3.6.74 \(\int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [574]

3.6.74.1 Optimal result
3.6.74.2 Mathematica [C] (warning: unable to verify)
3.6.74.3 Rubi [A] (verified)
3.6.74.4 Maple [B] (verified)
3.6.74.5 Fricas [C] (verification not implemented)
3.6.74.6 Sympy [F(-1)]
3.6.74.7 Maxima [F(-1)]
3.6.74.8 Giac [F]
3.6.74.9 Mupad [F(-1)]

3.6.74.1 Optimal result

Integrand size = 43, antiderivative size = 231 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {(9 A+B-C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(3 A+B+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}-\frac {(A-B+C) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {(6 A-B-4 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(3 A+B+C) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 d \left (a^3+a^3 \sec (c+d x)\right )} \]

output
-1/5*(A-B+C)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3+1/15*(6*A-B- 
4*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^2+1/6*(3*A+B+C)*sin( 
d*x+c)*sec(d*x+c)^(1/2)/d/(a^3+a^3*sec(d*x+c))-1/10*(9*A+B-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^3/d+1/6*(3*A+B+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^3/d
 
3.6.74.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.48 (sec) , antiderivative size = 1431, normalized size of antiderivative = 6.19 \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \]

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])^3,x]
 
output
(6*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]^6*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]*Sec[c + d*x]*(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])^3) + (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))]*Cos[c/2 + (d*x)/2]^6*Csc[c/ 
2]*(-3*Sqrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hy 
pergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])*Sec[c/2]*Sec[c + d*x 
]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(15*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])^3) - (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 
))]*Cos[c/2 + (d*x)/2]^6*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]*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/ 
(15*d*E^(I*d*x)*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + a*S 
ec[c + d*x])^3) + (4*A*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d*x]]*Csc[c/2]*El 
lipticF[(c + d*x)/2, 2]*Sec[c/2]*Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + 
C*Sec[c + d*x]^2)*Sin[c])/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d 
*x])*(a + a*Sec[c + d*x])^3) + (4*B*Cos[c/2 + (d*x)/2]^6*Sqrt[Cos[c + d...
 
3.6.74.3 Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.06, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.372, Rules used = {3042, 4572, 27, 3042, 4507, 25, 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.

\(\displaystyle \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a \sec (c+d x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\)

\(\Big \downarrow \) 4572

\(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x)} (a (9 A+B-C)-a (3 A-3 B-7 C) \sec (c+d x))}{2 (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {\sec (c+d x)} (a (9 A+B-C)-a (3 A-3 B-7 C) \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (a (9 A+B-C)-a (3 A-3 B-7 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4507

\(\displaystyle \frac {\frac {\int -\frac {a^2 (6 A-B-4 C)-3 a^2 (3 A+2 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}+\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\int \frac {a^2 (6 A-B-4 C)-3 a^2 (3 A+2 B+3 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} (\sec (c+d x) a+a)}dx}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\int \frac {a^2 (6 A-B-4 C)-3 a^2 (3 A+2 B+3 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {\int \frac {3 a^3 (9 A+B-C)-5 a^3 (3 A+B+C) \sec (c+d x)}{2 \sqrt {\sec (c+d x)}}dx}{a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {\int \frac {3 a^3 (9 A+B-C)-5 a^3 (3 A+B+C) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {\int \frac {3 a^3 (9 A+B-C)-5 a^3 (3 A+B+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {3 a^3 (9 A+B-C) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 a^3 (3 A+B+C) \int \sqrt {\sec (c+d x)}dx}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {3 a^3 (9 A+B-C) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 a^3 (3 A+B+C) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {3 a^3 (9 A+B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx-5 a^3 (3 A+B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {3 a^3 (9 A+B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 a^3 (3 A+B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {\frac {6 a^3 (9 A+B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-5 a^3 (3 A+B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {2 a (6 A-B-4 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {\frac {6 a^3 (9 A+B-C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}-\frac {10 a^3 (3 A+B+C) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}}{2 a^2}-\frac {5 a^2 (3 A+B+C) \sin (c+d x) \sqrt {\sec (c+d x)}}{d (a \sec (c+d x)+a)}}{3 a^2}}{10 a^2}-\frac {(A-B+C) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

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])^3,x]
 
output
-1/5*((A - B + C)*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(d*(a + a*Sec[c + d*x]) 
^3) + ((2*a*(6*A - B - 4*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*S 
ec[c + d*x])^2) - (((6*a^3*(9*A + B - C)*Sqrt[Cos[c + d*x]]*EllipticE[(c + 
 d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d - (10*a^3*(3*A + B + C)*Sqrt[Cos[c + d*x 
]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/(2*a^2) - (5*a^2*(3*A 
+ B + C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2 
))/(10*a^2)
 

3.6.74.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119
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]
 

rule 3120
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]
 

rule 4258
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]
 

rule 4274
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]
 

rule 4507
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]
 

rule 4508
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]
 

rule 4572
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)]
 
3.6.74.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(623\) vs. \(2(259)=518\).

Time = 3.79 (sec) , antiderivative size = 624, 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 (108 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+30 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 )+54 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 )+12 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+10 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 )+6 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 )-12 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+10 C \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 )-6 C \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 )-198 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-2 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+22 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+114 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-6 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-27 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+17 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+3 A -3 B +3 C \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}\) \(624\)

input
int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^3,x, 
method=_RETURNVERBOSE)
 
output
-1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(108*A*cos(1 
/2*d*x+1/2*c)^8+30*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))+54*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))+12*B*cos(1/2*d*x+1/2*c)^8+ 
10*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))+6*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)*Ellipti 
cE(cos(1/2*d*x+1/2*c),2^(1/2))-12*C*cos(1/2*d*x+1/2*c)^8+10*C*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)*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-6*C*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))-198*A*cos(1/2*d*x+1/2*c)^6-2*B*cos(1/2*d*x+1/2*c)^6+22*C*cos 
(1/2*d*x+1/2*c)^6+114*A*cos(1/2*d*x+1/2*c)^4-24*B*cos(1/2*d*x+1/2*c)^4-6*C 
*cos(1/2*d*x+1/2*c)^4-27*A*cos(1/2*d*x+1/2*c)^2+17*B*cos(1/2*d*x+1/2*c)^2- 
7*C*cos(1/2*d*x+1/2*c)^2+3*A-3*B+3*C)/a^3/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
 
3.6.74.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 529, normalized size of antiderivative = 2.29 \[ \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))^3} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (3 i \, A + i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (3 i \, A + i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (3 i \, A + i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + i \, B + 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 (-3 i \, A - i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-3 i \, A - i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-3 i \, A - i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - i \, B - 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 (9 i \, A + i \, B - i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (9 i \, A + i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (9 i \, A + i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (9 i \, A + 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 (-9 i \, A - i \, B + i \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, \sqrt {2} {\left (-9 i \, A - i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, \sqrt {2} {\left (-9 i \, A - i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-9 i \, A - 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 (3 \, {\left (9 \, A + B - C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (18 \, A + 7 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \, {\left (3 \, A + B + C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) 
)^3,x, algorithm="fricas")
 
output
-1/60*(5*(sqrt(2)*(3*I*A + I*B + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(3*I*A + 
I*B + I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(3*I*A + I*B + I*C)*cos(d*x + c) + s 
qrt(2)*(3*I*A + I*B + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*si 
n(d*x + c)) + 5*(sqrt(2)*(-3*I*A - I*B - I*C)*cos(d*x + c)^3 + 3*sqrt(2)*( 
-3*I*A - I*B - I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-3*I*A - I*B - I*C)*cos(d* 
x + c) + sqrt(2)*(-3*I*A - I*B - I*C))*weierstrassPInverse(-4, 0, cos(d*x 
+ c) - I*sin(d*x + c)) + 3*(sqrt(2)*(9*I*A + I*B - I*C)*cos(d*x + c)^3 + 3 
*sqrt(2)*(9*I*A + I*B - I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(9*I*A + I*B - I*C 
)*cos(d*x + c) + sqrt(2)*(9*I*A + I*B - I*C))*weierstrassZeta(-4, 0, weier 
strassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 3*(sqrt(2)*(-9*I*A 
 - I*B + I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-9*I*A - I*B + I*C)*cos(d*x + c) 
^2 + 3*sqrt(2)*(-9*I*A - I*B + I*C)*cos(d*x + c) + sqrt(2)*(-9*I*A - I*B + 
 I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I* 
sin(d*x + c))) - 2*(3*(9*A + B - C)*cos(d*x + c)^3 + 2*(18*A + 7*B - 2*C)* 
cos(d*x + c)^2 + 5*(3*A + B + C)*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)
 
3.6.74.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)*sec(d*x+c)**(1/2)/(a+a*sec(d*x+ 
c))**3,x)
 
output
Timed out
 
3.6.74.7 Maxima [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\text {Timed out} \]

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) 
)^3,x, algorithm="maxima")
 
output
Timed out
 
3.6.74.8 Giac [F]

\[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt {\sec \left (d x + c\right )}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

input
integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c) 
)^3,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)^3, x)
 
3.6.74.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\int \frac {\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3} \,d x \]

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))^3,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))^3, x)