Integrand size = 35, antiderivative size = 122 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=-\frac {(A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}+\frac {(3 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {(3 A+5 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a d}-\frac {(A+C) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{d (a+a \cos (c+d x))} \]
-(A+3*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/d+1/3*(3*A+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))/a/d-(A+C)*cos(d*x+c)^(3 /2)*sin(d*x+c)/d/(a+a*cos(d*x+c))+1/3*(3*A+5*C)*sin(d*x+c)*cos(d*x+c)^(1/2 )/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.87 (sec) , antiderivative size = 890, normalized size of antiderivative = 7.30 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {\cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \left (\frac {2 (A+C+2 C \cos (c)) \csc (c)}{d}+\frac {4 C \cos (d x) \sin (c)}{3 d}+\frac {2 \sec \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}+\frac {d x}{2}\right ) \left (A \sin \left (\frac {d x}{2}\right )+C \sin \left (\frac {d x}{2}\right )\right )}{d}+\frac {4 C \cos (c) \sin (d x)}{3 d}\right )}{a+a \cos (c+d x)}-\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d (a+a \cos (c+d x)) \sqrt {1+\cot ^2(c)}}-\frac {5 C \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec \left (\frac {c}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d (a+a \cos (c+d x)) \sqrt {1+\cot ^2(c)}}+\frac {A \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (a+a \cos (c+d x))}+\frac {3 C \cos ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d (a+a \cos (c+d x))} \]
(Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*((2*(A + C + 2*C*Cos[c])*Csc[c])/ d + (4*C*Cos[d*x]*Sin[c])/(3*d) + (2*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d *x)/2] + C*Sin[(d*x)/2]))/d + (4*C*Cos[c]*Sin[d*x])/(3*d)))/(a + a*Cos[c + d*x]) - (A*Cos[c/2 + (d*x)/2]^2*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]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(a + a*Cos[c + d *x])*Sqrt[1 + Cot[c]^2]) - (5*C*Cos[c/2 + (d*x)/2]^2*Csc[c/2]*Hypergeometr icPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[d*x - A rcTan[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])*Sqrt[1 + Cot[c]^2]) + (A*Cos[c/2 + (d*x)/2]^2*C sc[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[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan [Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan [c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*S qrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan [c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d*(a + a*Cos[c + d*x])) + (3*C*Cos[c/2 + (d *x)/2]^2*Csc[c/2]*Sec[c/2]*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos...
Time = 0.62 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 3521, 27, 3042, 3227, 3042, 3115, 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.
| \(\displaystyle \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a \cos (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a \sin \left (c+d x+\frac {\pi }{2}\right )+a}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int -\frac {1}{2} \sqrt {\cos (c+d x)} (a (A+3 C)-a (3 A+5 C) \cos (c+d x))dx}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \sqrt {\cos (c+d x)} (a (A+3 C)-a (3 A+5 C) \cos (c+d x))dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a (A+3 C)-a (3 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {a (A+3 C) \int \sqrt {\cos (c+d x)}dx-a (3 A+5 C) \int \cos ^{\frac {3}{2}}(c+d x)dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a (A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a (3 A+5 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {a (A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a (A+3 C) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {2 a (A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {2 a (A+3 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a (3 A+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\) |
-(((A + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x]))) - (( 2*a*(A + 3*C)*EllipticE[(c + d*x)/2, 2])/d - a*(3*A + 5*C)*((2*EllipticF[( c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/(2*a^2 )
3.2.53.3.1 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[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(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)) I nt[(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)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(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, 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)]
Time = 6.21 (sec) , antiderivative size = 262, normalized size of antiderivative = 2.15
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (3 A F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 C E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 A +18 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-3 A -7 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) | \(262\) |
-1/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(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)*(3*A* EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*EllipticE(cos(1/2*d*x+1/2*c),2^( 1/2))+5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*C*EllipticE(cos(1/2*d*x+ 1/2*c),2^(1/2)))-8*C*sin(1/2*d*x+1/2*c)^6+(6*A+18*C)*sin(1/2*d*x+1/2*c)^4+ (-3*A-7*C)*sin(1/2*d*x+1/2*c)^2)/a/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)/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 = 250, normalized size of antiderivative = 2.05 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\frac {2 \, {\left (2 \, C \cos \left (d x + c\right ) + 3 \, A + 5 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/6*(2*(2*C*cos(d*x + c) + 3*A + 5*C)*sqrt(cos(d*x + c))*sin(d*x + c) + (s qrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A - 5*I*C))*weierstra ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(3*I*A + 5*I*C )*cos(d*x + c) + sqrt(2)*(3*I*A + 5*I*C))*weierstrassPInverse(-4, 0, cos(d *x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + c) + sqrt(2 )*(I*A + 3*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c) + sqrt(2 )*(-I*A - 3*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* x + c) - I*sin(d*x + c))))/(a*d*cos(d*x + c) + a*d)
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{a \cos \left (d x + c\right ) + a} \,d x } \]
\[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {\cos \left (d x + c\right )}}{a \cos \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{a+a \cos (c+d x)} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{a+a\,\cos \left (c+d\,x\right )} \,d x \]