Integrand size = 35, antiderivative size = 170 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {4 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}-\frac {(5 A-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(5 A-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
-1/3*(5*A-C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*si n(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^2+4*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^2/d-1/3*(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 = 5.40 (sec) , antiderivative size = 298, normalized size of antiderivative = 1.75 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (\frac {8 i e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \left (-C e^{i (c+d x)} \left (-1+e^{i (c+d x)}\right )+A \left (3+16 e^{i (c+d x)}+20 e^{2 i (c+d x)}+9 e^{3 i (c+d x)}\right )\right ) \sqrt {\sec (c+d x)}}{d \left (1+e^{i (c+d x)}\right )^3}-\frac {8 (5 A-C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{d}-\frac {32 i A e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right ) \sqrt {\sec (c+d x)}}{d}\right ) \left (A+C \sec ^2(c+d x)\right )}{3 a^2 (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^2} \]
(Cos[(c + d*x)/2]^4*(((8*I)*(1 + E^((2*I)*(c + d*x)))*(-(C*E^(I*(c + d*x)) *(-1 + E^(I*(c + d*x)))) + A*(3 + 16*E^(I*(c + d*x)) + 20*E^((2*I)*(c + d* x)) + 9*E^((3*I)*(c + d*x))))*Sqrt[Sec[c + d*x]])/(d*E^(I*(c + d*x))*(1 + E^(I*(c + d*x)))^3) - (8*(5*A - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/ 2, 2]*Sqrt[Sec[c + d*x]])/d - ((32*I)*A*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)* (c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]*Sqrt[Se c[c + d*x]])/d)*(A + C*Sec[c + d*x]^2))/(3*a^2*(A + 2*C + A*Cos[2*(c + d*x )])*(1 + Sec[c + d*x])^2)
Time = 0.96 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4573, 27, 3042, 4508, 3042, 4274, 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.
| \(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4573 |
| \(\displaystyle -\frac {\int -\frac {a (7 A+C)-3 a (A-C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (7 A+C)-3 a (A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 A+C)-3 a (A-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {12 a^2 A-a^2 (5 A-C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {12 a^2 A-a^2 (5 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {12 a^2 A \int \frac {1}{\sqrt {\sec (c+d x)}}dx-a^2 (5 A-C) \int \sqrt {\sec (c+d x)}dx}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {12 a^2 A \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a^2 (5 A-C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {12 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-a^2 (5 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {12 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (5 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {\frac {24 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a^2 (5 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {24 a^2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 a^2 (5 A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}-\frac {2 (5 A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}\) |
-1/3*((A + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + (((24*a^2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d* x]])/d - (2*a^2*(5*A - C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqr t[Sec[c + d*x]])/d)/a^2 - (2*(5*A - C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d *(1 + Sec[c + d*x])))/(6*a^2)
3.3.40.3.1 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*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 = 3.02 (sec) , antiderivative size = 352, normalized size of antiderivative = 2.07
| method | result | size |
| default | \(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+10 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-38 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+15 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A -C \right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(352\) |
1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*A*cos(1/2* d*x+1/2*c)^6+10*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))*cos(1/2*d*x+1/2*c)^3+24*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))-2*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))-38*A*cos(1/2*d*x+1/2*c)^4-2*C*cos(1/2*d*x+1/2*c)^4+ 15*A*cos(1/2*d*x+1/2*c)^2+3*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 = 326, normalized size of antiderivative = 1.92 \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} {\left (5 i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-5 i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (-5 i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (5 i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (-i \, \sqrt {2} A \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} A \cos \left (d x + c\right ) - i \, \sqrt {2} A\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (i \, \sqrt {2} A \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} A \cos \left (d x + c\right ) + i \, \sqrt {2} A\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{2} + {\left (5 \, A - C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/6*((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)) + (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)) - 12*(-I*sqrt(2)*A*cos(d*x + c)^2 - 2*I* sqrt(2)*A*cos(d*x + c) - I*sqrt(2)*A)*weierstrassZeta(-4, 0, weierstrassPI nverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 12*(I*sqrt(2)*A*cos(d*x + c)^2 + 2*I*sqrt(2)*A*cos(d*x + c) + I*sqrt(2)*A)*weierstrassZeta(-4, 0, we ierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*A*cos(d*x + c)^2 + (5*A - C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*c os(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {5}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{2}} \]
(Integral(A/(sec(c + d*x)**(5/2) + 2*sec(c + d*x)**(3/2) + sqrt(sec(c + d* x))), x) + Integral(C*sec(c + d*x)**2/(sec(c + d*x)**(5/2) + 2*sec(c + d*x )**(3/2) + sqrt(sec(c + d*x))), x))/a**2
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sqrt {\sec \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \]