Integrand size = 33, antiderivative size = 205 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=-\frac {2 b (3 A+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 a (5 A+7 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 d}+\frac {2 b (3 A+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a (5 A+7 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{21 d}+\frac {2 A b \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 a A \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{7 d} \]
2/21*a*(5*A+7*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*A*b*sec(d*x+c)^(5/2)*si n(d*x+c)/d+2/7*a*A*sec(d*x+c)^(7/2)*sin(d*x+c)/d+2/5*b*(3*A+5*C)*sin(d*x+c )*sec(d*x+c)^(1/2)/d-2/5*b*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2* d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+ c)^(1/2)/d+2/21*a*(5*A+7*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c )*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/ d
Time = 3.39 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.76 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {\sec ^{\frac {7}{2}}(c+d x) \left (-168 b (3 A+5 C) \cos ^{\frac {7}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+40 a (5 A+7 C) \cos ^{\frac {7}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 (110 a A+70 a C+21 b (13 A+15 C) \cos (c+d x)+10 a (5 A+7 C) \cos (2 (c+d x))+63 A b \cos (3 (c+d x))+105 b C \cos (3 (c+d x))) \sin (c+d x)\right )}{420 d} \]
(Sec[c + d*x]^(7/2)*(-168*b*(3*A + 5*C)*Cos[c + d*x]^(7/2)*EllipticE[(c + d*x)/2, 2] + 40*a*(5*A + 7*C)*Cos[c + d*x]^(7/2)*EllipticF[(c + d*x)/2, 2] + 2*(110*a*A + 70*a*C + 21*b*(13*A + 15*C)*Cos[c + d*x] + 10*a*(5*A + 7*C )*Cos[2*(c + d*x)] + 63*A*b*Cos[3*(c + d*x)] + 105*b*C*Cos[3*(c + d*x)])*S in[c + d*x]))/(420*d)
Time = 0.94 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.89, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4709, 3042, 3511, 27, 3042, 3500, 27, 3042, 3227, 3042, 3116, 3042, 3119, 3120}
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.
| \(\displaystyle \int \sec ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{9/2} (a+b \cos (c+d x)) \left (A+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x)) \left (C \cos ^2(c+d x)+A\right )}{\cos ^{\frac {9}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{9/2}}dx\) |
| \(\Big \downarrow \) 3511 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2}{7} \int \frac {7 b C \cos ^2(c+d x)+a (5 A+7 C) \cos (c+d x)+7 A b}{2 \cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \frac {7 b C \cos ^2(c+d x)+a (5 A+7 C) \cos (c+d x)+7 A b}{\cos ^{\frac {7}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \int \frac {7 b C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (5 A+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )+7 A b}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {2}{5} \int \frac {5 a (5 A+7 C)+7 b (3 A+5 C) \cos (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \int \frac {5 a (5 A+7 C)+7 b (3 A+5 C) \cos (c+d x)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \int \frac {5 a (5 A+7 C)+7 b (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx+7 b (3 A+5 C) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx\right )+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+7 b (3 A+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx\right )+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{7} \left (\frac {1}{5} \left (5 a (5 A+7 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+7 b (3 A+5 C) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {14 A b \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sin[c + d*x])/(7*d*Cos[c + d *x]^(7/2)) + ((14*A*b*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (5*a*(5*A + 7*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2))) + 7*b*(3*A + 5*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin [c + d*x])/(d*Sqrt[Cos[c + d*x]])))/5)/7)
3.14.60.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1)) I nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2*n]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[ (-(b*c - a*d))*(A*b^2 + a^2*C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/( b^2*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*(a*C*(b*c - a*d) + A*b*(a*c - b*d )) - ((b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] + b *C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e , f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSineIntegrandQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(813\) vs. \(2(229)=458\).
Time = 314.25 (sec) , antiderivative size = 814, normalized size of antiderivative = 3.97
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(814\) |
| parts | \(\text {Expression too large to display}\) | \(1002\) |
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*C*b/sin(1/2* d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d *x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x +1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/ 2*c),2^(1/2)))+2*a*A*(-1/56*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+si n(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^4-5/42*cos(1/2*d*x+1/ 2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2 *c)^2-1/2)^2+5/21*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1) ^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos( 1/2*d*x+1/2*c),2^(1/2)))+2*C*a*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/ 2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(sin(1 /2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1 /2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))) +2/5*A*b/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c )^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6- 12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE (cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4* cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c) ^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin (1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.19 \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\frac {-5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} {\left (5 \, A + 7 \, C\right )} a \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} b \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (21 \, {\left (3 \, A + 5 \, C\right )} b \cos \left (d x + c\right )^{3} + 5 \, {\left (5 \, A + 7 \, C\right )} a \cos \left (d x + c\right )^{2} + 21 \, A b \cos \left (d x + c\right ) + 15 \, A a\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{105 \, d \cos \left (d x + c\right )^{3}} \]
1/105*(-5*I*sqrt(2)*(5*A + 7*C)*a*cos(d*x + c)^3*weierstrassPInverse(-4, 0 , cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqrt(2)*(5*A + 7*C)*a*cos(d*x + c)^ 3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 21*I*sqrt(2) *(3*A + 5*C)*b*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(- 4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*I*sqrt(2)*(3*A + 5*C)*b*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I *sin(d*x + c))) + 2*(21*(3*A + 5*C)*b*cos(d*x + c)^3 + 5*(5*A + 7*C)*a*cos (d*x + c)^2 + 21*A*b*cos(d*x + c) + 15*A*a)*sin(d*x + c)/sqrt(cos(d*x + c) ))/(d*cos(d*x + c)^3)
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {9}{2}} \,d x } \]
\[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {9}{2}} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+C \cos ^2(c+d x)\right ) \sec ^{\frac {9}{2}}(c+d x) \, dx=\int \left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{9/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \]