Integrand size = 33, antiderivative size = 228 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\frac {(49 A-9 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}-\frac {(13 A-3 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(8 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(13 A-3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
1/10*(49*A-9*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec (d*x+c)^(1/2)/a^3/d-1/6*(13*A-3*B)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d* x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/a^3/d-1/5*(A-B)*sec(d*x+c)^(1/2)*sin(d*x +c)/d/(a+a*sec(d*x+c))^3-1/15*(8*A-3*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d/(a +a*sec(d*x+c))^2-1/6*(13*A-3*B)*sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a^3+a^3*sec (d*x+c))
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 6.67 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.60 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=-\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)) (\cos (d x)+i \sin (d x)) \left (160 (13 A-3 B) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+i (49 A-9 B) e^{-\frac {3}{2} i (c+d x)} \left (1+e^{i (c+d x)}\right )^5 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+2 \cos (c+d x) \left (-30 i (49 A-9 B) \cos \left (\frac {1}{2} (c+d x)\right )-15 i (49 A-9 B) \cos \left (\frac {3}{2} (c+d x)\right )-147 i A \cos \left (\frac {5}{2} (c+d x)\right )+27 i B \cos \left (\frac {5}{2} (c+d x)\right )+142 A \sin \left (\frac {1}{2} (c+d x)\right )-42 B \sin \left (\frac {1}{2} (c+d x)\right )+205 A \sin \left (\frac {3}{2} (c+d x)\right )-45 B \sin \left (\frac {3}{2} (c+d x)\right )+87 A \sin \left (\frac {5}{2} (c+d x)\right )-27 B \sin \left (\frac {5}{2} (c+d x)\right )\right )\right )}{120 a^3 d (B+A \cos (c+d x)) (1+\sec (c+d x))^3} \] Input:
Integrate[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3) ,x]
Output:
-1/120*(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*(Cos[d*x] + I*Sin[d*x])*(160*(13*A - 3*B)*Cos[(c + d*x)/2]^5*Sqrt[Cos[c + d*x]]*Ell ipticF[(c + d*x)/2, 2] + (I*(49*A - 9*B)*(1 + E^(I*(c + d*x)))^5*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x)) ])/E^(((3*I)/2)*(c + d*x)) + 2*Cos[c + d*x]*((-30*I)*(49*A - 9*B)*Cos[(c + d*x)/2] - (15*I)*(49*A - 9*B)*Cos[(3*(c + d*x))/2] - (147*I)*A*Cos[(5*(c + d*x))/2] + (27*I)*B*Cos[(5*(c + d*x))/2] + 142*A*Sin[(c + d*x)/2] - 42*B *Sin[(c + d*x)/2] + 205*A*Sin[(3*(c + d*x))/2] - 45*B*Sin[(3*(c + d*x))/2] + 87*A*Sin[(5*(c + d*x))/2] - 27*B*Sin[(5*(c + d*x))/2])))/(a^3*d*E^(I*d* x)*(B + A*Cos[c + d*x])*(1 + Sec[c + d*x])^3)
Time = 1.42 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 4508, 27, 3042, 4508, 3042, 4508, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\int \frac {a (11 A-B)-5 a (A-B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (11 A-B)-5 a (A-B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (11 A-B)-5 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (41 A-6 B)-3 a^2 (8 A-3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {a^2 (41 A-6 B)-3 a^2 (8 A-3 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4508 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^3 (49 A-9 B)-5 a^3 (13 A-3 B) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx}{a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^3 (49 A-9 B)-5 a^3 (13 A-3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {3 a^3 (49 A-9 B)-5 a^3 (13 A-3 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {\frac {3 a^3 (49 A-9 B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 a^3 (13 A-3 B) \int \sqrt {\sec (c+d x)}dx}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 a^3 (49 A-9 B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 a^3 (13 A-3 B) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {\frac {3 a^3 (49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-5 a^3 (13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {3 a^3 (49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 a^3 (13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {\frac {\frac {6 a^3 (49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 a^3 (13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {\frac {\frac {6 a^3 (49 A-9 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 a^3 (13 A-3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{2 a^2}-\frac {5 a^2 (13 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}-\frac {2 a (8 A-3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{5 d (a \sec (c+d x)+a)^3}\) |
Input:
Int[(A + B*Sec[c + d*x])/(Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3),x]
Output:
-1/5*((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + ((-2*a*(8*A - 3*B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + (((6*a^3*(49*A - 9*B)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (10*a^3*(13*A - 3*B)*Sqrt[Cos[c + d*x]]*Ellipt icF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/(2*a^2) - (5*a^2*(13*A - 3*B)*S qrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(10*a^2 )
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_))^(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]
Leaf count of result is larger than twice the leaf count of optimal. \(450\) vs. \(2(207)=414\).
Time = 5.43 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.98
| 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 (348 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+130 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 )+294 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 )-578 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+198 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+264 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-114 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-37 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+27 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}\) | \(451\) |
Input:
int((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x,method=_RETURNV ERBOSE)
Output:
1/60/a^3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(348*A*co s(1/2*d*x+1/2*c)^8+130*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2 94*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)* EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-578*A*cos(1/2*d*x+1/2*c)^6+198*B*cos (1/2*d*x+1/2*c)^6+264*A*cos(1/2*d*x+1/2*c)^4-114*B*cos(1/2*d*x+1/2*c)^4-37 *A*cos(1/2*d*x+1/2*c)^2+27*B*cos(1/2*d*x+1/2*c)^2+3*A-3*B)/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 complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.10 \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorith m="fricas")
Output:
-1/60*(5*(sqrt(2)*(-13*I*A + 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-13*I*A + 3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-13*I*A + 3*I*B)*cos(d*x + c) + sqrt(2) *(-13*I*A + 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*(13*I*A - 3*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(13*I*A - 3*I *B)*cos(d*x + c)^2 + 3*sqrt(2)*(13*I*A - 3*I*B)*cos(d*x + c) + sqrt(2)*(13 *I*A - 3*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 3*(sqrt(2)*(-49*I*A + 9*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-49*I*A + 9*I*B) *cos(d*x + c)^2 + 3*sqrt(2)*(-49*I*A + 9*I*B)*cos(d*x + c) + sqrt(2)*(-49* I*A + 9*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(49*I*A - 9*I*B)*cos(d*x + c)^3 + 3*sqr t(2)*(49*I*A - 9*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(49*I*A - 9*I*B)*cos(d*x + c) + sqrt(2)*(49*I*A - 9*I*B))*weierstrassZeta(-4, 0, weierstrassPInvers e(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(29*A - 9*B)*cos(d*x + c)^ 3 + 2*(73*A - 18*B)*cos(d*x + c)^2 + 5*(13*A - 3*B)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(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)
\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {5}{2}}{\left (c + d x \right )} + 3 \sec ^{\frac {3}{2}}{\left (c + d x \right )} + \sqrt {\sec {\left (c + d x \right )}}}\, dx}{a^{3}} \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)**(1/2)/(a+a*sec(d*x+c))**3,x)
Output:
(Integral(A/(sec(c + d*x)**(7/2) + 3*sec(c + d*x)**(5/2) + 3*sec(c + d*x)* *(3/2) + sqrt(sec(c + d*x))), x) + Integral(B*sec(c + d*x)/(sec(c + d*x)** (7/2) + 3*sec(c + d*x)**(5/2) + 3*sec(c + d*x)**(3/2) + sqrt(sec(c + d*x)) ), x))/a**3
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorith m="maxima")
Output:
Timed out
\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\int { \frac {B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3} \sqrt {\sec \left (d x + c\right )}} \,d x } \] Input:
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, algorith m="giac")
Output:
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^3*sqrt(sec(d*x + c))) , x)
Timed out. \[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \] Input:
int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(1/2)),x )
Output:
int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(1/2)), x)
\[ \int \frac {A+B \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4}+3 \sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+\sec \left (d x +c \right )}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) b}{a^{3}} \] Input:
int((A+B*sec(d*x+c))/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x)
Output:
(int(sqrt(sec(c + d*x))/(sec(c + d*x)**4 + 3*sec(c + d*x)**3 + 3*sec(c + d *x)**2 + sec(c + d*x)),x)*a + int(sqrt(sec(c + d*x))/(sec(c + d*x)**3 + 3* sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*b)/a**3