Integrand size = 43, antiderivative size = 173 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {(A-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a^2 d}+\frac {(2 A+B+2 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {(A-C) \sqrt {\sec (c+d x)} \sin (c+d x)}{a^2 d (1+\sec (c+d x))}-\frac {(A-B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2} \] Output:
-(A-C)*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^2/d+1/3*(2*A+B+2*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c, 2^(1/2))*sec(d*x+c)^(1/2)/a^2/d+(A-C)*sec(d*x+c)^(1/2)*sin(d*x+c)/a^2/d/(1 +sec(d*x+c))-1/3*(A-B+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 7.55 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.95 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=-\frac {\cos ^4\left (\frac {1}{2} (c+d x)\right ) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\frac {8 (2 A+B+2 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{\sqrt {\cos (c+d x)}}-\frac {16 i (A-C) e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+\frac {16 e^{i c} \left (\cos \left (\frac {d x}{2}\right )+i \sin \left (\frac {d x}{2}\right )\right ) \left ((5 A+B-7 C) e^{i c}-\left (B+2 C+3 C e^{2 i c}-A \left (4+3 e^{2 i c}\right )\right ) \cos (d x)+i \left (-4 A+B+2 C+3 A e^{2 i c}-3 C e^{2 i c}\right ) \sin (d x)\right )}{\left (i \left (1+e^{i c}\right ) \cos \left (\frac {d x}{2}\right )-\left (-1+e^{i c}\right ) \sin \left (\frac {d x}{2}\right )\right )^3}\right )}{3 a^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) \sqrt {\sec (c+d x)} (1+\sec (c+d x))^2} \] Input:
Integrate[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^2,x]
Output:
-1/3*(Cos[(c + d*x)/2]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-8*(2*A + B + 2*C)*EllipticF[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] - ((16*I)*(A - C )*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x) )])/Sqrt[1 + E^((2*I)*(c + d*x))] + (16*E^(I*c)*(Cos[(d*x)/2] + I*Sin[(d*x )/2])*((5*A + B - 7*C)*E^(I*c) - (B + 2*C + 3*C*E^((2*I)*c) - A*(4 + 3*E^( (2*I)*c)))*Cos[d*x] + I*(-4*A + B + 2*C + 3*A*E^((2*I)*c) - 3*C*E^((2*I)*c ))*Sin[d*x]))/(I*(1 + E^(I*c))*Cos[(d*x)/2] - (-1 + E^(I*c))*Sin[(d*x)/2]) ^3))/(a^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sqrt[Sec[c + d*x]]*(1 + Sec[c + d*x])^2)
Time = 1.01 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.302, Rules used = {3042, 4572, 27, 3042, 4507, 25, 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.
| \(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x)} (a (5 A+B-C)-a (A-B-5 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x)} (a (5 A+B-C)-a (A-B-5 C) \sec (c+d x))}{\sec (c+d x) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (5 A+B-C)-a (A-B-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4507 |
| \(\displaystyle \frac {\frac {\int -\frac {3 a^2 (A-C)-a^2 (2 A+B+2 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}+\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {3 a^2 (A-C)-a^2 (2 A+B+2 C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\int \frac {3 a^2 (A-C)-a^2 (2 A+B+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 (A-C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-a^2 (2 A+B+2 C) \int \sqrt {\sec (c+d x)}dx}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 (A-C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-a^2 (2 A+B+2 C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-a^2 (2 A+B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {3 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-a^2 (2 A+B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\frac {6 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-a^2 (2 A+B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {\frac {6 (A-C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (\sec (c+d x)+1)}-\frac {\frac {6 a^2 (A-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {2 a^2 (2 A+B+2 C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{a^2}}{6 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
Input:
Int[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Se c[c + d*x])^2,x]
Output:
-1/3*((A - B + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) ^2) + (-(((6*a^2*(A - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt [Sec[c + d*x]])/d - (2*a^2*(2*A + B + 2*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2) + (6*(A - C)*Sqrt[Sec[c + d*x]]* Sin[c + d*x])/(d*(1 + Sec[c + d*x])))/(6*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[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. \(508\) vs. \(2(160)=320\).
Time = 4.29 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.94
| 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 (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 A \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 ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+6 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 )+2 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 )-12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+4 C \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 )-6 C \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 )-20 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+16 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+9 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-A +B -C \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}\) | \(509\) |
Input:
int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x, method=_RETURNVERBOSE)
Output:
-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*A*cos(1/2 *d*x+1/2*c)^6+4*A*(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))*cos(1/2*d*x+1/2*c)^3+6*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))+2*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))-12*C*cos(1/2*d*x+1/2*c)^6+4*C*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(co s(1/2*d*x+1/2*c),2^(1/2))-6*C*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))-20*A*cos(1/2*d*x+1/2*c)^4+2*B*cos(1/2*d*x+1/2*c)^4+16*C*cos(1/2*d*x+1/ 2*c)^4+9*A*cos(1/2*d*x+1/2*c)^2-3*B*cos(1/2*d*x+1/2*c)^2-3*C*cos(1/2*d*x+1 /2*c)^2-A+B-C)/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 complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.20 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {{\left (\sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} {\left (-2 i \, A - i \, B - 2 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 i \, A + i \, B + 2 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (i \, A - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A - i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-i \, A + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A + i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (3 \, {\left (A - C\right )} \cos \left (d x + c\right )^{2} + {\left (2 \, A + B - 4 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="fricas")
Output:
1/6*((sqrt(2)*(-2*I*A - I*B - 2*I*C)*cos(d*x + c)^2 - 2*sqrt(2)*(2*I*A + I *B + 2*I*C)*cos(d*x + c) + sqrt(2)*(-2*I*A - I*B - 2*I*C))*weierstrassPInv erse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(2*I*A + I*B + 2*I*C )*cos(d*x + c)^2 - 2*sqrt(2)*(-2*I*A - I*B - 2*I*C)*cos(d*x + c) + sqrt(2) *(2*I*A + I*B + 2*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* x + c)) - 3*(sqrt(2)*(I*A - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(I*A - I*C)*co s(d*x + c) + sqrt(2)*(I*A - I*C))*weierstrassZeta(-4, 0, weierstrassPInver se(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-I*A + I*C)*cos(d* x + c)^2 + 2*sqrt(2)*(-I*A + I*C)*cos(d*x + c) + sqrt(2)*(-I*A + I*C))*wei erstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2*(3*(A - C)*cos(d*x + c)^2 + (2*A + B - 4*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)
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {A \sqrt {\sec {\left (c + d x \right )}}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{\frac {3}{2}}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{\frac {5}{2}}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \] Input:
integrate(sec(d*x+c)**(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+ c))**2,x)
Output:
(Integral(A*sqrt(sec(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**(3/2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x ) + Integral(C*sec(c + d*x)**(5/2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x))/a**2
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a*se c(d*x + c) + a)^2, x)
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}} \,d x } \] Input:
integrate(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c) )^2,x, algorithm="giac")
Output:
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a*se c(d*x + c) + a)^2, x)
Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2} \,d x \] Input:
int(((1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2,x)
Output:
int(((1/cos(c + d*x))^(1/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^2, x)
\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx=\frac {\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) c +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right ) b}{a^{2}} \] Input:
int(sec(d*x+c)^(1/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^2,x)
Output:
(int(sqrt(sec(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*a + int( (sqrt(sec(c + d*x))*sec(c + d*x)**2)/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1 ),x)*c + int((sqrt(sec(c + d*x))*sec(c + d*x))/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1),x)*b)/a**2