Integrand size = 35, antiderivative size = 253 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 a^3 (9 A-5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 (3 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 d}-\frac {8 a^3 (3 A-10 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 d}+\frac {2 A (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 A \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{5 a d \sqrt {\sec (c+d x)}}-\frac {2 (9 A-5 C) \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d} \]
2/5*A*(a+a*sec(d*x+c))^3*sin(d*x+c)/d/sec(d*x+c)^(3/2)+4/5*A*(a^2+a^2*sec( d*x+c))^2*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)-8/15*a^3*(3*A-10*C)*sin(d*x+c)*s ec(d*x+c)^(1/2)/d-2/15*(9*A-5*C)*(a^3+a^3*sec(d*x+c))*sin(d*x+c)*sec(d*x+c )^(1/2)/d+4/5*a^3*(9*A-5*C)*(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^3*(3*A+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt icF(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 = 5.56 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.87 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {a^3 e^{-i d x} \sec ^{\frac {3}{2}}(c+d x) (\cos (d x)+i \sin (d x)) \left (216 i A-120 i C+216 i A \cos (2 (c+d x))-120 i C \cos (2 (c+d x))+80 (3 A+5 C) \cos ^{\frac {3}{2}}(c+d x) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-8 i (9 A-5 C) \left (1+e^{2 i (c+d x)}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+30 A \sin (c+d x)+40 C \sin (c+d x)+6 A \sin (2 (c+d x))+180 C \sin (2 (c+d x))+30 A \sin (3 (c+d x))+3 A \sin (4 (c+d x))\right )}{60 d} \]
(a^3*Sec[c + d*x]^(3/2)*(Cos[d*x] + I*Sin[d*x])*((216*I)*A - (120*I)*C + ( 216*I)*A*Cos[2*(c + d*x)] - (120*I)*C*Cos[2*(c + d*x)] + 80*(3*A + 5*C)*Co s[c + d*x]^(3/2)*EllipticF[(c + d*x)/2, 2] - (8*I)*(9*A - 5*C)*(1 + E^((2* I)*(c + d*x)))^(3/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x)) ] + 30*A*Sin[c + d*x] + 40*C*Sin[c + d*x] + 6*A*Sin[2*(c + d*x)] + 180*C*S in[2*(c + d*x)] + 30*A*Sin[3*(c + d*x)] + 3*A*Sin[4*(c + d*x)]))/(60*d*E^( I*d*x))
Time = 1.55 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.01, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.514, Rules used = {3042, 4575, 27, 3042, 4505, 27, 3042, 4506, 3042, 4485, 27, 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.
| \(\displaystyle \int \frac {(a \sec (c+d x)+a)^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {2 \int \frac {(\sec (c+d x) a+a)^3 (6 a A-a (3 A-5 C) \sec (c+d x))}{2 \sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^3 (6 a A-a (3 A-5 C) \sec (c+d x))}{\sec ^{\frac {3}{2}}(c+d x)}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3 \left (6 a A-a (3 A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {3 (\sec (c+d x) a+a)^2 \left (a^2 (11 A+5 C)-a^2 (9 A-5 C) \sec (c+d x)\right )}{2 \sqrt {\sec (c+d x)}}dx+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(\sec (c+d x) a+a)^2 \left (a^2 (11 A+5 C)-a^2 (9 A-5 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a^2 (11 A+5 C)-a^2 (9 A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4506 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {(\sec (c+d x) a+a) \left (a^3 (21 A+5 C)-2 a^3 (3 A-10 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}}dx-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2}{3} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^3 (21 A+5 C)-2 a^3 (3 A-10 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4485 |
| \(\displaystyle \frac {\frac {2}{3} \left (2 \int \frac {3 (9 A-5 C) a^4+5 (3 A+5 C) \sec (c+d x) a^4}{2 \sqrt {\sec (c+d x)}}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {2}{3} \left (\int \frac {3 (9 A-5 C) a^4+5 (3 A+5 C) \sec (c+d x) a^4}{\sqrt {\sec (c+d x)}}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2}{3} \left (\int \frac {3 (9 A-5 C) a^4+5 (3 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^4}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {2}{3} \left (3 a^4 (9 A-5 C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx+5 a^4 (3 A+5 C) \int \sqrt {\sec (c+d x)}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2}{3} \left (3 a^4 (9 A-5 C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+5 a^4 (3 A+5 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {2}{3} \left (5 a^4 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+3 a^4 (9 A-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2}{3} \left (5 a^4 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+3 a^4 (9 A-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {2}{3} \left (5 a^4 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {6 a^4 (9 A-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {-\frac {2 (9 A-5 C) \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^4 \sec (c+d x)+a^4\right )}{3 d}+\frac {2}{3} \left (-\frac {4 a^4 (3 A-10 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {10 a^4 (3 A+5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+\frac {6 a^4 (9 A-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )+\frac {4 A \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{d \sqrt {\sec (c+d x)}}}{5 a}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^3}{5 d \sec ^{\frac {3}{2}}(c+d x)}\) |
(2*A*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + ((4*A *(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]) - (2*(9*A - 5*C)*Sqrt[Sec[c + d*x]]*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(3*d) + (2*((6*a^4*(9*A - 5*C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[S ec[c + d*x]])/d + (10*a^4*(3*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d* x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (4*a^4*(3*A - 10*C)*Sqrt[Sec[c + d*x]]*Si n[c + d*x])/d))/3)/(5*a)
3.3.26.3.1 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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B* Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x] )^n*Simp[a*A*d*(m + n) + B*(b*d*n) + (A*b*d*(m + n) + a*B*d*(2*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] && 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])
Leaf count of result is larger than twice the leaf count of optimal. \(703\) vs. \(2(277)=554\).
Time = 3.92 (sec) , antiderivative size = 704, normalized size of antiderivative = 2.78
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(704\) |
| parts | \(\text {Expression too large to display}\) | \(1078\) |
-4/15*(24*A*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*cos(1/2*d *x+1/2*c)*sin(1/2*d*x+1/2*c)^8-96*A*(-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)^6*cos(1/2*d*x+1/2*c)+6*(-2*sin(1/2*d*x+1 /2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(13*A+15*C)*cos(1/2*d*x+1/2*c)*sin(1/2 *d*x+1/2*c)^4-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(9*A+ 25*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-2*(sin(1/2*d*x+1/2*c)^2)^(1/ 2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1 /2*c)^2)^(1/2)*(15*A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-27*A*EllipticE( cos(1/2*d*x+1/2*c),2^(1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+15* C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*sin(1/2*d*x+1/2*c)^2+15*A*(-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)^(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))-27 *A*(-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)^(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))+25*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))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d* x+1/2*c)^2)^(1/2)+15*C*(-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)^(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)))*a^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2* c)^2)^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^(3/2)/sin(1/2*d*x+1/2*c)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.94 \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (3 \, A + 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 i \, \sqrt {2} {\left (9 \, A - 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 3 i \, \sqrt {2} {\left (9 \, A - 5 \, C\right )} a^{3} \cos \left (d x + c\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, A a^{3} \cos \left (d x + c\right )^{3} + 15 \, A a^{3} \cos \left (d x + c\right )^{2} + 45 \, C a^{3} \cos \left (d x + c\right ) + 5 \, C a^{3}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d \cos \left (d x + c\right )} \]
-2/15*(5*I*sqrt(2)*(3*A + 5*C)*a^3*cos(d*x + c)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*I*sqrt(2)*(3*A + 5*C)*a^3*cos(d*x + c) *weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 3*I*sqrt(2)*( 9*A - 5*C)*a^3*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*I*sqrt(2)*(9*A - 5*C)*a^3*cos(d*x + c)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*si n(d*x + c))) - (3*A*a^3*cos(d*x + c)^3 + 15*A*a^3*cos(d*x + c)^2 + 45*C*a^ 3*cos(d*x + c) + 5*C*a^3)*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) )
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=a^{3} \left (\int \frac {A}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {3 A}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {3 A}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int A \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \frac {C}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int 3 C \sqrt {\sec {\left (c + d x \right )}}\, dx + \int 3 C \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx + \int C \sec ^{\frac {5}{2}}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(A/sec(c + d*x)**(5/2), x) + Integral(3*A/sec(c + d*x)**(3/2 ), x) + Integral(3*A/sqrt(sec(c + d*x)), x) + Integral(A*sqrt(sec(c + d*x) ), x) + Integral(C/sqrt(sec(c + d*x)), x) + Integral(3*C*sqrt(sec(c + d*x) ), x) + Integral(3*C*sec(c + d*x)**(3/2), x) + Integral(C*sec(c + d*x)**(5 /2), x))
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} {\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]