Integrand size = 43, antiderivative size = 207 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {(B-4 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(A+2 B-5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {(B-4 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d}+\frac {(A+2 B-5 C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
1/3*(A+2*B-5*C)*sec(d*x+c)^(3/2)*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A-B+ C)*sec(d*x+c)^(5/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2-(B-4*C)*sin(d*x+c)*sec (d*x+c)^(1/2)/a^2/d+(B-4*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)/ a^2/d+1/3*(A+2*B-5*C)*(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))*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 = 4.33 (sec) , antiderivative size = 567, normalized size of antiderivative = 2.74 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 \sqrt {2} B e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+8 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+4 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+8 B \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-20 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}-2 \sqrt {\sec (c+d x)} \left (6 (B-4 C) \cos (d x) \csc (c)-2 (A+2 B-5 C) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )+(A-B+C) \sec \left (\frac {c}{2}\right ) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sin \left (\frac {d x}{2}\right )-2 (A+2 B-5 C) \tan \left (\frac {c}{2}\right )+(A-B+C) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )\right )}{3 a^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (1+\sec (c+d x))^2} \]
Integrate[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^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)*((-2*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 ) + (8*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) + 4*A*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Se c[c + d*x]] + 8*B*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] - 20*C*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d *x]] - 2*Sqrt[Sec[c + d*x]]*(6*(B - 4*C)*Cos[d*x]*Csc[c] - 2*(A + 2*B - 5* 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] - 2*(A + 2*B - 5*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 = 1.17 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4572, 27, 3042, 4507, 3042, 4274, 3042, 4255, 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.
| \(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (3 a (A+B-C)+a (A-B+7 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sec ^{\frac {3}{2}}(c+d x) (3 a (A+B-C)+a (A-B+7 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (3 a (A+B-C)+a (A-B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int \sqrt {\sec (c+d x)} \left (a^2 (A+2 B-5 C)-3 a^2 (B-4 C) \sec (c+d x)\right )dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a^2 (A+2 B-5 C)-3 a^2 (B-4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {a^2 (A+2 B-5 C) \int \sqrt {\sec (c+d x)}dx-3 a^2 (B-4 C) \int \sec ^{\frac {3}{2}}(c+d x)dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (A+2 B-5 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^2 (B-4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {\frac {a^2 (A+2 B-5 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^2 (B-4 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (A+2 B-5 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx-3 a^2 (B-4 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {a^2 (A+2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx-3 a^2 (B-4 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {a^2 (A+2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^2 (B-4 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {a^2 (A+2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-3 a^2 (B-4 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {2 a^2 (A+2 B-5 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}-3 a^2 (B-4 C) \left (\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}+\frac {2 (A+2 B-5 C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
-1/3*((A - B + C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) ^2) + ((2*(A + 2*B - 5*C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(1 + Sec[c + d*x])) + ((2*a^2*(A + 2*B - 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2 , 2]*Sqrt[Sec[c + d*x]])/d - 3*a^2*(B - 4*C)*((-2*Sqrt[Cos[c + d*x]]*Ellip ticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d))/a^2)/(6*a^2)
3.6.66.3.1 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)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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[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[((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)]
Leaf count of result is larger than twice the leaf count of optimal. \(558\) vs. \(2(239)=478\).
Time = 2.68 (sec) , antiderivative size = 559, normalized size of antiderivative = 2.70
| 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 {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 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 {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+2 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 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 )+12 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (B -4 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A -10 B +43 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A -7 B +37 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{6 a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-1+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(559\) |
int(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^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)/a^2*(-2*(si n(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)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1 /2))+2*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*EllipticE(cos(1/2*d*x+1 /2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*EllipticE(co s(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*(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)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2 ))+2*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-3*B*EllipticE(cos(1/2*d*x+1/2 *c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*EllipticE(cos( 1/2*d*x+1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)+12*(-2*sin(1/2*d*x+1/2*c)^4+si n(1/2*d*x+1/2*c)^2)^(1/2)*(B-4*C)*sin(1/2*d*x+1/2*c)^6+2*(-2*sin(1/2*d*x+1 /2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A-10*B+43*C)*sin(1/2*d*x+1/2*c)^4-(-2 *sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A-7*B+37*C)*sin(1/2*d*x +1/2*c)^2)/sin(1/2*d*x+1/2*c)^3/(2*sin(1/2*d*x+1/2*c)^2-1)/cos(1/2*d*x+1/2 *c)/(-1+sin(1/2*d*x+1/2*c)^2)/(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 = 385, normalized size of antiderivative = 1.86 \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} {\left (-i \, A - 2 i \, B + 5 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (i \, A + 2 i \, B - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 2 i \, B + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A + 2 i \, B - 5 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-i \, A - 2 i \, B + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 2 i \, B - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, B + 4 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, B + 4 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, B + 4 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, B - 4 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, B - 4 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, B - 4 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (3 \, {\left (B - 4 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (A - 4 \, B + 19 \, C\right )} \cos \left (d x + c\right ) - 6 \, C\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="fricas")
1/6*((sqrt(2)*(-I*A - 2*I*B + 5*I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(I*A + 2*I *B - 5*I*C)*cos(d*x + c) + sqrt(2)*(-I*A - 2*I*B + 5*I*C))*weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A + 2*I*B - 5*I*C )*cos(d*x + c)^2 - 2*sqrt(2)*(-I*A - 2*I*B + 5*I*C)*cos(d*x + c) + sqrt(2) *(I*A + 2*I*B - 5*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c)) - 3*(sqrt(2)*(-I*B + 4*I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-I*B + 4*I *C)*cos(d*x + c) + sqrt(2)*(-I*B + 4*I*C))*weierstrassZeta(-4, 0, weierstr assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(I*B - 4*I *C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*B - 4*I*C)*cos(d*x + c) + sqrt(2)*(I*B - 4*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(3*(B - 4*C)*cos(d*x + c)^2 - (A - 4*B + 19*C)*cos(d* x + c) - 6*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)
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{\frac {5}{2}}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{\frac {7}{2}}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
(Integral(A*sec(c + d*x)**(3/2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**(5/2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(C*sec(c + d*x)**(7/2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1) , x))/a**2
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="maxima")
\[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \]
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*se c(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\sec ^{\frac {3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \]
int(((1/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2,x)