Integrand size = 43, antiderivative size = 133 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx=\frac {(A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(2 A+B+2 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {(A-C) \sqrt {\cos (c+d x)} \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
(A-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))/a^2/d+1/3*(2*A+B+2*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))/a^2/d-(A-C)*sin(d*x+c) *cos(d*x+c)^(1/2)/a^2/d/(1+cos(d*x+c))-1/3*(A-B+C)*sin(d*x+c)*cos(d*x+c)^( 1/2)/d/(a+a*cos(d*x+c))^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.12 (sec) , antiderivative size = 1101, normalized size of antiderivative = 8.28 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx =\text {Too large to display} \]
Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2),x]
(-4*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Si n[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan [Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x]) ^2*Sqrt[1 + Cot[c]^2]) - (2*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Hypergeometric PFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - Arc Tan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2] *Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/( 3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) - (4*C*Cos[c/2 + (d*x)/2]^4 *Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2 ]*Sec[c/2]*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*S qrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d *x - ArcTan[Cot[c]]]])/(3*d*(a + a*Cos[c + d*x])^2*Sqrt[1 + Cot[c]^2]) + ( Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*((-4*(A - C)*Csc[c])/d - (4*Sec[c/ 2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] - C*Sin[(d*x)/2]))/d - (2*Sec[c/2]*S ec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] - B*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(3 *d) - (2*(A - B + C)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/(3*d)))/(a + a*Cos[c + d*x])^2 - (A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{ -1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]] *Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan...
Time = 0.84 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {3042, 3520, 27, 3042, 3457, 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.
| \(\displaystyle \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3520 |
| \(\displaystyle \frac {\int \frac {a (5 A+B-C)-a (A-B-5 C) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (5 A+B-C)-a (A-B-5 C) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (5 A+B-C)-a (A-B-5 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {(2 A+B+2 C) a^2+3 (A-C) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(2 A+B+2 C) a^2+3 (A-C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {a^2 (2 A+B+2 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^2 (A-C) \int \sqrt {\cos (c+d x)}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (2 A+B+2 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 (A-C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {a^2 (2 A+B+2 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (2 A+B+2 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^2 (A-C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}-\frac {6 (A-C) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]) ^2) + (((6*a^2*(A - C)*EllipticE[(c + d*x)/2, 2])/d + (2*a^2*(2*A + B + 2* C)*EllipticF[(c + d*x)/2, 2])/d)/a^2 - (6*(A - C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*(1 + Cos[c + d*x])))/(6*a^2)
3.5.66.3.1 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[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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] && (IntegerQ[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)]
Leaf count of result is larger than twice the leaf count of optimal. \(506\) vs. \(2(177)=354\).
Time = 7.50 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.81
| method | result | size |
| default | \(\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (12 A \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 A \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 )+6 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}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 B \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 )-12 C \left (\cos ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 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 )-6 \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 )-16 A \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 B \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 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 )+3 B \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 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 {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(507\) |
int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+cos(d*x+c)*a)^2/cos(d*x+c)^(1/2),x, method=_RETURNVERBOSE)
1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2* d*x+1/2*c)^6-4*A*(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+6*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-2*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))* cos(1/2*d*x+1/2*c)^3-12*C*cos(1/2*d*x+1/2*c)^6-4*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)^3-6*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 ))-16*A*cos(1/2*d*x+1/2*c)^4-2*B*cos(1/2*d*x+1/2*c)^4+20*C*cos(1/2*d*x+1/2 *c)^4+3*A*cos(1/2*d*x+1/2*c)^2+3*B*cos(1/2*d*x+1/2*c)^2-9*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)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.82 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, {\left (A - C\right )} \cos \left (d x + c\right ) + 4 \, A - B - 2 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - {\left (\sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - {\left (\sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/cos(d*x+c)^(1 /2),x, algorithm="fricas")
-1/6*(2*(3*(A - C)*cos(d*x + c) + 4*A - B - 2*C)*sqrt(cos(d*x + c))*sin(d* x + c) - (sqrt(2)*(-2*I*A - I*B - 2*I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(2*I*A + I*B + 2*I*C)*cos(d*x + c) + sqrt(2)*(-2*I*A - I*B - 2*I*C))*weierstrass PInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - (sqrt(2)*(2*I*A + I*B + 2 *I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(-2*I*A - I*B - 2*I*C)*cos(d*x + c) + sqr t(2)*(2*I*A + I*B + 2*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*si n(d*x + c)) + 3*(sqrt(2)*(-I*A + I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-I*A + I *C)*cos(d*x + c) + sqrt(2)*(-I*A + I*C))*weierstrassZeta(-4, 0, weierstras sPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A - I*C)* cos(d*x + c)^2 + 2*sqrt(2)*(I*A - I*C)*cos(d*x + c) + sqrt(2)*(I*A - I*C)) *weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/cos(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(cos(d*x + c))), x)
\[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\cos \left (d x + c\right )}} \,d x } \]
integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))^2/cos(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(cos(d*x + c))), x)
Timed out. \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{\sqrt {\cos \left (c+d\,x\right )}\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]