Integrand size = 35, antiderivative size = 196 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {16 a^2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {2 a^2 (7 A-15 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 A \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d \sqrt {\sec (c+d x)}} \]
2/5*A*(a+a*sec(d*x+c))^2*sin(d*x+c)/d/sec(d*x+c)^(3/2)+8/15*A*(a^2+a^2*sec (d*x+c))*sin(d*x+c)/d/sec(d*x+c)^(1/2)-2/15*a^2*(7*A-15*C)*sin(d*x+c)*sec( d*x+c)^(1/2)/d+16/5*a^2*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)/d+ 4/3*a^2*(A+3*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)/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 5.35 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.07 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^2 \left (\cos \left (\frac {c}{2}\right )-i \sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right ) \left (-96 i A+\frac {192 i A \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}-\frac {80 i (A+3 C) e^{i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+40 A \sin (c+d x)+3 A \sec (c+d x) \sin (3 (c+d x))+3 A \tan (c+d x)+60 C \tan (c+d x)\right )}{30 d \sqrt {\sec (c+d x)}} \]
(a^2*(Cos[c/2] - I*Sin[c/2])*(Cos[c/2] + I*Sin[c/2])*((-96*I)*A + ((192*I) *A*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2 *I)*(c + d*x))] - ((80*I)*(A + 3*C)*E^(I*(c + d*x))*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 40*A*Sin [c + d*x] + 3*A*Sec[c + d*x]*Sin[3*(c + d*x)] + 3*A*Tan[c + d*x] + 60*C*Ta n[c + d*x]))/(30*d*Sqrt[Sec[c + d*x]])
Time = 1.16 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 4575, 27, 3042, 4505, 27, 3042, 4485, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^2 (4 a A-a (A-5 C) \sec (c+d x))}{2 \sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^2 (4 a A-a (A-5 C) \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (4 a A-a (A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left (a^2 (17 A+15 C)-a^2 (7 A-15 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {(\sec (c+d x) a+a) \left (a^2 (17 A+15 C)-a^2 (7 A-15 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (17 A+15 C)-a^2 (7 A-15 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {12 A a^3+5 (A+3 C) \sec (c+d x) a^3}{\sqrt {\sec (c+d x)}}dx-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \int \frac {12 A a^3+5 (A+3 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \int \sqrt {\sec (c+d x)}dx+12 a^3 A \int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+12 a^3 A \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+12 a^3 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^3 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \left (5 a^3 (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {24 a^3 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {1}{3} \left (2 \left (\frac {10 a^3 (A+3 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {24 a^3 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 a^3 (7 A-15 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )+\frac {8 A \sin (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{3 d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^2}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
(2*A*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((8*A *(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*((24 *a^3*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (10*a^3*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[ c + d*x]])/d) - (2*a^3*(7*A - 15*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d)/3) /(5*a)
3.3.18.3.1 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_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1) Int[(d*Csc [e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x ], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && !LeQ[ n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + 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 ] && GtQ[m, 1/2] && LtQ[n, -1]
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*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 3.24 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.37
| method | result | size |
| default | \(\frac {4 a^{2} \left (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-32 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+13 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) A -5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C -15 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(268\) |
| parts | \(\text {Expression too large to display}\) | \(853\) |
4/15*a^2*(12*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-32*A*cos(1/2*d*x+1/ 2*c)*sin(1/2*d*x+1/2*c)^4+13*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*A-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(c os(1/2*d*x+1/2*c),2^(1/2))+12*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))+15*sin(1/2*d*x+1 /2*c)^2*cos(1/2*d*x+1/2*c)*C-15*C*(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)))/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 = 0.10 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.90 \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 3 \, C\right )} a^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 i \, \sqrt {2} A a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 12 i \, \sqrt {2} A a^{2} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, A a^{2} \cos \left (d x + c\right )^{2} + 10 \, A a^{2} \cos \left (d x + c\right ) + 15 \, C a^{2}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d} \]
-2/15*(5*I*sqrt(2)*(A + 3*C)*a^2*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(A + 3*C)*a^2*weierstrassPInverse(-4, 0, co s(d*x + c) - I*sin(d*x + c)) - 12*I*sqrt(2)*A*a^2*weierstrassZeta(-4, 0, w eierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 12*I*sqrt(2)*A *a^2*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si n(d*x + c))) - (3*A*a^2*cos(d*x + c)^2 + 10*A*a^2*cos(d*x + c) + 15*C*a^2) *sin(d*x + c)/sqrt(cos(d*x + c)))/d
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=a^{2} \left (\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {2 A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {C}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int 2 C \sqrt {\sec {\left (c + d x \right )}}\, dx + \int C \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A/sec(c + d*x)**(5/2), x) + Integral(2*A/sec(c + d*x)**(3/2 ), x) + Integral(A/sqrt(sec(c + d*x)), x) + Integral(C/sqrt(sec(c + d*x)), x) + Integral(2*C*sqrt(sec(c + d*x)), x) + Integral(C*sec(c + d*x)**(3/2) , x))
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]