Integrand size = 43, antiderivative size = 179 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=-\frac {(B-4 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(A+2 B-5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A-B+C) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x)}+\frac {(A+2 B-5 C) \sin (c+d x)}{3 a^2 d (1+\cos (c+d x)) \sqrt {\sec (c+d x)}} \]
-1/3*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^2/sec(d*x+c)^(3/2)+1/3*(A+2*B-5 *C)*sin(d*x+c)/a^2/d/(1+cos(d*x+c))/sec(d*x+c)^(1/2)-(B-4*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))*c os(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d+1/3*(A+2*B-5*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^2/d
Time = 3.72 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (6 (B-4 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )-2 (A+2 B-5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\frac {1}{2} (A+2 B-5 C+3 (B-2 C) \cos (c+d x)) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + a*Cos[c + d*x])^2* Sqrt[Sec[c + d*x]]),x]
(-2*Cos[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]]*(6*(B - 4*C)*Sqrt[Cos[c + d*x]]* EllipticE[(c + d*x)/2, 2] - 2*(A + 2*B - 5*C)*Sqrt[Cos[c + d*x]]*EllipticF [(c + d*x)/2, 2] + ((A + 2*B - 5*C + 3*(B - 2*C)*Cos[c + d*x])*Sec[(c + d* x)/2]^3*(Sin[(c + d*x)/2] - Sin[(3*(c + d*x))/2]))/2))/(3*a^2*d*(1 + Cos[c + d*x])^2)
Time = 1.03 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4709, 3042, 3520, 27, 3042, 3456, 3042, 3227, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos (c+d x)^2}{\sqrt {\sec (c+d x)} (a \cos (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)} \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )}{(\cos (c+d x) a+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a (A+B-C)+a (A-B+7 C) \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\cos (c+d x)} (3 a (A+B-C)+a (A-B+7 C) \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (3 a (A+B-C)+a (A-B+7 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (A+2 B-5 C)-3 a^2 (B-4 C) \cos (c+d x)}{\sqrt {\cos (c+d x)}}dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\int \frac {a^2 (A+2 B-5 C)-3 a^2 (B-4 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (A+2 B-5 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^2 (B-4 C) \int \sqrt {\cos (c+d x)}dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (A+2 B-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^2 (B-4 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {a^2 (A+2 B-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {6 a^2 (B-4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\frac {2 a^2 (A+2 B-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-\frac {6 a^2 (B-4 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/3*((A - B + C)*Cos[c + d*x]^(3/2 )*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) + (((-6*a^2*(B - 4*C)*EllipticE [(c + d*x)/2, 2])/d + (2*a^2*(A + 2*B - 5*C)*EllipticF[(c + d*x)/2, 2])/d) /a^2 + (2*(A + 2*B - 5*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(1 + Cos[c + d*x])))/(6*a^2))
3.13.100.3.1 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[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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{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[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (In tegerQ[2*n] || EqQ[c, 0])
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. \(506\) vs. \(2(213)=426\).
Time = 3.72 (sec) , antiderivative size = 507, normalized size of antiderivative = 2.83
| method | result | size |
| default | \(-\frac {\sqrt {\left (-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 A \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 B \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+4 B \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 B \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-24 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-10 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-24 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A -20 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+38 C \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 A \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+9 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-15 C \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+A -B +C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-1+2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(507\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1/2),x, method=_RETURNVERBOSE)
-1/6*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A*cos(1/2 *d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*B*cos(1/2*d*x+1/2*c)^6+4*B*cos (1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*B*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1 /2*d*x+1/2*c),2^(1/2))-24*C*cos(1/2*d*x+1/2*c)^6-10*C*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^ (1/2))*cos(1/2*d*x+1/2*c)^3-24*cos(1/2*d*x+1/2*c)^3*C*(sin(1/2*d*x+1/2*c)^ 2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^ (1/2))+2*cos(1/2*d*x+1/2*c)^4*A-20*B*cos(1/2*d*x+1/2*c)^4+38*C*cos(1/2*d*x +1/2*c)^4-3*A*cos(1/2*d*x+1/2*c)^2+9*B*cos(1/2*d*x+1/2*c)^2-15*C*cos(1/2*d *x+1/2*c)^2+A-B+C)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(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 = 381, normalized size of antiderivative = 2.13 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {{\left (\sqrt {2} {\left (-i \, A - 2 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (i \, A + 2 i \, B - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 2 i \, B + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A + 2 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-i \, A - 2 i \, B + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 2 i \, B - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, B - 4 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, B - 4 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, B - 4 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, B + 4 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, B + 4 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, B + 4 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, {\left (B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (A + 2 \, B - 5 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1 /2),x, algorithm="fricas")
1/6*((sqrt(2)*(-I*A - 2*I*B + 5*I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(I*A + 2*I *B - 5*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 2*I*B + 5*I*C))*weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A + 2*I*B - 5*I*C )*cos(d*x + c)^2 - 2*sqrt(2)*(-I*A - 2*I*B + 5*I*C)*cos(d*x + c) + sqrt(2) *(I*A + 2*I*B - 5*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c)) - 3*(sqrt(2)*(I*B - 4*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*B - 4*I*C )*cos(d*x + c) + sqrt(2)*(I*B - 4*I*C))*weierstrassZeta(-4, 0, weierstrass PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-I*B + 4*I*C )*cos(d*x + c)^2 + 2*sqrt(2)*(-I*B + 4*I*C)*cos(d*x + c) + sqrt(2)*(-I*B + 4*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(B - 2*C)*cos(d*x + c)^2 + (A + 2*B - 5*C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*co s(d*x + c) + a^2*d)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\int \frac {A}{\cos ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + 2 \cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \cos {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + 2 \cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \cos ^{2}{\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + 2 \cos {\left (c + d x \right )} \sqrt {\sec {\left (c + d x \right )}} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{2}} \]
(Integral(A/(cos(c + d*x)**2*sqrt(sec(c + d*x)) + 2*cos(c + d*x)*sqrt(sec( c + d*x)) + sqrt(sec(c + d*x))), x) + Integral(B*cos(c + d*x)/(cos(c + d*x )**2*sqrt(sec(c + d*x)) + 2*cos(c + d*x)*sqrt(sec(c + d*x)) + sqrt(sec(c + d*x))), x) + Integral(C*cos(c + d*x)**2/(cos(c + d*x)**2*sqrt(sec(c + d*x )) + 2*cos(c + d*x)*sqrt(sec(c + d*x)) + sqrt(sec(c + d*x))), x))/a**2
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1 /2),x, algorithm="maxima")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2* sqrt(sec(d*x + c))), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/sec(d*x+c)^(1 /2),x, algorithm="giac")
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((a*cos(d*x + c) + a)^2* sqrt(sec(d*x + c))), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+a \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]
int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(1/2)*(a + a *cos(c + d*x))^2),x)