Integrand size = 33, antiderivative size = 242 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\frac {2 \left (3 a^2 A+5 A b^2-5 a b B\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^3 d}-\frac {2 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^4 d}+\frac {2 b^3 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^4 (a+b) d}+\frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}} \]
2/5*A*sin(d*x+c)/a/d/sec(d*x+c)^(3/2)-2/3*(A*b-B*a)*sin(d*x+c)/a^2/d/sec(d *x+c)^(1/2)+2/5*(3*A*a^2+5*A*b^2-5*B*a*b)*(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^3/d-2/3*(a^2+3*b^2)*(A*b-B*a)*(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^4/d+2*b^3*(A*b-B*a)*(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^4/(a+b)/d
Leaf count is larger than twice the leaf count of optimal. \(612\) vs. \(2(242)=484\).
Time = 7.87 (sec) , antiderivative size = 612, normalized size of antiderivative = 2.53 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\frac {\frac {2 \left (9 a^2 A+5 A b^2-5 a b B\right ) \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )-\operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {2 \left (8 a A b+10 a^2 B\right ) \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) (a+b \sec (c+d x)) \sqrt {1-\sec ^2(c+d x)} \sin (c+d x)}{a (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right )}+\frac {\left (9 a^2 A+15 A b^2-15 a b B\right ) \cos (2 (c+d x)) (a+b \sec (c+d x)) \left (-4 a b+4 a b \sec ^2(c+d x)-4 a b E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-2 a (a-2 b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}+2 a^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}-4 b^2 \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)}\right ) \sin (c+d x)}{a^2 b (b+a \cos (c+d x)) \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right )}}{30 a^2 d}+\frac {\sqrt {\sec (c+d x)} \left (\frac {A \sin (c+d x)}{10 a}+\frac {(-A b+a B) \sin (2 (c+d x))}{3 a^2}+\frac {A \sin (3 (c+d x))}{10 a}\right )}{d} \]
((2*(9*a^2*A + 5*A*b^2 - 5*a*b*B)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[Se c[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*(8*a*A*b + 10*a^2*B)*Cos[c + d*x]^2*Ell ipticPi[-(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)) + ((9*a^2*A + 15*A*b^2 - 15*a*b*B)*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)))/(30*a^2*d) + (Sqrt[Sec[c + d*x]]*((A*Sin[c + d*x ])/(10*a) + ((-(A*b) + a*B)*Sin[2*(c + d*x)])/(3*a^2) + (A*Sin[3*(c + d*x) ])/(10*a)))/d
Time = 1.91 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.05, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4522, 27, 3042, 4592, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 4522 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \int \frac {-3 A b \sec ^2(c+d x)-3 a A \sec (c+d x)+5 (A b-a B)}{2 \sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}dx}{5 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 A b \sec ^2(c+d x)-3 a A \sec (c+d x)+5 (A b-a B)}{\sec ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))}dx}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\int \frac {-3 A b \csc \left (c+d x+\frac {\pi }{2}\right )^2-3 a A \csc \left (c+d x+\frac {\pi }{2}\right )+5 (A b-a B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{5 a}\) |
| \(\Big \downarrow \) 4592 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {2 \int \frac {-5 b (A b-a B) \sec ^2(c+d x)+a (4 A b+5 a B) \sec (c+d x)+3 \left (3 A a^2-5 b B a+5 A b^2\right )}{2 \sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-5 b (A b-a B) \sec ^2(c+d x)+a (4 A b+5 a B) \sec (c+d x)+3 \left (3 A a^2-5 b B a+5 A b^2\right )}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))}dx}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\int \frac {-5 b (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A b+5 a B) \csc \left (c+d x+\frac {\pi }{2}\right )+3 \left (3 A a^2-5 b B a+5 A b^2\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 4594 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}+\frac {\int \frac {3 a \left (3 A a^2-5 b B a+5 A b^2\right )-5 \left (a^2+3 b^2\right ) (A b-a B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\int \frac {3 a \left (3 A a^2-5 b B a+5 A b^2\right )-5 \left (a^2+3 b^2\right ) (A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 \left (a^2+3 b^2\right ) (A b-a B) \int \sqrt {\sec (c+d x)}dx}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 \left (a^2+3 b^2\right ) (A b-a B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-5 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 4336 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}+\frac {\frac {6 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^3 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}+\frac {\frac {6 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{3 a}}{5 a}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {30 b^3 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}+\frac {\frac {6 a \left (3 a^2 A-5 a b B+5 A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 \left (a^2+3 b^2\right ) (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{3 a}}{5 a}\) |
(2*A*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (-1/3*(((6*a*(3*a^2*A + 5* A*b^2 - 5*a*b*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (10*(a^2 + 3*b^2)*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + (30*b^3*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2* (a + b)*d))/a + (10*(A*b - a*B)*Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]))/ (5*a)
3.5.21.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*n)), x] + Sim p[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*B* n - A*b*(m + n + 1) + A*a*(n + 1)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f* x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
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[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d *Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n) Int[(a + b*Csc[e + f*x])^m *(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d , e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
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. \(1073\) vs. \(2(296)=592\).
Time = 9.50 (sec) , antiderivative size = 1074, normalized size of antiderivative = 4.44
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((-24*A*a^4+ 24*A*a^3*b)*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(24*A*a^4-44*A*a^3*b+2 0*A*a^2*b^2+20*B*a^4-20*B*a^3*b)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+( -6*A*a^4+16*A*a^3*b-10*A*a^2*b^2-10*B*a^4+10*B*a^3*b)*sin(1/2*d*x+1/2*c)^2 *cos(1/2*d*x+1/2*c)-5*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c) ^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b+5*A*(sin(1/2*d*x+1 /2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2* c),2^(1/2))*a^2*b^2-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c )^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^3+15*A*(sin(1/2*d*x +1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/ 2*c),2^(1/2))*b^4-9*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2 -1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4+9*A*(sin(1/2*d*x+1/2*c )^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2 ^(1/2))*a^3*b-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1) ^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^2+15*A*(sin(1/2*d*x+1/2 *c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c) ,2^(1/2))*a*b^3-15*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2- 1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))*b^4+5*B*(sin(1/2 *d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d* x+1/2*c),2^(1/2))*a^4-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1...
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int \frac {A + B \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]