Integrand size = 41, antiderivative size = 192 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {2 (3 a A+5 b B+5 a C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {2 (A b+a B+3 b C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 (3 a A+5 b B+5 a C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (A b+a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d}+\frac {2 a A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 d} \]
2/3*(A*b+B*a)*sec(d*x+c)^(3/2)*sin(d*x+c)/d+2/5*a*A*sec(d*x+c)^(5/2)*sin(d *x+c)/d+2/5*(3*A*a+5*B*b+5*C*a)*sin(d*x+c)*sec(d*x+c)^(1/2)/d-2/5*(3*A*a+5 *B*b+5*C*a)*(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+2/3*(A*b+B*a+3 *C*b)*(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 = 2.05 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.78 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-3 (3 a A+5 b B+5 a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 (A b+a B+3 b C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {(10 (A b+a B) \cos (c+d x)+3 (5 b B+5 a (A+C)+(3 a A+5 b B+5 a C) \cos (2 (c+d x)))) \sin (c+d x)}{2 \cos ^{\frac {5}{2}}(c+d x)}\right )}{15 d} \]
(2*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-3*(3*a*A + 5*b*B + 5*a*C)*Ellip ticE[(c + d*x)/2, 2] + 5*(A*b + a*B + 3*b*C)*EllipticF[(c + d*x)/2, 2] + ( (10*(A*b + a*B)*Cos[c + d*x] + 3*(5*b*B + 5*a*(A + C) + (3*a*A + 5*b*B + 5 *a*C)*Cos[2*(c + d*x)]))*Sin[c + d*x])/(2*Cos[c + d*x]^(5/2))))/(15*d)
Time = 0.94 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.87, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 4709, 3042, 3510, 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.
| \(\displaystyle \int \sec ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sec (c+d x)^{7/2} (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {(a+b \cos (c+d x)) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{\cos ^{\frac {7}{2}}(c+d x)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int -\frac {5 b C \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+5 (A b+a B)}{2 \cos ^{\frac {5}{2}}(c+d x)}dx\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {5 b C \cos ^2(c+d x)+(3 a A+5 b B+5 a C) \cos (c+d x)+5 (A b+a B)}{\cos ^{\frac {5}{2}}(c+d x)}dx+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \int \frac {5 b C \sin \left (c+d x+\frac {\pi }{2}\right )^2+(3 a A+5 b B+5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+5 (A b+a B)}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {2}{3} \int \frac {3 (3 a A+5 b B+5 a C)+5 (A b+3 C b+a B) \cos (c+d x)}{2 \cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {3 (3 a A+5 b B+5 a C)+5 (A b+3 C b+a B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \int \frac {3 (3 a A+5 b B+5 a C)+5 (A b+3 C b+a B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (3 (3 a A+5 a C+5 b B) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+5 (a B+A b+3 b C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx\right )+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (3 (3 a A+5 a C+5 b B) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+5 (a B+A b+3 b C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 (a B+A b+3 b C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 (3 a A+5 a C+5 b B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )\right )+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 (a B+A b+3 b C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 (3 a A+5 a C+5 b B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )\right )+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (5 (a B+A b+3 b C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 (3 a A+5 a C+5 b B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {1}{5} \left (\frac {1}{3} \left (\frac {10 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (a B+A b+3 b C)}{d}+3 (3 a A+5 a C+5 b B) \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )+\frac {10 (a B+A b) \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )+\frac {2 a A \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*((2*a*A*Sin[c + d*x])/(5*d*Cos[c + d *x]^(5/2)) + ((10*(A*b + a*B)*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)) + ((1 0*(A*b + a*B + 3*b*C)*EllipticF[(c + d*x)/2, 2])/d + 3*(3*a*A + 5*b*B + 5* a*C)*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/3)/5)
3.15.52.3.1 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + 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] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (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] /; F reeQ[{a, b, c, d, e, f, A, B, 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. \(714\) vs. \(2(220)=440\).
Time = 342.28 (sec) , antiderivative size = 715, normalized size of antiderivative = 3.72
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(715\) |
| parts | \(\text {Expression too large to display}\) | \(902\) |
int((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2),x,me thod=_RETURNVERBOSE)
-(-(-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)^(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*A/(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)/sin(1/2*d*x+1/2*c)^2*(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(co s(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*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*sin(1/2 *d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+2*(B*b+C*a)/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*b+B*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))))/sin(1/2*d*x+1/ 2*c)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.35 \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=-\frac {5 \, \sqrt {2} {\left (i \, B a + i \, {\left (A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, \sqrt {2} {\left (-i \, B a - i \, {\left (A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, \sqrt {2} {\left (i \, {\left (3 \, A + 5 \, C\right )} a + 5 i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, \sqrt {2} {\left (-i \, {\left (3 \, A + 5 \, C\right )} a - 5 i \, B b\right )} \cos \left (d x + c\right )^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left ({\left (3 \, A + 5 \, C\right )} a + 5 \, B b\right )} \cos \left (d x + c\right )^{2} + 3 \, A a + 5 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{15 \, d \cos \left (d x + c\right )^{2}} \]
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2 ),x, algorithm="fricas")
-1/15*(5*sqrt(2)*(I*B*a + I*(A + 3*C)*b)*cos(d*x + c)^2*weierstrassPInvers e(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*sqrt(2)*(-I*B*a - I*(A + 3*C)* b)*cos(d*x + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c) ) + 3*sqrt(2)*(I*(3*A + 5*C)*a + 5*I*B*b)*cos(d*x + c)^2*weierstrassZeta(- 4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*sqrt( 2)*(-I*(3*A + 5*C)*a - 5*I*B*b)*cos(d*x + c)^2*weierstrassZeta(-4, 0, weie rstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*((3*A + 5*C) *a + 5*B*b)*cos(d*x + c)^2 + 3*A*a + 5*(B*a + A*b)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2)
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\text {Timed out} \]
\[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2 ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)*sec (d*x + c)^(7/2), x)
\[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {7}{2}} \,d x } \]
integrate((a+b*cos(d*x+c))*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2 ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)*sec (d*x + c)^(7/2), x)
Timed out. \[ \int (a+b \cos (c+d x)) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]