Integrand size = 35, antiderivative size = 277 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 \left (a^2-b^2\right ) d}+\frac {a \left (A b^2-3 a^2 C+4 b^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^4-3 a^4 C+a^2 b^2 (A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{(a-b) b^3 (a+b)^2 d}-\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \]
-(A*b^2+C*a^2)*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2)+ (A*b^2+3*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)*El lipticE(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*(A*b^2-3*C*a^2+4*C*b^2)*(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)/b^3/(a^2-b^2)/d-(A*b^4-3*a^4*C+a^2*b^2*(A+5*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-b)/b^3/(a+b)^2/d
Leaf count is larger than twice the leaf count of optimal. \(657\) vs. \(2(277)=554\).
Time = 7.31 (sec) , antiderivative size = 657, normalized size of antiderivative = 2.37 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\frac {\frac {2 \left (-A b^2+a^2 C-2 b^2 C\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 (4 a A b+4 a b C) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (b+a \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) \cos (2 (c+d x)) (b+a \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 (2 a-b) b \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 a^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a b^2 (a+b \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{4 (a-b) b (a+b) d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {\left (A b^2+a^2 C\right ) \sin (c+d x)}{b^2 \left (-a^2+b^2\right )}+\frac {-a A b^2 \sin (c+d x)-a^3 C \sin (c+d x)}{b^2 \left (-a^2+b^2\right ) (a+b \cos (c+d x))}\right )}{d} \]
((2*(-(A*b^2) + a^2*C - 2*b^2*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec [c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(4*a*A*b + 4*a*b*C)*Cos[c + d*x]^2*Ellip ticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^ 2)) + ((A*b^2 + 3*a^2*C - 2*b^2*C)*Cos[2*(c + d*x)]*(b + a*Sec[c + d*x])*( -4*a*b + 4*a*b*Sec[c + d*x]^2 - 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]] , -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*(2*a - b)*b*Elliptic F[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x] ^2] - 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Se c[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x ])/(a*b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(4*(a - b)*b*(a + b)*d) + (Sqrt[Sec[c + d*x]]*(((A*b^2 + a^2*C)*Sin[c + d*x])/(b^2*(-a^2 + b^2)) + (-(a*A*b^2*Sin[c + d*x]) - a^ 3*C*Sin[c + d*x])/(b^2*(-a^2 + b^2)*(a + b*Cos[c + d*x]))))/d
Time = 1.57 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.85, 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, 3527, 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.
| \(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \cos (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \cos (c+d x)^2}{\sqrt {\sec (c+d x)} (a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 4709 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\cos (c+d x)} \left (C \cos ^2(c+d x)+A\right )}{(a+b \cos (c+d x))^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+A\right )}{\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {C a^2-2 b (A+C) \cos (c+d x) a+A b^2-\left (3 C a^2+A b^2-2 b^2 C\right ) \cos ^2(c+d x)}{2 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {C a^2-2 b (A+C) \cos (c+d x) a+A b^2-\left (3 C a^2+A b^2-2 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {C a^2-2 b (A+C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+A b^2+\left (-3 C a^2-A b^2+2 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3538 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {-\frac {\left (3 a^2 C+A b^2-2 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {b \left (C a^2+A b^2\right )-a \left (-3 C a^2+A b^2+4 b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {b \left (C a^2+A b^2\right )-a \left (-3 C a^2+A b^2+4 b^2 C\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {\left (3 a^2 C+A b^2-2 b^2 C\right ) \int \sqrt {\cos (c+d x)}dx}{b}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {b \left (C a^2+A b^2\right )-a \left (-3 C a^2+A b^2+4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {\left (3 a^2 C+A b^2-2 b^2 C\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int \frac {b \left (C a^2+A b^2\right )-a \left (-3 C a^2+A b^2+4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3481 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\frac {\left (-3 a^4 C+a^2 b^2 (A+5 C)+A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {a \left (-3 a^2 C+A b^2+4 b^2 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\frac {\left (-3 a^4 C+a^2 b^2 (A+5 C)+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {a \left (-3 a^2 C+A b^2+4 b^2 C\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\frac {\left (-3 a^4 C+a^2 b^2 (A+5 C)+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 a \left (-3 a^2 C+A b^2+4 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\right )\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {2 \left (-3 a^4 C+a^2 b^2 (A+5 C)+A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}-\frac {2 a \left (-3 a^2 C+A b^2+4 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/2*((-2*(A*b^2 + 3*a^2*C - 2*b^2* C)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((-2*a*(A*b^2 - 3*a^2*C + 4*b^2*C)*E llipticF[(c + d*x)/2, 2])/(b*d) + (2*(A*b^4 - 3*a^4*C + a^2*b^2*(A + 5*C)) *EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d))/b)/(b*(a^2 - b^ 2)) - ((A*b^2 + a^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*( a + b*Cos[c + d*x])))
3.15.1.3.1 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[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
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. \(833\) vs. \(2(341)=682\).
Time = 5.08 (sec) , antiderivative size = 834, normalized size of antiderivative = 3.01
-(-(-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^3/(-2*s in(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)^(1/ 2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*a*EllipticF(cos(1/2*d*x+1/2*c),2^(1 /2))+b*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-4/b^2*(A*b^2+3*C*a^2)/(-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)*EllipticPi(cos(1/2*d*x+ 1/2*c),-2*b/(a-b),2^(1/2))-2*a*(A*b^2+C*a^2)/b^3*(-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+si n(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(cos(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)*Ellipti...
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
integral((C*cos(d*x + c)^2 + A)/((b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)*sqrt(sec(d*x + c))), x)
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int \frac {A + C \cos ^{2}{\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2} \sqrt {\sec {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x))^2 \sqrt {\sec (c+d x)}} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]