 - 107*sqrt(2)*sqrt(a)*sin(d*x + c)^11/(c os(d*x + c) + 1)^11)*(sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)^5/((sin(...
Timed out. \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\text {Timed out} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*(a+a*cos(d*x+c))^(1/2)/cos(d*x+c )^(11/2),x, algorithm="giac")
Time = 8.34 (sec) , antiderivative size = 629, normalized size of antiderivative = 2.78 \[ \int \frac {\sqrt {a+a \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {11}{2}}(c+d x)} \, dx=\frac {\sqrt {a+a\,\left (\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}\right )}\,\left (\frac {\left (256\,A+288\,B+336\,C\right )\,1{}\mathrm {i}}{315\,d}-\frac {C\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,8{}\mathrm {i}}{3\,d}+\frac {C\,{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}\,8{}\mathrm {i}}{3\,d}-\frac {{\mathrm {e}}^{c\,9{}\mathrm {i}+d\,x\,9{}\mathrm {i}}\,\left (256\,A+288\,B+336\,C\right )\,1{}\mathrm {i}}{315\,d}+\frac {{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\left (1152\,A+1296\,B+1512\,C\right )\,1{}\mathrm {i}}{315\,d}-\frac {{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\left (1152\,A+1296\,B+1512\,C\right )\,1{}\mathrm {i}}{315\,d}+\frac {{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\left (2016\,A+1008\,B+2016\,C\right )\,1{}\mathrm {i}}{315\,d}-\frac {{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\left (2016\,A+1008\,B+2016\,C\right )\,1{}\mathrm {i}}{315\,d}\right )}{\sqrt {\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}}+{\mathrm {e}}^{c\,1{}\mathrm {i}+d\,x\,1{}\mathrm {i}}\,\sqrt {\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}}+4\,{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,\sqrt {\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}}+4\,{\mathrm {e}}^{c\,3{}\mathrm {i}+d\,x\,3{}\mathrm {i}}\,\sqrt {\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}}+6\,{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,\sqrt {\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}}+6\,{\mathrm {e}}^{c\,5{}\mathrm {i}+d\,x\,5{}\mathrm {i}}\,\sqrt {\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}}+4\,{\mathrm {e}}^{c\,6{}\mathrm {i}+d\,x\,6{}\mathrm {i}}\,\sqrt {\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}}+4\,{\mathrm {e}}^{c\,7{}\mathrm {i}+d\,x\,7{}\mathrm {i}}\,\sqrt {\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}}+{\mathrm {e}}^{c\,8{}\mathrm {i}+d\,x\,8{}\mathrm {i}}\,\sqrt {\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}}+{\mathrm {e}}^{c\,9{}\mathrm {i}+d\,x\,9{}\mathrm {i}}\,\sqrt {\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}}} \]
int(((a + a*cos(c + d*x))^(1/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/c os(c + d*x)^(11/2),x)
((a + a*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2))^(1/2)*(((256*A + 288*B + 336*C)*1i)/(315*d) - (C*exp(c*3i + d*x*3i)*8i)/(3*d) + (C*exp(c*6i + d*x*6i)*8i)/(3*d) - (exp(c*9i + d*x*9i)*(256*A + 288*B + 336*C)*1i)/(31 5*d) + (exp(c*2i + d*x*2i)*(1152*A + 1296*B + 1512*C)*1i)/(315*d) - (exp(c *7i + d*x*7i)*(1152*A + 1296*B + 1512*C)*1i)/(315*d) + (exp(c*4i + d*x*4i) *(2016*A + 1008*B + 2016*C)*1i)/(315*d) - (exp(c*5i + d*x*5i)*(2016*A + 10 08*B + 2016*C)*1i)/(315*d)))/((exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i) /2)^(1/2) + exp(c*1i + d*x*1i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i )/2)^(1/2) + 4*exp(c*2i + d*x*2i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x *1i)/2)^(1/2) + 4*exp(c*3i + d*x*3i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 6*exp(c*4i + d*x*4i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + 6*exp(c*5i + d*x*5i)*(exp(- c*1i - d*x*1i)/2 + exp(c *1i + d*x*1i)/2)^(1/2) + 4*exp(c*6i + d*x*6i)*(exp(- c*1i - d*x*1i)/2 + ex p(c*1i + d*x*1i)/2)^(1/2) + 4*exp(c*7i + d*x*7i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*8i + d*x*8i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2) + exp(c*9i + d*x*9i)*(exp(- c*1i - d*x*1i)/2 + exp(c*1i + d*x*1i)/2)^(1/2))