\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\) [1017]

Optimal result

Integrand size = 43, antiderivative size = 269 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^3 d}-\frac {2 \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^4 d}+\frac {2 b^2 \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a^4 (a+b) d}+\frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b-a B) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}} \] Output:

2/5*(5*A*b^2-5*B*a*b+a^2*(3*A+5*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^3/d-2/3*(3*A*b^3-B*a^3-3*B*a*b^2+a^2*b 
*(A+3*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^4/d+2*b^2*(A*b^2-a*(B*b-C*a))*cos(d*x+c)^(1/2)*EllipticPi(sin(1 
/2*d*x+1/2*c),2*a/(a+b),2^(1/2))*sec(d*x+c)^(1/2)/a^4/(a+b)/d+2/5*A*sin(d* 
x+c)/a/d/sec(d*x+c)^(3/2)-2/3*(A*b-B*a)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx \] Input:

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + 
 b*Sec[c + d*x])),x]
 

Output:

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + 
 b*Sec[c + d*x])), x]
 

Rubi [A] (verified)

Time = 2.08 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.04, number of steps used = 18, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.419, Rules used = {3042, 4592, 27, 3042, 4592, 27, 3042, 4594, 3042, 4274, 3042, 4258, 3042, 3119, 3120, 4336, 3042, 3284}

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.

\(\displaystyle \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4592

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4594

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{a+b \sec (c+d x)}dx}{a^2}+\frac {\int \frac {3 a \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )-5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)}}dx}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\int \frac {3 a \left ((3 A+5 C) a^2-5 b B a+5 A b^2\right )-5 \left (-B a^3+b (A+3 C) a^2-3 b^2 B a+3 A b^3\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \frac {1}{\sqrt {\sec (c+d x)}}dx-5 \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right ) \int \sqrt {\sec (c+d x)}dx}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx-5 \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right ) \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\cos (c+d x)}dx-5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right ) \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {3 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right ) \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx-5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}-5 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right )}{d}}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (b+a \cos (c+d x))}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right )}{d}}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {15 b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right )}{d}}{a^2}}{3 a}}{5 a}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 A \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {\frac {10 (A b-a B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}}-\frac {\frac {30 b^2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (c+d x),2\right )}{a^2 d (a+b)}+\frac {\frac {6 a \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{d}-\frac {10 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \left (a^3 (-B)+a^2 b (A+3 C)-3 a b^2 B+3 A b^3\right )}{d}}{a^2}}{3 a}}{5 a}\)

Input:

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + b*Sec 
[c + d*x])),x]
 

Output:

(2*A*Sin[c + d*x])/(5*a*d*Sec[c + d*x]^(3/2)) - (-1/3*(((6*a*(5*A*b^2 - 5* 
a*b*B + a^2*(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt 
[Sec[c + d*x]])/d - (10*(3*A*b^3 - a^3*B - 3*a*b^2*B + a^2*b*(A + 3*C))*Sq 
rt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d)/a^2 + (3 
0*b^2*(A*b^2 - a*(b*B - a*C))*Sqrt[Cos[c + d*x]]*EllipticPi[(2*a)/(a + b), 
 (c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(a^2*(a + b)*d))/a + (10*(A*b - a*B)* 
Sin[c + d*x])/(3*a*d*Sqrt[Sec[c + d*x]]))/(5*a)
 

Defintions of rubi rules used

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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

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 4336

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]]   Int[ 
1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4592

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d 
*Csc[e + f*x])^n/(a*f*n)), x] + Simp[1/(a*d*n)   Int[(a + b*Csc[e + f*x])^m 
*(d*Csc[e + f*x])^(n + 1)*Simp[a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)* 
Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d 
, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
 

rule 4594

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2)   Int[(d*Csc[e + 
f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2   Int[(a*A - (A*b - a 
*B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, 
B, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(800\) vs. \(2(250)=500\).

Time = 8.12 (sec) , antiderivative size = 801, normalized size of antiderivative = 2.98

method	result	size
default	\(\text {Expression too large to display}\)	\(801\)

Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c)),x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b^2*(A*b^2- 
B*a*b+C*a^2)/a^3/(a^2-a*b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(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)*Ellip 
ticPi(cos(1/2*d*x+1/2*c),2*a/(a-b),2^(1/2))-2*(A*a^3+A*a^2*b+A*a*b^2+A*b^3 
-B*a^3-B*a^2*b-B*a*b^2+C*a^3+C*a^2*b)/a^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2 
*cos(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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-4/5*A/a/(-2*sin(1/2*d*x+1/ 
2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2 
*c)-14*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2*cos( 
1/2*d*x+1/2*c)-5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^( 
1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+9*(2*sin(1/2*d*x+1/2*c)^2-1)^(1 
/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2))+2/ 
a^3*(3*A*a^2+2*A*a*b+A*b^2-2*B*a^2-B*a*b+C*a^2)*(sin(1/2*d*x+1/2*c)^2)^(1/ 
2)*(-2*cos(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)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d 
*x+1/2*c),2^(1/2)))-4/3/a^2*(3*A*a+A*b-B*a)*(2*cos(1/2*d*x+1/2*c)*sin(1/2* 
d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+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)*EllipticF(cos(1/2*d*x+1/2*c),2^ 
(1/2))-3*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^( 
1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*...
 

Fricas [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) 
),x, algorithm="fricas")
 

Output:

integral((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(b*sec 
(d*x + c)^4 + a*sec(d*x + c)^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\text {Timed out} \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+b*sec(d*x+ 
c)),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) 
),x, algorithm="maxima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*se 
c(d*x + c)^(5/2)), x)
 

Giac [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c) 
),x, algorithm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)*se 
c(d*x + c)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + b/cos(c + d*x))*(1/cos(c 
 + d*x))^(5/2)),x)
 

Output:

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + b/cos(c + d*x))*(1/cos(c 
 + d*x))^(5/2)), x)
 

Reduce [F]

\[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))} \, dx=\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{4} b +\sec \left (d x +c \right )^{3} a}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{3} b +\sec \left (d x +c \right )^{2} a}d x \right ) b +\left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{2} b +\sec \left (d x +c \right ) a}d x \right ) c \] Input:

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c)),x)
 

Output:

int(sqrt(sec(c + d*x))/(sec(c + d*x)**4*b + sec(c + d*x)**3*a),x)*a + int( 
sqrt(sec(c + d*x))/(sec(c + d*x)**3*b + sec(c + d*x)**2*a),x)*b + int(sqrt 
(sec(c + d*x))/(sec(c + d*x)**2*b + sec(c + d*x)*a),x)*c