Integrand size = 43, antiderivative size = 214 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {3 (7 A-5 B+5 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(5 A-5 B+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}+\frac {(7 A-5 B+5 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(5 A-5 B+3 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))} \]
1/5*(7*A-5*B+5*C)*sin(d*x+c)/a/d/sec(d*x+c)^(3/2)-(A-B+C)*sin(d*x+c)/d/sec (d*x+c)^(3/2)/(a+a*sec(d*x+c))-1/3*(5*A-5*B+3*C)*sin(d*x+c)/a/d/sec(d*x+c) ^(1/2)+3/5*(7*A-5*B+5*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E llipticE(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d -1/3*(5*A-5*B+3*C)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipti cF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.78 (sec) , antiderivative size = 1350, normalized size of antiderivative = 6.31 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx =\text {Too large to display} \]
Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])),x]
(-7*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^( (2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-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))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec [c + d*x]^2))/(5*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d *x])*(a + a*Sec[c + d*x])) + (Sqrt[2]*B*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I) *(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/2 + (d*x)/2]^2*Cos[c + d *x]*Csc[c/2]*(-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))])*Sec[c/2]*( A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (Sqrt[2]*C*Sqrt[E^(I* (c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[c/ 2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*(-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))])*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(d*E^(I*d* x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) - (10*A*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d *x)/2, 2]*Sec[c/2]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sin[c])/(3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + a* Sec[c + d*x])) + (10*B*Cos[c/2 + (d*x)/2]^2*Sqrt[Cos[c + d*x]]*Csc[c/2]...
Time = 0.86 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.92, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.279, Rules used = {3042, 4572, 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.
| \(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\) |
| \(\Big \downarrow \) 4572 |
| \(\displaystyle \frac {\int \frac {a (7 A-5 B+5 C)-a (5 A-5 B+3 C) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {a (7 A-5 B+5 C)-a (5 A-5 B+3 C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (7 A-5 B+5 C)-a (5 A-5 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)}dx-a (5 A-5 B+3 C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-a (5 A-5 B+3 C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4256 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-a (5 A-5 B+3 C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
| \(\Big \downarrow \) 3120 |
| \(\displaystyle \frac {a (7 A-5 B+5 C) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-a (5 A-5 B+3 C) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )}{2 a^2}-\frac {(A-B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)}\) |
-(((A - B + C)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x]))) + (a*(7*A - 5*B + 5*C)*((6*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sq rt[Sec[c + d*x]])/(5*d) + (2*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2))) - a*( 5*A - 5*B + 3*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]])))/(2*a^2)
3.6.63.3.1 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[((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 = 3.86 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.50
| method | result | size |
| default | \(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (-\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (25 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-25 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-45 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+15 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+48 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-56 A -40 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-30 A +90 B -30 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (23 A -35 B +15 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\) | \(320\) |
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x,me thod=_RETURNVERBOSE)
-1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*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)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(25 *A*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+63*A*EllipticE(cos(1/2*d*x+1/2*c) ,2^(1/2))-25*B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-45*B*EllipticE(cos(1/ 2*d*x+1/2*c),2^(1/2))+15*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*C*Elli pticE(cos(1/2*d*x+1/2*c),2^(1/2)))+48*A*sin(1/2*d*x+1/2*c)^8+(-56*A-40*B)* sin(1/2*d*x+1/2*c)^6+(-30*A+90*B-30*C)*sin(1/2*d*x+1/2*c)^4+(23*A-35*B+15* C)*sin(1/2*d*x+1/2*c)^2)/a/cos(1/2*d*x+1/2*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*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 = 0.12 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.43 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-5 i \, A + 5 i \, B - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A + 5 i \, B - 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (5 i \, A - 5 i \, B + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A - 5 i \, B + 3 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-7 i \, A + 5 i \, B - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-7 i \, A + 5 i \, B - 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (7 i \, A - 5 i \, B + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (7 i \, A - 5 i \, B + 5 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, A \cos \left (d x + c\right )^{3} - 2 \, {\left (2 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} - 5 \, {\left (5 \, A - 5 \, B + 3 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c) ),x, algorithm="fricas")
-1/30*(5*(sqrt(2)*(-5*I*A + 5*I*B - 3*I*C)*cos(d*x + c) + sqrt(2)*(-5*I*A + 5*I*B - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c) ) + 5*(sqrt(2)*(5*I*A - 5*I*B + 3*I*C)*cos(d*x + c) + sqrt(2)*(5*I*A - 5*I *B + 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9 *(sqrt(2)*(-7*I*A + 5*I*B - 5*I*C)*cos(d*x + c) + sqrt(2)*(-7*I*A + 5*I*B - 5*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(sqrt(2)*(7*I*A - 5*I*B + 5*I*C)*cos(d*x + c) + sqrt (2)*(7*I*A - 5*I*B + 5*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4 , 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*A*cos(d*x + c)^3 - 2*(2*A - 5* B)*cos(d*x + c)^2 - 5*(5*A - 5*B + 3*C)*cos(d*x + c))*sin(d*x + c)/sqrt(co s(d*x + c)))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {\int \frac {A}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {B \sec {\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {C \sec ^{2}{\left (c + d x \right )}}{\sec ^{\frac {7}{2}}{\left (c + d x \right )} + \sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx}{a} \]
(Integral(A/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x) + Integral(B*s ec(c + d*x)/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x) + Integral(C*s ec(c + d*x)**2/(sec(c + d*x)**(7/2) + sec(c + d*x)**(5/2)), x))/a
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c) ),x, algorithm="maxima")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*se c(d*x + c)^(5/2)), x)
\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (a \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c) ),x, algorithm="giac")
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)*se c(d*x + c)^(5/2)), x)
Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]