Integrand size = 35, antiderivative size = 201 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {(7 A+C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {2 (5 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {2 (5 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \]
2/3*(5*A+C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)-1/3*(7*A+C)*sin(d*x+c)/a^2/d /(1+sec(d*x+c))/sec(d*x+c)^(1/2)-1/3*(A+C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2 /sec(d*x+c)^(1/2)-(7*A+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*(5*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/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.90 (sec) , antiderivative size = 912, normalized size of antiderivative = 4.54 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {14 \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 {2 \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 {40 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 {8 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 (5 A+C+2 A \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )}{d}+\frac {8 A \cos (2 d x) \sin (2 c)}{3 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 (5 A \sin \left (\frac {d x}{2}\right )+2 C \sin \left (\frac {d x}{2}\right )\right )}{3 d}-\frac {32 A \cos (c) \sin (d x)}{d}+\frac {8 A \cos (2 c) \sin (2 d x)}{3 d}-\frac {16 (5 A+2 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} \]
(14*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) + (2*Sqrt[2]*C*Sqrt[ E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*C os[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) + (40*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) + (8*C*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*Ell ipticF[(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) + (Cos [c/2 + (d*x)/2]^4*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*((4*(5*A + C + 2*A*Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2])/d + (8*A*Cos[2*d*x]*Sin[2*c])/( 3*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]*(5*A*Sin[(d*x)/2] + 2*C*Sin[(d*x) /2]))/(3*d) - (32*A*Cos[c]*Sin[d*x])/d + (8*A*Cos[2*c]*Sin[2*d*x])/(3*d...
Time = 1.15 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.02, 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, 4508, 27, 3042, 4274, 3042, 4256, 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.
| \(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {3 a (3 A+C)-a (5 A-C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {3 a (3 A+C)-a (5 A-C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 a (3 A+C)-a (5 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {3 \left (2 a^2 (5 A+C)-a^2 (7 A+C) \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \frac {2 a^2 (5 A+C)-a^2 (7 A+C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \frac {2 a^2 (5 A+C)-a^2 (7 A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx-a^2 (7 A+C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-a^2 (7 A+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-\frac {2 a^2 (7 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {3 \left (2 a^2 (5 A+C) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )-\frac {2 a^2 (7 A+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}-\frac {2 (7 A+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
-1/3*((A + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^2) + ((-2*(7*A + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) + (3*((-2*a^2*(7*A + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[S ec[c + d*x]])/d + 2*a^2*(5*A + C)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d* x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d* x]]))))/a^2)/(6*a^2)
3.3.41.3.1 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[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B , 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[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)]
Time = 3.24 (sec) , antiderivative size = 437, normalized size of antiderivative = 2.17
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (16 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+20 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+42 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+6 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-48 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-20 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+21 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+9 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A -C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(437\) |
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*A*cos(1/2 *d*x+1/2*c)^8+12*A*cos(1/2*d*x+1/2*c)^6+20*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*co s(1/2*d*x+1/2*c)^3+42*A*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+12 *C*cos(1/2*d*x+1/2*c)^6+4*C*cos(1/2*d*x+1/2*c)^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )+6*C*cos(1/2*d*x+1/2*c)^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-48*A*cos(1/2*d*x+1/2 *c)^4-20*C*cos(1/2*d*x+1/2*c)^4+21*A*cos(1/2*d*x+1/2*c)^2+9*C*cos(1/2*d*x+ 1/2*c)^2-A-C)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(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*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 = 375, normalized size of antiderivative = 1.87 \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (5 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (5 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (\sqrt {2} {\left (-5 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-5 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (\sqrt {2} {\left (7 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (7 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A + i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (\sqrt {2} {\left (-7 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-7 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A - i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (2 \, A \cos \left (d x + c\right )^{3} + {\left (13 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-1/6*(2*(sqrt(2)*(5*I*A + I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(5*I*A + I*C)*co s(d*x + c) + sqrt(2)*(5*I*A + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c ) + I*sin(d*x + c)) + 2*(sqrt(2)*(-5*I*A - I*C)*cos(d*x + c)^2 + 2*sqrt(2) *(-5*I*A - I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A - I*C))*weierstrassPInverse (-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(7*I*A + I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(7*I*A + I*C)*cos(d*x + c) + sqrt(2)*(7*I*A + I*C))*wei erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-7*I*A - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-7*I*A - I*C) *cos(d*x + c) + sqrt(2)*(-7*I*A - I*C))*weierstrassZeta(-4, 0, weierstrass PInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(2*A*cos(d*x + c)^3 + (13*A + 3*C)*cos(d*x + c)^2 + 2*(5*A + C)*cos(d*x + c))*sin(d*x + c)/sqrt( cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a^{2}} \]
(Integral(A/(sec(c + d*x)**(7/2) + 2*sec(c + d*x)**(5/2) + sec(c + d*x)**( 3/2)), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)**(7/2) + 2*sec(c + d* x)**(5/2) + sec(c + d*x)**(3/2)), x))/a**2
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]