Integrand size = 33, antiderivative size = 180 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {(A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}+\frac {(A-B) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {(A-6 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {(A+9 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \]
1/10*(A+9*B)*(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^3/d+1/6*(A+3*B)*(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))/a^3/d+1/5*(A-B)*sin (d*x+c)*cos(d*x+c)^(1/2)/d/(a+a*cos(d*x+c))^3+1/15*(A-6*B)*sin(d*x+c)*cos( d*x+c)^(1/2)/a/d/(a+a*cos(d*x+c))^2-1/10*(A+9*B)*sin(d*x+c)*cos(d*x+c)^(1/ 2)/d/(a^3+a^3*cos(d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.23 (sec) , antiderivative size = 1171, normalized size of antiderivative = 6.51 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \]
(-2*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*S ec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[ Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x ])^3) - (2*B*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, { 5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[- (Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^3) + (Cos[c/2 + (d*x)/2]^6*(A + B*Sec[c + d*x])*((-4*(A + 9*B)*Csc [c])/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^5*(-(A*Sin[(d*x)/2]) + B*Sin[( d*x)/2]))/(5*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(-(A*Sin[(d*x)/2]) + 6* B*Sin[(d*x)/2]))/(15*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + 9*B*Sin[(d*x)/2]))/(5*d) - (4*(-A + 6*B)*Sec[c/2 + (d*x)/2]^2*Tan[c/2])/( 15*d) - (2*(-A + B)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d)))/(Cos[c + d*x]^( 3/2)*(B + A*Cos[c + d*x])*(a + a*Sec[c + d*x])^3) - (A*Cos[c/2 + (d*x)/2]^ 6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((HypergeometricPF Q[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c ]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + Arc...
Time = 1.19 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.08, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3433, 3042, 3457, 27, 3042, 3457, 3042, 3457, 27, 3042, 3227, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {A \cos (c+d x)+B}{\sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )+B}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\int \frac {a (A+9 B)+3 a (A-B) \cos (c+d x)}{2 \sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^2}dx}{5 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (A+9 B)+3 a (A-B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)^2}dx}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (A+9 B)+3 a (A-B) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\int \frac {(4 A+21 B) a^2+(A-6 B) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(4 A+21 B) a^2+(A-6 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3457 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {5 (A+3 B) a^3+3 (A+9 B) \cos (c+d x) a^3}{2 \sqrt {\cos (c+d x)}}dx}{a^2}-\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {5 (A+3 B) a^3+3 (A+9 B) \cos (c+d x) a^3}{\sqrt {\cos (c+d x)}}dx}{2 a^2}-\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {5 (A+3 B) a^3+3 (A+9 B) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^3 (A+9 B) \int \sqrt {\cos (c+d x)}dx}{2 a^2}-\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^3 (A+9 B) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {\frac {5 a^3 (A+3 B) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {6 a^3 (A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {\frac {10 a^3 (A+3 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^3 (A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{2 a^2}-\frac {3 a^2 (A+9 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{d (a \cos (c+d x)+a)}}{3 a^2}+\frac {2 a (A-6 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}+\frac {(A-B) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\) |
((A - B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + ( (2*a*(A - 6*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^ 2) + (((6*a^3*(A + 9*B)*EllipticE[(c + d*x)/2, 2])/d + (10*a^3*(A + 3*B)*E llipticF[(c + d*x)/2, 2])/d)/(2*a^2) - (3*a^2*(A + 9*B)*Sqrt[Cos[c + d*x]] *Sin[c + d*x])/(d*(a + a*Cos[c + d*x])))/(3*a^2))/(10*a^2)
3.6.13.3.1 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[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]* (d_.) + (c_))^(n_.)*((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Sim p[g^(m + n) Int[(g*Sin[e + f*x])^(p - m - n)*(b + a*Sin[e + f*x])^m*(d + c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(216)=432\).
Time = 10.54 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.51
| 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 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-10 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 )+108 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-30 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 )+54 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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 )-22 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-138 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+6 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+7 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 A +3 B \right )}{60 a^{3} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} \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}\) | \(451\) |
1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2 *d*x+1/2*c)^8-10*A*cos(1/2*d*x+1/2*c)^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*c os(1/2*d*x+1/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+6*A*cos (1/2*d*x+1/2*c)^5*(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))+108*B*cos(1/2*d*x+1/2*c)^8-30 *B*cos(1/2*d*x+1/2*c)^5*(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))+54*B*cos(1/2*d*x+1/2*c) ^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*Elliptic E(cos(1/2*d*x+1/2*c),2^(1/2))-22*A*cos(1/2*d*x+1/2*c)^6-138*B*cos(1/2*d*x+ 1/2*c)^6+6*A*cos(1/2*d*x+1/2*c)^4+24*B*cos(1/2*d*x+1/2*c)^4+7*A*cos(1/2*d* x+1/2*c)^2+3*B*cos(1/2*d*x+1/2*c)^2-3*A+3*B)/a^3/cos(1/2*d*x+1/2*c)^5/(-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 = 465, normalized size of antiderivative = 2.58 \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {2 \, {\left (3 \, {\left (A + 9 \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A + 33 \, B\right )} \cos \left (d x + c\right ) - 5 \, A + 45 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 5 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A + 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, B\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\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A - 3 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, B\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\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-i \, A - 9 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 9 i \, B\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\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (i \, A + 9 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 9 i \, B\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 )}{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 )}} \]
-1/60*(2*(3*(A + 9*B)*cos(d*x + c)^2 + 2*(2*A + 33*B)*cos(d*x + c) - 5*A + 45*B)*sqrt(cos(d*x + c))*sin(d*x + c) + 5*(sqrt(2)*(I*A + 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A + 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 3*I*B)* cos(d*x + c) + sqrt(2)*(I*A + 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(-I*A - 3*I*B)*cos(d*x + c)^3 + 3*sqrt( 2)*(-I*A - 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 3*I*B)*cos(d*x + c) + sqrt(2)*(-I*A - 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d *x + c)) + 3*(sqrt(2)*(-I*A - 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-I*A - 9* I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 9*I*B)*cos(d*x + c) + sqrt(2)*(-I* A - 9*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(I*A + 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)* (I*A + 9*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 9*I*B)*cos(d*x + c) + sqrt (2)*(I*A + 9*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d *x + c) - I*sin(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. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]
\[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \cos \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]