Integrand size = 35, antiderivative size = 229 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {(A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 (A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(A+7 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {2 (A+5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}-\frac {(A+7 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
2/3*(A+5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-1/3*(A+7*C)*sec(d*x+c)^(5/2) *sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/( a+a*sec(d*x+c))^2-(A+7*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^2/d+(A+7*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d+2/3*(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))*cos( d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^2/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.71 (sec) , antiderivative size = 884, normalized size of antiderivative = 3.86 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right )}{3 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}-\frac {14 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \csc \left (\frac {c}{2}\right ) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right ) \sec \left (\frac {c}{2}\right ) \left (A+C \sec ^2(c+d x)\right )}{3 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {8 A \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{3 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {40 C \cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\cos (c+d x)} \csc \left (\frac {c}{2}\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sec \left (\frac {c}{2}\right ) \sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \sin (c)}{3 d (A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2}+\frac {\cos ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sqrt {\sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \left (-\frac {4 (A+7 C) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}+\frac {4 \sec \left (\frac {c}{2}\right ) \sec ^3\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 )}{3 d}+\frac {16 \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 )+4 C \sin \left (\frac {d x}{2}\right )\right )}{3 d}+\frac {16 C \sec (c) \sec (c+d x) \sin (d x)}{3 d}+\frac {16 (C+A \cos (c)+5 C \cos (c)) \sec (c) \tan \left (\frac {c}{2}\right )}{3 d}+\frac {4 (A+C) \sec ^2\left (\frac {c}{2}+\frac {d x}{2}\right ) \tan \left (\frac {c}{2}\right )}{3 d}\right )}{(A+2 C+A \cos (2 c+2 d x)) (a+a \sec (c+d x))^2} \]
(-2*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( (2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4 , -E^((2*I)*(c + d*x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*( A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) - (14*Sqrt[2]*C*Sqrt [E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]* Cos[c/2 + (d*x)/2]^4*Csc[c/2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I) *d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d *x))])*Sec[c/2]*(A + C*Sec[c + d*x]^2))/(3*d*E^(I*d*x)*(A + 2*C + A*Cos[2* c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (8*A*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Se c[c + d*x])^2) + (40*C*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*El lipticF[(c + d*x)/2, 2]*Sec[c/2]*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2) *Sin[c])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^2) + (Co s[c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((-4*(A + 7*C )*Cos[d*x]*Csc[c/2]*Sec[c/2])/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[ (d*x)/2] + C*Sin[(d*x)/2]))/(3*d) + (16*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin [(d*x)/2] + 4*C*Sin[(d*x)/2]))/(3*d) + (16*C*Sec[c]*Sec[c + d*x]*Sin[d*x]) /(3*d) + (16*(C + A*Cos[c] + 5*C*Cos[c])*Sec[c]*Tan[c/2])/(3*d) + (4*(A...
Time = 1.17 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4573, 27, 3042, 4507, 27, 3042, 4274, 3042, 4255, 3042, 4258, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {\sec ^{\frac {5}{2}}(c+d x) (a (A-5 C)+3 a (A+3 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (a (A-5 C)+3 a (A+3 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a (A-5 C)+3 a (A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int -3 \sec ^{\frac {3}{2}}(c+d x) \left (a^2 (A+7 C)-2 a^2 (A+5 C) \sec (c+d x)\right )dx}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {3 \int \sec ^{\frac {3}{2}}(c+d x) \left (a^2 (A+7 C)-2 a^2 (A+5 C) \sec (c+d x)\right )dx}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a^2 (A+7 C)-2 a^2 (A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \int \sec ^{\frac {3}{2}}(c+d x)dx-2 a^2 (A+5 C) \int \sec ^{\frac {5}{2}}(c+d x)dx\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-2 a^2 (A+5 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )-2 a^2 (A+5 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-2 a^2 (A+5 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-2 a^2 (A+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-2 a^2 (A+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-2 a^2 (A+5 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}\right )\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {-\frac {3 \left (a^2 (A+7 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-2 a^2 (A+5 C) \left (\frac {2 \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )\right )}{a^2}-\frac {2 (A+7 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
-1/3*((A + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + ((-2*(A + 7*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])) - (3*(a^2*(A + 7*C)*((-2*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[ Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d) - 2*a^2*(A + 5*C )*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3* d) + (2*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(3*d))))/a^2)/(6*a^2)
3.3.37.3.1 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[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
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_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*( 2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)* (d*Csc[e + f*x])^(n - 1)*Simp[A*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && G tQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. ))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) *(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(710\) vs. \(2(257)=514\).
Time = 22.31 (sec) , antiderivative size = 711, normalized size of antiderivative = 3.10
-1/2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a^2*(4*C*(c os(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)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c), 2^(1/2)))-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^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)+1/3*(A+C)*(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*EllipticF(cos(1/2 *d*x+1/2*c),2^(1/2))-3*EllipticE(cos(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*(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*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*EllipticE(cos (1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-12*sin(1/2*d*x+1/2*c)^6+20*si n(1/2*d*x+1/2*c)^4-7*sin(1/2*d*x+1/2*c)^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)/(-1+sin(1/2*d*x+1/2*c)^2)+4*C*(-1 /6*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) /(cos(1/2*d*x+1/2*c)^2-1/2)^2+1/3*(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 )*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))-8*C/sin(1/2*d*x+1/2*c)^2/(2*sin(1 /2*d*x+1/2*c)^2-1)*(-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)*sin(1/2*d*x+1/2*c)^2-(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli pticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(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 = 417, normalized size of antiderivative = 1.82 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (\sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 5 i \, C\right )} \cos \left (d x + 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 - 7 i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A - 7 i \, C\right )} \cos \left (d x + 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 + 7 i \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, \sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A + 7 i \, C\right )} \cos \left (d x + 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 ) + \frac {2 \, {\left (3 \, {\left (A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (A + 8 \, C\right )} \cos \left (d x + c\right )^{2} + 8 \, C \cos \left (d x + c\right ) - 2 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \]
-1/6*(2*(sqrt(2)*(I*A + 5*I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(I*A + 5*I*C)*co s(d*x + c)^2 + sqrt(2)*(I*A + 5*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*(sqrt(2)*(-I*A - 5*I*C)*cos(d*x + c )^3 + 2*sqrt(2)*(-I*A - 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 5*I*C)*cos (d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*( sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A - 7*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weiers trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A + 7 *I*C)*cos(d*x + c)^3 + 2*sqrt(2)*(I*A + 7*I*C)*cos(d*x + c)^2 + sqrt(2)*(I *A + 7*I*C)*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0 , cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(A + 7*C)*cos(d*x + c)^3 + 4*(A + 8*C)*cos(d*x + c)^2 + 8*C*cos(d*x + c) - 2*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]