Integrand size = 33, antiderivative size = 204 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {7 (8 A-5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a^2 d}-\frac {5 (3 A-2 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}-\frac {5 (3 A-2 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{3 a^2 d}+\frac {7 (8 A-5 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{15 a^2 d}-\frac {(3 A-2 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{a^2 d (1+\cos (c+d x))}-\frac {(A-B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
7/5*(8*A-5*B)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(si n(1/2*d*x+1/2*c),2^(1/2))/a^2/d-5/3*(3*A-2*B)*(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))/a^2/d+7/15*(8*A- 5*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/a^2/d-(3*A-2*B)*cos(d*x+c)^(5/2)*sin(d*x+ c)/a^2/d/(1+cos(d*x+c))-1/3*(A-B)*cos(d*x+c)^(7/2)*sin(d*x+c)/d/(a+a*cos(d *x+c))^2-5/3*(3*A-2*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/a^2/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 8.81 (sec) , antiderivative size = 1158, normalized size of antiderivative = 5.68 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \]
(10*A*Cos[c/2 + (d*x)/2]^4*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]*(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[Co t[c]]]])/(d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d*x])^2 ) - (20*B*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4 }, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(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 - ArcTa n[Cot[c]]]])/(3*d*(B + A*Cos[c + d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d *x])^2) + (Cos[c/2 + (d*x)/2]^4*(A + B*Sec[c + d*x])*((4*(-20*A + 15*B - 3 6*A*Cos[c] + 20*B*Cos[c])*Csc[c])/(5*d) + (8*(-2*A + B)*Cos[d*x]*Sin[c])/( 3*d) + (4*A*Cos[2*d*x]*Sin[2*c])/(5*d) - (2*Sec[c/2]*Sec[c/2 + (d*x)/2]^3* (-(A*Sin[(d*x)/2]) + B*Sin[(d*x)/2]))/(3*d) + (4*Sec[c/2]*Sec[c/2 + (d*x)/ 2]*(-4*A*Sin[(d*x)/2] + 3*B*Sin[(d*x)/2]))/d + (8*(-2*A + B)*Cos[c]*Sin[d* x])/(3*d) + (4*A*Cos[2*c]*Sin[2*d*x])/(5*d) - (2*(-A + B)*Sec[c/2 + (d*x)/ 2]^2*Tan[c/2])/(3*d)))/(Sqrt[Cos[c + d*x]]*(B + A*Cos[c + d*x])*(a + a*Sec [c + d*x])^2) - (56*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*Sec[c/2]*Sec[c + d*x]* (A + B*Sec[c + d*x])*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + Ar cTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ...
Time = 1.10 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.98, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3433, 3042, 3456, 27, 3042, 3456, 3042, 3227, 3042, 3115, 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.
| \(\displaystyle \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3433 |
| \(\displaystyle \int \frac {\cos ^{\frac {7}{2}}(c+d x) (A \cos (c+d x)+B)}{(a \cos (c+d x)+a)^2}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (A \sin \left (c+d x+\frac {\pi }{2}\right )+B\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle \frac {\int -\frac {\cos ^{\frac {5}{2}}(c+d x) (7 a (A-B)-a (11 A-5 B) \cos (c+d x))}{2 (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\cos ^{\frac {5}{2}}(c+d x) (7 a (A-B)-a (11 A-5 B) \cos (c+d x))}{\cos (c+d x) a+a}dx}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (7 a (A-B)-a (11 A-5 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3456 |
| \(\displaystyle -\frac {\frac {\int \cos ^{\frac {3}{2}}(c+d x) \left (15 a^2 (3 A-2 B)-7 a^2 (8 A-5 B) \cos (c+d x)\right )dx}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (15 a^2 (3 A-2 B)-7 a^2 (8 A-5 B) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle -\frac {\frac {15 a^2 (3 A-2 B) \int \cos ^{\frac {3}{2}}(c+d x)dx-7 a^2 (8 A-5 B) \int \cos ^{\frac {5}{2}}(c+d x)dx}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {15 a^2 (3 A-2 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx-7 a^2 (8 A-5 B) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle -\frac {\frac {15 a^2 (3 A-2 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-7 a^2 (8 A-5 B) \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {15 a^2 (3 A-2 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-7 a^2 (8 A-5 B) \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {\frac {15 a^2 (3 A-2 B) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-7 a^2 (8 A-5 B) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle -\frac {\frac {15 a^2 (3 A-2 B) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-7 a^2 (8 A-5 B) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{a^2}+\frac {6 (3 A-2 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (\cos (c+d x)+1)}}{6 a^2}-\frac {(A-B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*((A - B)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^2) - ((6*(3*A - 2*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(d*(1 + Cos[c + d*x])) + (15*a^2*(3*A - 2*B)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) - 7*a^2*(8*A - 5*B)*((6*EllipticE[(c + d*x)/ 2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(5*d)))/a^2)/(6*a^2)
3.6.2.3.1 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[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
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[(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/( a*f*(2*m + 1))), x] - Simp[1/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x], x] /; Fre eQ[{a, b, c, d, e, f, A, B}, 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] && (In tegerQ[2*n] || EqQ[c, 0])
Time = 13.44 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.28
| 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 (96 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-352 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+80 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+120 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-150 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 {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-336 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 )+60 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+100 B \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 )+210 B \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 )+266 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-240 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-135 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+105 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+5 A -5 B \right )}{30 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}\) | \(465\) |
-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(96*A*cos(1/ 2*d*x+1/2*c)^10-352*A*cos(1/2*d*x+1/2*c)^8+80*B*cos(1/2*d*x+1/2*c)^8+120*A *cos(1/2*d*x+1/2*c)^6-150*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2) )-336*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))+60*B*cos(1/2*d*x+1 /2*c)^6+100*B*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))+210*B*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))+266*A*cos(1/2*d*x+1/2*c)^4-240*B *cos(1/2*d*x+1/2*c)^4-135*A*cos(1/2*d*x+1/2*c)^2+105*B*cos(1/2*d*x+1/2*c)^ 2+5*A-5*B)/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 = 0.12 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.88 \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{3} - 2 \, {\left (4 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (94 \, A - 65 \, B\right )} \cos \left (d x + c\right ) - 75 \, A + 50 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 25 \, {\left (\sqrt {2} {\left (-3 i \, A + 2 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-3 i \, A + 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 25 \, {\left (\sqrt {2} {\left (3 i \, A - 2 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (3 i \, A - 2 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 2 i \, B\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 21 \, {\left (\sqrt {2} {\left (-8 i \, A + 5 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-8 i \, A + 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-8 i \, A + 5 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 \, {\left (\sqrt {2} {\left (8 i \, A - 5 i \, B\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (8 i \, A - 5 i \, B\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (8 i \, A - 5 i \, B\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/30*(2*(6*A*cos(d*x + c)^3 - 2*(4*A - 5*B)*cos(d*x + c)^2 - (94*A - 65*B) *cos(d*x + c) - 75*A + 50*B)*sqrt(cos(d*x + c))*sin(d*x + c) - 25*(sqrt(2) *(-3*I*A + 2*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I*A + 2*I*B)*cos(d*x + c) + sqrt(2)*(-3*I*A + 2*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s in(d*x + c)) - 25*(sqrt(2)*(3*I*A - 2*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(3*I *A - 2*I*B)*cos(d*x + c) + sqrt(2)*(3*I*A - 2*I*B))*weierstrassPInverse(-4 , 0, cos(d*x + c) - I*sin(d*x + c)) - 21*(sqrt(2)*(-8*I*A + 5*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(-8*I*A + 5*I*B)*cos(d*x + c) + sqrt(2)*(-8*I*A + 5*I* B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin (d*x + c))) - 21*(sqrt(2)*(8*I*A - 5*I*B)*cos(d*x + c)^2 + 2*sqrt(2)*(8*I* A - 5*I*B)*cos(d*x + c) + sqrt(2)*(8*I*A - 5*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(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. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]