Integrand size = 35, antiderivative size = 274 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=-\frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a b \left (a^2-b^2\right ) d}-\frac {\left (A b^2-a^2 C+2 b^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}-\frac {\left (A b^4+a^4 C-3 a^2 b^2 (A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a (a-b) b^2 (a+b)^2 d}+\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \]
(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2)-( A*b^2+C*a^2)*(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/b/(a^2-b^2)/d -(A*b^2-C*a^2+2*C*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ell ipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^2/( a^2-b^2)/d-(A*b^4+a^4*C-3*a^2*b^2*(A+C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos( 1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*cos(d*x+c) ^(1/2)*sec(d*x+c)^(1/2)/a/(a-b)/b^2/(a+b)^2/d
Time = 5.56 (sec) , antiderivative size = 488, normalized size of antiderivative = 1.78 \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {4 a b^2 \left (A b^2+a^2 C\right ) \sin (c+d x)}{a^2-b^2}+\frac {\cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \sec (2 (c+d x)) \left (-8 a^2 b^2 (A+C) \cos (2 (c+d x)) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+\frac {2 b^2 \left (3 A b^2-a^2 (4 A+C)\right ) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) \left (-2+\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{\sec ^{\frac {3}{2}}(c+d x)}+\left (A b^2+a^2 C\right ) \cos (2 (c+d x)) \left (4 a b-4 a b \sec ^2(c+d x)+4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{(a-b) (a+b)}}{4 a^2 b^2 d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \]
((4*a*b^2*(A*b^2 + a^2*C)*Sin[c + d*x])/(a^2 - b^2) + (Cos[c + d*x]*Cot[c + d*x]*(b + a*Sec[c + d*x])*Sec[2*(c + d*x)]*(-8*a^2*b^2*(A + C)*Cos[2*(c + d*x)]*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d* x]]*Sqrt[-Tan[c + d*x]^2] + (2*b^2*(3*A*b^2 - a^2*(4*A + C))*(EllipticF[Ar cSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x ]]], -1])*(-2 + Sec[c + d*x]^2)*Sqrt[-Tan[c + d*x]^2])/Sec[c + d*x]^(3/2) + (A*b^2 + a^2*C)*Cos[2*(c + d*x)]*(4*a*b - 4*a*b*Sec[c + d*x]^2 + 4*a*b*E llipticE[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec [c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Se c[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 2*b^2*Ellipti cPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2])))/((a - b)*(a + b)))/(4*a^2*b^2*d*(a + b*Cos[c + d*x])*Sqrt[Se c[c + d*x]])
Time = 1.53 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.82, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4709, 3042, 3535, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}
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.
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+C \cos (c+d x)^2\right )}{(a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \cos ^2(c+d x)+A}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3535 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int -\frac {-\left ((2 A+C) a^2\right )+2 b (A+C) \cos (c+d x) a+A b^2+\left (C a^2+A b^2\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((2 A+C) a^2\right )+2 b (A+C) \cos (c+d x) a+A b^2+\left (C a^2+A b^2\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\int \frac {-\left ((2 A+C) a^2\right )+2 b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+A b^2+\left (C a^2+A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (a^2 C+A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {b \left (A b^2-a^2 (2 A+C)\right )+a \left (-C a^2+A b^2+2 b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (a^2 C+A b^2\right ) \int \sqrt {\cos (c+d x)}dx}{b}+\frac {\int \frac {b \left (A b^2-a^2 (2 A+C)\right )+a \left (-C a^2+A b^2+2 b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\left (a^2 C+A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {\int \frac {b \left (A b^2-a^2 (2 A+C)\right )+a \left (-C a^2+A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\int \frac {b \left (A b^2-a^2 (2 A+C)\right )+a \left (-C a^2+A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 \left (a^2 C+A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}+\frac {\left (a^4 C-3 a^2 b^2 (A+C)+A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{b}+\frac {2 \left (a^2 C+A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}+\frac {\left (a^4 C-3 a^2 b^2 (A+C)+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}}{b}+\frac {2 \left (a^2 C+A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {\left (a^4 C-3 a^2 b^2 (A+C)+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}+\frac {2 \left (a^2 C+A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 a \left (a^2-b^2\right )}\right )\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {2 \left (a^2 C+A b^2\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}+\frac {\frac {2 a \left (a^2 (-C)+A b^2+2 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}+\frac {2 \left (a^4 C-3 a^2 b^2 (A+C)+A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}}{b}}{2 a \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((2*(A*b^2 + a^2*C)*EllipticE[ (c + d*x)/2, 2])/(b*d) + ((2*a*(A*b^2 - a^2*C + 2*b^2*C)*EllipticF[(c + d* x)/2, 2])/(b*d) + (2*(A*b^4 + a^4*C - 3*a^2*b^2*(A + C))*EllipticPi[(2*b)/ (a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/(a*(a^2 - b^2)) + ((A*b^2 + a^ 2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x] )))
3.14.100.3.1 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[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ B/d Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d Int[(a + b* Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
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*b^2 + a^2*C))*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*((c + d*S in[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin [e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d *(A*b^2 + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d) Int[Sqrt[a + b*Sin[e + f*x]], x] , x] - Simp[1/(b*d) Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 ] && NeQ[c^2 - d^2, 0]
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. \(803\) vs. \(2(338)=676\).
Time = 4.42 (sec) , antiderivative size = 804, normalized size of antiderivative = 2.93
-(-(-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^2*(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))+ 8/b*a*C/(-2*a*b+2*b^2)*(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)*EllipticP i(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2/b^2*(A*b^2+C*a^2)*(-b^2/a/(a^2- b^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*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/2/(a+b)/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)/(-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))-1/2*b/a/(a^2-b^2)*(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) )+1/2*b/a/(a^2-b^2)*(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)*EllipticE(co s(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^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)/(-2*sin(1/2*d*x+1/2*c)^4+sin (1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)) +1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^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 )*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c...
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]