Integrand size = 43, antiderivative size = 299 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a b \left (a^2-b^2\right ) d}-\frac {\left (A b^2+a b B-a^2 (2 A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 \left (a^2-b^2\right ) d}+\frac {\left (A b^4+a^3 b B+a b^3 B+a^4 C-3 a^2 b^2 (A+C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^2 (a-b) b (a+b)^2 d}-\frac {\left (A b^2-a (b B-a C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
-(A*b^2-a*(B*b-C*a))*sin(d*x+c)*sec(d*x+c)^(1/2)/b/(a^2-b^2)/d/(a+b*sec(d* x+c))+(A*b^2-a*(B*b-C*a))*(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+B*a*b-a^2*(2*A+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/(a^2-b^2)/d+(A*b^4+B*a^3*b+B*a*b^3+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*a/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/(a-b)/b/(a+b) ^2/d
Leaf count is larger than twice the leaf count of optimal. \(824\) vs. \(2(299)=598\).
Time = 15.91 (sec) , antiderivative size = 824, normalized size of antiderivative = 2.76 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {(b+a \cos (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (-A b^2+a b B+3 a^2 C-4 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 {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (4 a A b-4 b^2 B+4 a b C\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (A b^2-a b B+a^2 C\right ) \cos (2 (c+d x)) (a+b \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 a (a-2 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)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\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^2 b (b+a \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 )}\right )}{2 (a-b) b (a+b) d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^2}+\frac {(b+a \cos (c+d x))^2 \sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {2 \left (A b^2-a b B+a^2 C\right ) \sin (c+d x)}{a b \left (-a^2+b^2\right )}+\frac {2 \left (A b^2 \sin (c+d x)-a b B \sin (c+d x)+a^2 C \sin (c+d x)\right )}{a \left (a^2-b^2\right ) (b+a \cos (c+d x))}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^2} \]
Integrate[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^2,x]
((b + a*Cos[c + d*x])^2*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(-(A*b ^2) + a*b*B + 3*a^2*C - 4*b^2*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Sec [c + d*x]]], -1] - EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(a + b*Sec[c + d*x])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(4*a*A*b - 4*b^2*B + 4*a*b*C)*Cos[c + d* x]^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(a + b*Sec[c + d*x ])*Sqrt[1 - Sec[c + d*x]^2]*Sin[c + d*x])/(a*(b + a*Cos[c + d*x])*(1 - Cos [c + d*x]^2)) + ((A*b^2 - a*b*B + a^2*C)*Cos[2*(c + d*x)]*(a + b*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*a*(a - 2*b)*El lipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*a^2*EllipticPi[-(b/a), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[ Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*b^2*EllipticPi[-(b/a), ArcSin[S qrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a^2*b*(b + a*Cos[c + d*x])*(1 - Cos[c + d*x]^2)*Sqrt[Sec[c + d*x ]]*(2 - Sec[c + d*x]^2))))/(2*(a - b)*b*(a + b)*d*(A + 2*C + 2*B*Cos[c + d *x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^2) + ((b + a*Cos[c + d*x])^ 2*Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((2*(A*b^2 - a*b*B + a^2*C)*Sin[c + d*x])/(a*b*(-a^2 + b^2)) + (2*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] + a^2*C*Sin[c + d*x]))/(a*(a^2 - b^2)*(b + a*Cos[c ...
Time = 1.82 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.96, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 4586, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 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.
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4586 |
| \(\displaystyle -\frac {\int -\frac {A b^2-2 (b B-a (A+C)) \sec (c+d x) b-\left (-C a^2-b B a+A b^2+2 b^2 C\right ) \sec ^2(c+d x)-a (b B-a C)}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {A b^2-2 (b B-a (A+C)) \sec (c+d x) b-\left (-C a^2-b B a+A b^2+2 b^2 C\right ) \sec ^2(c+d x)-a (b B-a C)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A b^2-2 (b B-a (A+C)) \csc \left (c+d x+\frac {\pi }{2}\right ) b+\left (C a^2+b B a-A b^2-2 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-a (b B-a C)}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4594 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (A b^2-a (b B-a C)\right )-b \left (-\left ((2 A+C) a^2\right )+b B a+A b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}+\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a \left (A b^2-a (b B-a C)\right )-b \left (-\left ((2 A+C) a^2\right )+b B a+A b^2\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-b \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}+\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-b \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \int \sqrt {\cos (c+d x)}dx-b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}+\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{d}-b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}+\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{d}-\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right )}{d}}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4336 |
| \(\displaystyle \frac {\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}+\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{d}-\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right )}{d}}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}+\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{d}-\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right )}{d}}{a^2}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {\frac {\frac {2 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (A b^2-a (b B-a C)\right )}{d}-\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (-\left (a^2 (2 A+C)\right )+a b B+A b^2\right )}{d}}{a^2}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^4 C+a^3 b B-3 a^2 b^2 (A+C)+a b^3 B+A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}}{2 b \left (a^2-b^2\right )}-\frac {\sin (c+d x) \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
(((2*a*(A*b^2 - a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2 ]*Sqrt[Sec[c + d*x]])/d - (2*b*(A*b^2 + a*b*B - a^2*(2*A + C))*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + (2*(A*b^4 + a^3*b*B + a*b^3*B + a^4*C - 3*a^2*b^2*(A + C))*Sqrt[Cos[c + d*x]]*Ellipti cPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*(a + b)*d))/(2 *b*(a^2 - b^2)) - ((A*b^2 - a*(b*B - a*C))*Sqrt[Sec[c + d*x]]*Sin[c + d*x] )/(b*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))
3.11.21.3.1 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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]] Int[ 1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-d)*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 1)/(b*f*(a^2 - b^2)*(m + 1)) ), x] + Simp[d/(b*(a^2 - b^2)*(m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*( d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1) + b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C }, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2) Int[(d*Csc[e + f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2 Int[(a*A - (A*b - a *B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(808\) vs. \(2(363)=726\).
Time = 3.86 (sec) , antiderivative size = 809, normalized size of antiderivative = 2.71
int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,x, method=_RETURNVERBOSE)
-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*A/a^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))- 2*(-2*A*b+B*a)/a/(a^2-a*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)*Ellip ticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))+2/a^2*(A*b^2-B*a*b+C*a^2)*(a^2 /b/(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*a-a+b)-1/2/(a+b)/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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*a/b/(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*a/b/(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)*Elli pticE(cos(1/2*d*x+1/2*c),2^(1/2))-1/2/b/(a^2-b^2)/(a^2-a*b)*a^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*a/(a-b),2^ (1/2))+3/2*b/(a^2-b^2)/(a^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)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c...
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="fricas")
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sqrt(sec(c + d*x))/(a + b*sec(c + d*x))**2, x)
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="maxima")
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c) )^2,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(b*se c(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
int(((1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^2,x)