Integrand size = 23, antiderivative size = 109 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {\sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)} (1+\sec (c+d x))}-\frac {\sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \]
(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))/a^2/d+2/3*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Elli pticF(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d-1/3*sin(d*x+c)/d/cos(d*x+c)^(3/2)/ (a+a*sec(d*x+c))^2-sin(d*x+c)/a^2/d/(1+sec(d*x+c))/cos(d*x+c)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 4.43 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.86 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (-\frac {\left (7 \cos \left (\frac {1}{2} (c-d x)\right )+2 \cos \left (\frac {1}{2} (3 c+d x)\right )+3 \cos \left (\frac {1}{2} (c+3 d x)\right )\right ) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right )}{2 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 i \sqrt {2} e^{-i (c+d x)} \left (3 \left (1+e^{2 i (c+d x)}\right )+3 \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )-2 e^{i (c+d x)} \left (-1+e^{2 i c}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^2(c+d x)}{d \left (-1+e^{2 i c}\right ) \sqrt {e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{3 a^2 (1+\sec (c+d x))^2} \]
(Cos[(c + d*x)/2]^4*(-1/2*((7*Cos[(c - d*x)/2] + 2*Cos[(3*c + d*x)/2] + 3* Cos[(c + 3*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^3)/(d*Cos[c + d*x]^ (3/2)) + ((4*I)*Sqrt[2]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*c) )*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I )*(c + d*x))] - 2*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + d *x]^2)/(d*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)) )/E^(I*(c + d*x))])))/(3*a^2*(1 + Sec[c + d*x])^2)
Time = 0.96 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.65, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3042, 4752, 3042, 4303, 27, 3042, 4507, 25, 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.
| \(\displaystyle \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4752 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\sec ^{\frac {5}{2}}(c+d x)}{(\sec (c+d x) a+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\sec (c+d x)} (a-5 a \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\sec (c+d x)} (a-5 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a-5 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {\int -\frac {2 \sec (c+d x) a^2+3 a^2}{\sqrt {\sec (c+d x)}}dx}{a^2}+\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {2 \sec (c+d x) a^2+3 a^2}{\sqrt {\sec (c+d x)}}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {2 \csc \left (c+d x+\frac {\pi }{2}\right ) a^2+3 a^2}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 \int \frac {1}{\sqrt {\sec (c+d x)}}dx+2 a^2 \int \sqrt {\sec (c+d x)}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+2 a^2 \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {2 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {\frac {6 \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\frac {4 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a^2}}{6 a^2}-\frac {\sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )\) |
Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-1/3*(Sec[c + d*x]^(3/2)*Sin[c + d* x])/(d*(a + a*Sec[c + d*x])^2) - (-(((6*a^2*Sqrt[Cos[c + d*x]]*EllipticE[( c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (4*a^2*Sqrt[Cos[c + d*x]]*EllipticF [(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2) + (6*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(1 + Sec[c + d*x])))/(6*a^2))
3.4.86.3.1 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_), x_Symbol] :> Simp[(-d^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d *Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Simp[d^2/(a*b*(2*m + 1)) Int[ (a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
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[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m Int[ActivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[u, x ]
Time = 5.58 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.36
| method | result | size |
| default | \(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (12 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+3 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )}{6 a^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(257\) |
1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*cos(1/2*d* x+1/2*c)^6-4*(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+6*(sin(1/2*d* x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^3*E llipticE(cos(1/2*d*x+1/2*c),2^(1/2))-16*cos(1/2*d*x+1/2*c)^4+3*cos(1/2*d*x +1/2*c)^2+1)/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 = 0.10 (sec) , antiderivative size = 268, normalized size of antiderivative = 2.46 \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \, {\left (3 \, \cos \left (d x + c\right ) + 4\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 2 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 \, {\left (-i \, \sqrt {2} \cos \left (d x + c\right )^{2} - 2 i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 \, {\left (i \, \sqrt {2} \cos \left (d x + c\right )^{2} + 2 i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
-1/6*(2*(3*cos(d*x + c) + 4)*sqrt(cos(d*x + c))*sin(d*x + c) + 2*(I*sqrt(2 )*cos(d*x + c)^2 + 2*I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 2*(-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(-I*sqrt(2)*cos(d*x + c)^2 - 2*I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d *x + c) + I*sin(d*x + c))) + 3*(I*sqrt(2)*cos(d*x + c)^2 + 2*I*sqrt(2)*cos (d*x + c) + I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c os(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int { \frac {1}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]