Integrand size = 43, antiderivative size = 250 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {3 (7 A-5 B+5 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {(5 A-5 B+3 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a d}+\frac {(7 A-5 B+5 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]
-1/3*(5*A-5*B+3*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d+1/5*(7*A-5*B+5*C)*sec(d *x+c)^(5/2)*sin(d*x+c)/a/d-(A-B+C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*cos( d*x+c))+3/5*(7*A-5*B+5*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d-3/5*(7*A-5*B+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)/a/d-1/3*(5*A-5*B+3*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)/a/d
Time = 4.94 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.80 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) \left (36 (7 A-5 B+5 C) \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+20 (5 A-5 B+3 C) \cos ^{\frac {5}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-(100 A-40 B+60 C+(173 A-95 B+135 C) \cos (c+d x)+(76 A-40 B+60 C) \cos (2 (c+d x))+63 A \cos (3 (c+d x))-45 B \cos (3 (c+d x))+45 C \cos (3 (c+d x))) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{30 a d (1+\cos (c+d x))} \]
Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^(7/2))/(a + a*Cos[c + d*x]),x]
-1/30*(Cos[(c + d*x)/2]^2*Sec[c + d*x]^(5/2)*(36*(7*A - 5*B + 5*C)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 20*(5*A - 5*B + 3*C)*Cos[c + d*x]^ (5/2)*EllipticF[(c + d*x)/2, 2] - (100*A - 40*B + 60*C + (173*A - 95*B + 1 35*C)*Cos[c + d*x] + (76*A - 40*B + 60*C)*Cos[2*(c + d*x)] + 63*A*Cos[3*(c + d*x)] - 45*B*Cos[3*(c + d*x)] + 45*C*Cos[3*(c + d*x)])*Tan[(c + d*x)/2] ))/(a*d*(1 + Cos[c + d*x]))
Time = 1.04 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.81, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4709, 3042, 3520, 27, 3042, 3227, 3042, 3116, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {7}{2}}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a \cos (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (c+d x)^{7/2} \left (A+B \cos (c+d x)+C \cos (c+d x)^2\right )}{a \cos (c+d x)+a}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+B \cos (c+d x)+A}{\cos ^{\frac {7}{2}}(c+d x) (\cos (c+d x) a+a)}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-5 B+5 C)-a (5 A-5 B+3 C) \cos (c+d x)}{2 \cos ^{\frac {7}{2}}(c+d x)}dx}{a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-5 B+5 C)-a (5 A-5 B+3 C) \cos (c+d x)}{\cos ^{\frac {7}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (7 A-5 B+5 C)-a (5 A-5 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)}dx-a (5 A-5 B+3 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx-a (5 A-5 B+3 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3116 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \left (\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (7 A-5 B+5 C) \left (\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-a (5 A-5 B+3 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A - B + C)*Sin[c + d*x])/(d*Cos [c + d*x]^(5/2)*(a + a*Cos[c + d*x]))) + (-(a*(5*A - 5*B + 3*C)*((2*Ellipt icF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3*d*Cos[c + d*x]^(3/2)))) + a*(7*A - 5*B + 5*C)*((2*Sin[c + d*x])/(5*d*Cos[c + d*x]^(5/2)) + (3*((-2* EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]])))/5 ))/(2*a^2))
3.13.90.3.1 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_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(a*A - b*B + a*C)*Cos[e + f*x]*(a + b* Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x ] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) + B*(b*c*m + a *d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(m + n + 2) + C*(b*c *(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c ^2 - d^2, 0] && LtQ[m, -2^(-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. \(784\) vs. \(2(276)=552\).
Time = 55.01 (sec) , antiderivative size = 785, normalized size of antiderivative = 3.14
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c)),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)/a*((-A+B-C)*(co s(1/2*d*x+1/2*c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^( 1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(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)/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)+2/5*A/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*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*A+2*B)*(-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*A-2*B+2*C)/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...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.48 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-5 i \, A + 5 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-5 i \, A + 5 i \, B - 3 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (5 i \, A - 5 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (5 i \, A - 5 i \, B + 3 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (7 i \, A - 5 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (7 i \, A - 5 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-7 i \, A + 5 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )^{3} + \sqrt {2} {\left (-7 i \, A + 5 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )^{2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (9 \, {\left (7 \, A - 5 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (19 \, A - 10 \, B + 15 \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right ) + 6 \, A\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c) ),x, algorithm="fricas")
-1/30*(5*(sqrt(2)*(-5*I*A + 5*I*B - 3*I*C)*cos(d*x + c)^3 + sqrt(2)*(-5*I* A + 5*I*B - 3*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(5*I*A - 5*I*B + 3*I*C)*cos(d*x + c)^3 + s qrt(2)*(5*I*A - 5*I*B + 3*I*C)*cos(d*x + c)^2)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*(sqrt(2)*(7*I*A - 5*I*B + 5*I*C)*cos(d* x + c)^3 + sqrt(2)*(7*I*A - 5*I*B + 5*I*C)*cos(d*x + c)^2)*weierstrassZeta (-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(sq rt(2)*(-7*I*A + 5*I*B - 5*I*C)*cos(d*x + c)^3 + sqrt(2)*(-7*I*A + 5*I*B - 5*I*C)*cos(d*x + c)^2)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c os(d*x + c) - I*sin(d*x + c))) - 2*(9*(7*A - 5*B + 5*C)*cos(d*x + c)^3 + 2 *(19*A - 10*B + 15*C)*cos(d*x + c)^2 - 2*(2*A - 5*B)*cos(d*x + c) + 6*A)*s in(d*x + c)/sqrt(cos(d*x + c)))/(a*d*cos(d*x + c)^3 + a*d*cos(d*x + c)^2)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c) ),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/(a*co s(d*x + c) + a), x)
\[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {7}{2}}}{a \cos \left (d x + c\right ) + a} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+a*cos(d*x+c) ),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^(7/2)/(a*co s(d*x + c) + a), x)
Timed out. \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+a \cos (c+d x)} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{7/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{a+a\,\cos \left (c+d\,x\right )} \,d x \]