Integrand size = 43, antiderivative size = 269 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {(A+9 B-49 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(A+3 B-13 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A+9 B-49 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a^3 d}-\frac {(A-B+C) \sec ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(2 A+3 B-8 C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(A+3 B-13 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \]
-1/5*(A-B+C)*sec(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(2*A+3* B-8*C)*sec(d*x+c)^(5/2)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2+1/6*(A+3*B-13*C) *sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))-1/10*(A+9*B-49*C)*sin( d*x+c)*sec(d*x+c)^(1/2)/a^3/d+1/10*(A+9*B-49*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^3/d+1/6*(A+3*B-13*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co s(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)*se c(d*x+c)^(1/2)/a^3/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 9.40 (sec) , antiderivative size = 1430, normalized size of antiderivative = 5.32 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \]
Integrate[(Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]
(-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]^6*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]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec [c + d*x]^2))/(15*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2* d*x])*(a + a*Sec[c + d*x])^3) - (6*Sqrt[2]*B*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]^6*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]*Sec[c + d *x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(5*d*E^(I*d*x)*(A + 2*C + 2*B *Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (98*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]^6*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]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2) )/(15*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a *Sec[c + d*x])^3) + (4*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]* EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])^3) + (4*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c...
Time = 1.59 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.02, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 4572, 27, 3042, 4507, 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.
\(\displaystyle \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 4572 |
\(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a (A+B-C)+a (A-B+11 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sec ^{\frac {5}{2}}(c+d x) (5 a (A+B-C)+a (A-B+11 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (5 a (A+B-C)+a (A-B+11 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle \frac {\frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (3 (2 A+3 B-8 C) a^2+(A-6 B+41 C) \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 (2 A+3 B-8 C) a^2+(A-6 B+41 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4507 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{2} \sqrt {\sec (c+d x)} \left (5 a^3 (A+3 B-13 C)-3 a^3 (A+9 B-49 C) \sec (c+d x)\right )dx}{a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\frac {\int \sqrt {\sec (c+d x)} \left (5 a^3 (A+3 B-13 C)-3 a^3 (A+9 B-49 C) \sec (c+d x)\right )dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (5 a^3 (A+3 B-13 C)-3 a^3 (A+9 B-49 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4274 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \int \sqrt {\sec (c+d x)}dx-3 a^3 (A+9 B-49 C) \int \sec ^{\frac {3}{2}}(c+d x)dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^3 (A+9 B-49 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4255 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4258 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3119 |
\(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}+\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle \frac {\frac {\frac {5 a^2 (A+3 B-13 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (a \sec (c+d x)+a)}+\frac {\frac {10 a^3 (A+3 B-13 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-3 a^3 (A+9 B-49 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}}{3 a^2}+\frac {2 a (2 A+3 B-8 C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {7}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*((A - B + C)*Sec[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) ^3) + ((2*a*(2*A + 3*B - 8*C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(3*d*(a + a *Sec[c + d*x])^2) + ((5*a^2*(A + 3*B - 13*C)*Sec[c + d*x]^(3/2)*Sin[c + d* x])/(d*(a + a*Sec[c + d*x])) + ((10*a^3*(A + 3*B - 13*C)*Sqrt[Cos[c + d*x] ]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - 3*a^3*(A + 9*B - 49*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))/(3*a^2))/(10*a^2)
3.6.72.3.1 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_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + 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*Csc[e + f*x])^n*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, 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. \(788\) vs. \(2(293)=586\).
Time = 78.91 (sec) , antiderivative size = 789, normalized size of antiderivative = 2.93
int(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, method=_RETURNVERBOSE)
1/60*(-2*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)^(1/2)*(5*A*EllipticF(cos(1/2*d *x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*Elliptic F(cos(1/2*d*x+1/2*c),2^(1/2))-27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-6 5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*C*EllipticE(cos(1/2*d*x+1/2* c),2^(1/2)))*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+4*(2*sin(1/2*d*x+1/2* c)^2-1)^(1/2)*(sin(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)^(1/2)*(5*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*Ellip ticE(cos(1/2*d*x+1/2*c),2^(1/2))+15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )-27*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/ 2*c),2^(1/2))+147*C*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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(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)^(1/2)*(5*A*E llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1 /2))+15*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*B*EllipticE(cos(1/2*d*x +1/2*c),2^(1/2))-65*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+147*C*Elliptic E(cos(1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2*c )^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+9*B-49*C)*sin(1/2*d*x+1/2*c)^8-2*(-2*si n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(13*A+147*B-817*C)*sin(1/2* d*x+1/2*c)^6+6*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.99 \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, B - 13 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, B + 13 i \, 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 - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 9 i \, B + 49 i \, 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 + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 9 i \, B - 49 i \, 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 + 9 \, B - 49 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (2 \, A + 33 \, B - 188 \, C\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (A - 9 \, B + 59 \, C\right )} \cos \left (d x + c\right ) - 60 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^3,x, algorithm="fricas")
-1/60*(5*(sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 3*I*B - 13*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 3*I*B - 13*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c)^ 3 + 3*sqrt(2)*(-I*A - 3*I*B + 13*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 3 *I*B + 13*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 3*I*B + 13*I*C))*weierstrass PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c)^ 2 + 3*sqrt(2)*(-I*A - 9*I*B + 49*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 9*I*B + 49*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c)^3 + 3 *sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 9*I*B - 49*I*C)*cos(d*x + c) + sqrt(2)*(I*A + 9*I*B - 49*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(A + 9*B - 49*C)*cos(d*x + c)^3 + 2*(2*A + 33*B - 188*C)*cos(d*x + c)^2 - 5*(A - 9*B + 59*C)*cos(d*x + c) - 60*C)*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^3* d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^3,x, algorithm="maxima")
\[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]
integrate(sec(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^3,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^(5/2)/(a*se c(d*x + c) + a)^3, x)
Timed out. \[ \int \frac {\sec ^{\frac {5}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]
int(((1/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^3,x)