\(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [222]

Optimal result

Integrand size = 33, antiderivative size = 261 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {7 (17 A-7 B) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{10 a^3 d}+\frac {(33 A-13 B) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{6 a^3 d}+\frac {(33 A-13 B) \sin (c+d x)}{6 a^3 d \sqrt {\sec (c+d x)}}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^3}-\frac {(2 A-B) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^2}-\frac {7 (17 A-7 B) \sin (c+d x)}{30 d \sqrt {\sec (c+d x)} \left (a^3+a^3 \sec (c+d x)\right )} \] Output:

-7/10*(17*A-7*B)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*se 
c(d*x+c)^(1/2)/a^3/d+1/6*(33*A-13*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+1/6*(33*A-13*B)*sin(d*x+c)/a^3/d 
/sec(d*x+c)^(1/2)-1/5*(A-B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c)) 
^3-1/3*(2*A-B)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^2-7/30*(17 
*A-7*B)*sin(d*x+c)/d/sec(d*x+c)^(1/2)/(a^3+a^3*sec(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 7.35 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.44 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\frac {e^{-i d x} \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)) (\cos (d x)+i \sin (d x)) \left (160 (33 A-13 B) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+7 i (17 A-7 B) e^{-\frac {3}{2} i (c+d x)} \left (1+e^{i (c+d x)}\right )^5 \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+2 \cos (c+d x) \left (-210 i (17 A-7 B) \cos \left (\frac {1}{2} (c+d x)\right )-105 i (17 A-7 B) \cos \left (\frac {3}{2} (c+d x)\right )-357 i A \cos \left (\frac {5}{2} (c+d x)\right )+147 i B \cos \left (\frac {5}{2} (c+d x)\right )+352 A \sin \left (\frac {1}{2} (c+d x)\right )-142 B \sin \left (\frac {1}{2} (c+d x)\right )+545 A \sin \left (\frac {3}{2} (c+d x)\right )-205 B \sin \left (\frac {3}{2} (c+d x)\right )+227 A \sin \left (\frac {5}{2} (c+d x)\right )-87 B \sin \left (\frac {5}{2} (c+d x)\right )+10 A \sin \left (\frac {7}{2} (c+d x)\right )\right )\right )}{120 a^3 d (B+A \cos (c+d x)) (1+\sec (c+d x))^3} \] Input:

Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^3) 
,x]
 

Output:

(Cos[(c + d*x)/2]*Sec[c + d*x]^(5/2)*(A + B*Sec[c + d*x])*(Cos[d*x] + I*Si 
n[d*x])*(160*(33*A - 13*B)*Cos[(c + d*x)/2]^5*Sqrt[Cos[c + d*x]]*EllipticF 
[(c + d*x)/2, 2] + ((7*I)*(17*A - 7*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]*((-210*I)*(17*A - 7*B)*Cos[(c + 
d*x)/2] - (105*I)*(17*A - 7*B)*Cos[(3*(c + d*x))/2] - (357*I)*A*Cos[(5*(c 
+ d*x))/2] + (147*I)*B*Cos[(5*(c + d*x))/2] + 352*A*Sin[(c + d*x)/2] - 142 
*B*Sin[(c + d*x)/2] + 545*A*Sin[(3*(c + d*x))/2] - 205*B*Sin[(3*(c + d*x)) 
/2] + 227*A*Sin[(5*(c + d*x))/2] - 87*B*Sin[(5*(c + d*x))/2] + 10*A*Sin[(7 
*(c + d*x))/2])))/(120*a^3*d*E^(I*d*x)*(B + A*Cos[c + d*x])*(1 + Sec[c + d 
*x])^3)
 

Rubi [A] (verified)

Time = 1.55 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.04, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.515, Rules used = {3042, 4508, 27, 3042, 4508, 3042, 4508, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\int \frac {a (13 A-3 B)-7 a (A-B) \sec (c+d x)}{2 \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (13 A-3 B)-7 a (A-B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (13 A-3 B)-7 a (A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {3 a^2 (23 A-8 B)-25 a^2 (2 A-B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {3 a^2 (23 A-8 B)-25 a^2 (2 A-B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\frac {\int \frac {3 \left (5 a^3 (33 A-13 B)-7 a^3 (17 A-7 B) \sec (c+d x)\right )}{2 \sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (33 A-13 B)-7 a^3 (17 A-7 B) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x)}dx}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \int \frac {5 a^3 (33 A-13 B)-7 a^3 (17 A-7 B) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx-7 a^3 (17 A-7 B) \int \frac {1}{\sqrt {\sec (c+d x)}}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx-7 a^3 (17 A-7 B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-7 a^3 (17 A-7 B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-7 a^3 (17 A-7 B) \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-7 a^3 (17 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-7 a^3 (17 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )-\frac {14 a^3 (17 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {3 \left (5 a^3 (33 A-13 B) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )-\frac {14 a^3 (17 A-7 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{2 a^2}-\frac {7 a^2 (17 A-7 B) \sin (c+d x)}{d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)}}{3 a^2}-\frac {10 a (2 A-B) \sin (c+d x)}{3 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^2}}{10 a^2}-\frac {(A-B) \sin (c+d x)}{5 d \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^3}\)

Input:

Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^3),x]
 

Output:

-1/5*((A - B)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^3) 
+ ((-10*a*(2*A - B)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d 
*x])^2) + ((-7*a^2*(17*A - 7*B)*Sin[c + d*x])/(d*Sqrt[Sec[c + d*x]]*(a + a 
*Sec[c + d*x])) + (3*((-14*a^3*(17*A - 7*B)*Sqrt[Cos[c + d*x]]*EllipticE[( 
c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + 5*a^3*(33*A - 13*B)*((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*a^2))/(10*a^2)
 

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 4256

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]
 

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 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]
 

Maple [A] (verified)

Time = 6.02 (sec) , antiderivative size = 465, normalized size of antiderivative = 1.78

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 (160 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+468 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+330 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 )+714 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 )-348 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-130 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 )-294 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 )-1058 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+578 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+474 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-264 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-47 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+37 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}\)	\(465\)

Input:

int((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(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)*(160*A*c 
os(1/2*d*x+1/2*c)^10+468*A*cos(1/2*d*x+1/2*c)^8+330*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))+714*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))-348*B*cos(1/2*d*x+1/2*c)^8-130*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))-294*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))-1058* 
A*cos(1/2*d*x+1/2*c)^6+578*B*cos(1/2*d*x+1/2*c)^6+474*A*cos(1/2*d*x+1/2*c) 
^4-264*B*cos(1/2*d*x+1/2*c)^4-47*A*cos(1/2*d*x+1/2*c)^2+37*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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 489, normalized size of antiderivative = 1.87 \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algorith 
m="fricas")
 

Output:

-1/60*(5*(sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(33*I*A - 1 
3*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(33*I*A - 13*I*B)*cos(d*x + c) + sqrt(2) 
*(33*I*A - 13*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + 
c)) + 5*(sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(-33*I*A + 
13*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-33*I*A + 13*I*B)*cos(d*x + c) + sqrt( 
2)*(-33*I*A + 13*I*B))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
 + c)) + 21*(sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c)^3 + 3*sqrt(2)*(17*I*A - 
 7*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(17*I*A - 7*I*B)*cos(d*x + c) + sqrt(2) 
*(17*I*A - 7*I*B))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d 
*x + c) + I*sin(d*x + c))) + 21*(sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^3 
+ 3*sqrt(2)*(-17*I*A + 7*I*B)*cos(d*x + c)^2 + 3*sqrt(2)*(-17*I*A + 7*I*B) 
*cos(d*x + c) + sqrt(2)*(-17*I*A + 7*I*B))*weierstrassZeta(-4, 0, weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(20*A*cos(d*x + c)^ 
4 + 3*(79*A - 29*B)*cos(d*x + c)^3 + 2*(188*A - 73*B)*cos(d*x + c)^2 + 5*( 
33*A - 13*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)
 

Sympy [F(-1)]

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

integrate((A+B*sec(d*x+c))/sec(d*x+c)**(3/2)/(a+a*sec(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algorith 
m="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

integrate((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(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*sec(d*x + c)^(3/2)) 
, x)
 

Mupad [F(-1)]

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

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(3/2)),x 
)
 

Output:

int((A + B/cos(c + d*x))/((a + a/cos(c + d*x))^3*(1/cos(c + d*x))^(3/2)), 
x)
 

Reduce [F]

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

int((A+B*sec(d*x+c))/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x)
 

Output:

(int(sqrt(sec(c + d*x))/(sec(c + d*x)**5 + 3*sec(c + d*x)**4 + 3*sec(c + d 
*x)**3 + sec(c + d*x)**2),x)*a + 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)*b)/a**3