Integrand size = 43, antiderivative size = 220 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {(7 A-4 B+C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(10 A-5 B+2 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {(10 A-5 B+2 C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(7 A-4 B+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)} (1+\sec (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2} \]
1/3*(10*A-5*B+2*C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)-1/3*(7*A-4*B+C)*sin(d *x+c)/a^2/d/(1+sec(d*x+c))/sec(d*x+c)^(1/2)-1/3*(A-B+C)*sin(d*x+c)/d/(a+a* sec(d*x+c))^2/sec(d*x+c)^(1/2)-(7*A-4*B+C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/co s(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)*se c(d*x+c)^(1/2)/a^2/d+1/3*(10*A-5*B+2*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)/a^2/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.66 (sec) , antiderivative size = 762, normalized size of antiderivative = 3.46 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-14 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+8 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-2 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )-40 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+20 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-8 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-2 \sqrt {\sec (c+d x)} \left (3 (5 A-3 B+C+(2 A-B) \cos (2 c)) \cos (d x) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right )+2 A \cos (2 d x) \sin (2 c)-2 (10 A-7 B+4 C) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )+(A-B+C) \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )-12 (2 A-B) \cos (c) \sin (d x)+2 A \cos (2 c) \sin (2 d x)-2 (10 A-7 B+4 C) \tan \left (\frac {c}{2}\right )+(A-B+C) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )\right )}{3 a^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (1+\sec (c+d x))^2} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^2),x]
(-2*Cos[(c + d*x)/2]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-14*Sqrt[ 2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^ ((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d *x) + (8*Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I) *d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d *x))]))/E^(I*d*x) - (2*Sqrt[2]*C*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d *x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqrt[1 + E^((2*I)*(c + d*x ))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometric2F1[1/2, 3/4, 7/4, -E ^((2*I)*(c + d*x))]))/E^(I*d*x) - 40*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d *x)/2, 2]*Sqrt[Sec[c + d*x]] + 20*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x) /2, 2]*Sqrt[Sec[c + d*x]] - 8*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 2*Sqrt[Sec[c + d*x]]*(3*(5*A - 3*B + C + (2*A - B) *Cos[2*c])*Cos[d*x]*Csc[c/2]*Sec[c/2] + 2*A*Cos[2*d*x]*Sin[2*c] - 2*(10*A - 7*B + 4*C)*Sec[c/2]*Sec[(c + d*x)/2]*Sin[(d*x)/2] + (A - B + C)*Sec[c/2] *Sec[(c + d*x)/2]^3*Sin[(d*x)/2] - 12*(2*A - B)*Cos[c]*Sin[d*x] + 2*A*Cos[ 2*c]*Sin[2*d*x] - 2*(10*A - 7*B + 4*C)*Tan[c/2] + (A - B + C)*Sec[(c + d*x )/2]^2*Tan[c/2])))/(3*a^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x )])*(1 + Sec[c + d*x])^2)
Time = 1.20 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.349, Rules used = {3042, 4572, 27, 3042, 4508, 27, 3042, 4274, 3042, 4256, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {3 a (3 A-B+C)-a (5 A-5 B-C) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {3 a (3 A-B+C)-a (5 A-5 B-C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {3 a (3 A-B+C)-a (5 A-5 B-C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {3 \left (a^2 (10 A-5 B+2 C)-a^2 (7 A-4 B+C) \sec (c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \frac {a^2 (10 A-5 B+2 C)-a^2 (7 A-4 B+C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \frac {a^2 (10 A-5 B+2 C)-a^2 (7 A-4 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx-a^2 (7 A-4 B+C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-a^2 (7 A-4 B+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A-4 B+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A-4 B+C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A-4 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-a^2 (7 A-4 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-\frac {2 a^2 (7 A-4 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {3 \left (a^2 (10 A-5 B+2 C) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )-\frac {2 a^2 (7 A-4 B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}-\frac {2 (7 A-4 B+C) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x]) ^2) + ((-2*(7*A - 4*B + C)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])) + (3*((-2*a^2*(7*A - 4*B + C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d *x)/2, 2]*Sqrt[Sec[c + d*x]])/d + a^2*(10*A - 5*B + 2*C)*((2*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*Sin[c + d*x ])/(3*d*Sqrt[Sec[c + d*x]]))))/a^2)/(6*a^2)
3.6.69.3.1 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[Cos[c + d*x]*(( b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n) Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* n]
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_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Cs c[e + f*x])^n*Simp[b*B*n - a*A*(2*m + n + 1) + (A*b - a*B)*(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] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[(-(a*A - b*B + a*C))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(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*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m - n)))*Csc[ e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 3.65 (sec) , antiderivative size = 472, normalized size of antiderivative = 2.15
| 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 (-2 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (10 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+2 \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (10 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-12 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+16 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-76 A +24 B -12 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (84 A -34 B +16 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-25 A +11 B -5 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\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}\) | \(472\) |
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2,x, method=_RETURNVERBOSE)
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*(2*sin(1/ 2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*A*EllipticF(cos(1 /2*d*x+1/2*c),2^(1/2))+21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*B*Elli pticF(cos(1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2 ))+2*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(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)^2+2*(2*sin(1/2*d*x+1/2 *c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(10*A*EllipticF(cos(1/2*d*x+1/ 2*c),2^(1/2))+21*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-5*B*EllipticF(cos (1/2*d*x+1/2*c),2^(1/2))-12*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*C*El lipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/ 2)))*cos(1/2*d*x+1/2*c)+16*A*sin(1/2*d*x+1/2*c)^8+(-76*A+24*B-12*C)*sin(1/ 2*d*x+1/2*c)^6+(84*A-34*B+16*C)*sin(1/2*d*x+1/2*c)^4+(-25*A+11*B-5*C)*sin( 1/2*d*x+1/2*c)^2)/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 = 0.11 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.89 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-10 i \, A + 5 i \, B - 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (10 i \, A - 5 i \, B + 2 i \, C\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 (7 i \, A - 4 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (7 i \, A - 4 i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A - 4 i \, B + i \, C\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 (-7 i \, A + 4 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-7 i \, A + 4 i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A + 4 i \, B - i \, C\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 ) + \frac {2 \, {\left (2 \, A \cos \left (d x + c\right )^{3} + {\left (13 \, A - 6 \, B + 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (10 \, A - 5 \, B + 2 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\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 )}} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c) )^2,x, algorithm="fricas")
1/6*((sqrt(2)*(-10*I*A + 5*I*B - 2*I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(10*I*A - 5*I*B + 2*I*C)*cos(d*x + c) + sqrt(2)*(-10*I*A + 5*I*B - 2*I*C))*weiers trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(10*I*A - 5 *I*B + 2*I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(-10*I*A + 5*I*B - 2*I*C)*cos(d*x + c) + sqrt(2)*(10*I*A - 5*I*B + 2*I*C))*weierstrassPInverse(-4, 0, cos(d *x + c) - I*sin(d*x + c)) - 3*(sqrt(2)*(7*I*A - 4*I*B + I*C)*cos(d*x + c)^ 2 + 2*sqrt(2)*(7*I*A - 4*I*B + I*C)*cos(d*x + c) + sqrt(2)*(7*I*A - 4*I*B + I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I *sin(d*x + c))) - 3*(sqrt(2)*(-7*I*A + 4*I*B - I*C)*cos(d*x + c)^2 + 2*sqr t(2)*(-7*I*A + 4*I*B - I*C)*cos(d*x + c) + sqrt(2)*(-7*I*A + 4*I*B - I*C)) *weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c))) + 2*(2*A*cos(d*x + c)^3 + (13*A - 6*B + 3*C)*cos(d*x + c)^2 + (10 *A - 5*B + 2*C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos( d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 2 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + \sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx}{a^{2}} \]
(Integral(A/(sec(c + d*x)**(7/2) + 2*sec(c + d*x)**(5/2) + sec(c + d*x)**( 3/2)), x) + Integral(B*sec(c + d*x)/(sec(c + d*x)**(7/2) + 2*sec(c + d*x)* *(5/2) + sec(c + d*x)**(3/2)), x) + Integral(C*sec(c + d*x)**2/(sec(c + d* x)**(7/2) + 2*sec(c + d*x)**(5/2) + sec(c + d*x)**(3/2)), x))/a**2
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c) )^2,x, algorithm="maxima")
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2} \sec \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c) )^2,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^2* sec(d*x + c)^(3/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]
int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^2*(1/cos (c + d*x))^(3/2)),x)