\(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [570]

Optimal result

Integrand size = 43, antiderivative size = 254 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {(56 A-35 B+20 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a^2 d}-\frac {5 (3 A-2 B+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}+\frac {(56 A-35 B+20 C) \sin (c+d x)}{15 a^2 d \sec ^{\frac {3}{2}}(c+d x)}-\frac {5 (3 A-2 B+C) \sin (c+d x)}{3 a^2 d \sqrt {\sec (c+d x)}}-\frac {(3 A-2 B+C) \sin (c+d x)}{a^2 d \sec ^{\frac {3}{2}}(c+d x) (1+\sec (c+d x))}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \] Output:

1/5*(56*A-35*B+20*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-5/3*(3*A-2*B+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+1/15*(56*A-35*B+20*C)*sin(d* 
x+c)/a^2/d/sec(d*x+c)^(3/2)-5/3*(3*A-2*B+C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(1 
/2)-(3*A-2*B+C)*sin(d*x+c)/a^2/d/sec(d*x+c)^(3/2)/(1+sec(d*x+c))-1/3*(A-B+ 
C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^2
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 9.40 (sec) , antiderivative size = 1442, normalized size of antiderivative = 5.68 \[ \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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])^2),x]
 

Output:

(-112*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]^4*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 
))/(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])^2) + (14*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]^4*Csc[c/2]*(-3*S 
qrt[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeome 
tric2F1[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))/(3*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])^2) - (8*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] 
^4*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))/(3*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])^2) - (20*A*Cos[c/2 + (d 
*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/2, 2]*Sec[c/2]*Sq 
rt[Sec[c + d*x]]*(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])^2) + (40* 
B*Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*Csc[c/2]*EllipticF[(c + d*x)/...
 

Rubi [A] (verified)

Time = 1.28 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.96, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4572, 27, 3042, 4508, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4572

\(\displaystyle \frac {\int \frac {a (11 A-5 B+5 C)-a (7 A-7 B+C) \sec (c+d x)}{2 \sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{3 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (11 A-5 B+5 C)-a (7 A-7 B+C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (\sec (c+d x) a+a)}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (11 A-5 B+5 C)-a (7 A-7 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2} \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4508

\(\displaystyle \frac {\frac {\int \frac {a^2 (56 A-35 B+20 C)-15 a^2 (3 A-2 B+C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x)}dx}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (56 A-35 B+20 C)-15 a^2 (3 A-2 B+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)}dx-15 a^2 (3 A-2 B+C) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)}dx}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx-15 a^2 (3 A-2 B+C) \int \frac {1}{\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \left (\frac {3}{5} \int \frac {1}{\sqrt {\sec (c+d x)}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B+C) \left (\frac {1}{3} \int \sqrt {\sec (c+d x)}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \left (\frac {3}{5} \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B+C) \left (\frac {1}{3} \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B+C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \left (\frac {3}{5} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}\right )-15 a^2 (3 A-2 B+C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-15 a^2 (3 A-2 B+C) \left (\frac {1}{3} \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\right )}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {a^2 (56 A-35 B+20 C) \left (\frac {2 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {6 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}\right )-15 a^2 (3 A-2 B+C) \left (\frac {2 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}\right )}{a^2}-\frac {6 (3 A-2 B+C) \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x) (\sec (c+d x)+1)}}{6 a^2}-\frac {(A-B+C) \sin (c+d x)}{3 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((A - B + C)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x]) 
^2) + ((-6*(3*A - 2*B + C)*Sin[c + d*x])/(d*Sec[c + d*x]^(3/2)*(1 + Sec[c 
+ d*x])) + (a^2*(56*A - 35*B + 20*C)*((6*Sqrt[Cos[c + d*x]]*EllipticE[(c + 
 d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (2*Sin[c + d*x])/(5*d*Sec[c + d*x] 
^(3/2))) - 15*a^2*(3*A - 2*B + 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]])))/a^2)/(6*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]
 

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)]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(490\) vs. \(2(231)=462\).

Time = 7.87 (sec) , antiderivative size = 491, normalized size of antiderivative = 1.93

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 (2 \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 (75 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+168 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-50 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \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 (75 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+168 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-50 B \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-105 B \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+60 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-96 A \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (128 A +80 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (328 A -380 B +120 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-526 A +420 B -170 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (171 A -125 B +55 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{30 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}\)	\(491\)

Input:

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))^2,x, 
method=_RETURNVERBOSE)
 

Output:

-1/30*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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)*(75*A*EllipticF(cos(1 
/2*d*x+1/2*c),2^(1/2))+168*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*El 
lipticF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2^( 
1/2))+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+60*C*EllipticE(cos(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)^2-2*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(75*A*EllipticF(cos(1/2*d 
*x+1/2*c),2^(1/2))+168*A*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-50*B*Ellipt 
icF(cos(1/2*d*x+1/2*c),2^(1/2))-105*B*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2) 
)+25*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+60*C*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-96*A*sin(1/2*d*x+1/2*c)^10+(128*A+80*B)* 
sin(1/2*d*x+1/2*c)^8+(328*A-380*B+120*C)*sin(1/2*d*x+1/2*c)^6+(-526*A+420* 
B-170*C)*sin(1/2*d*x+1/2*c)^4+(171*A-125*B+55*C)*sin(1/2*d*x+1/2*c)^2)/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
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 435, normalized size of antiderivative = 1.71 \[ \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))^2} \, dx=-\frac {25 \, {\left (\sqrt {2} {\left (-3 i \, A + 2 i \, B - i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-3 i \, A + 2 i \, B - i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A + 2 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 ) + 25 \, {\left (\sqrt {2} {\left (3 i \, A - 2 i \, B + i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (3 i \, A - 2 i \, B + i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A - 2 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 (-56 i \, A + 35 i \, B - 20 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (-56 i \, A + 35 i \, B - 20 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-56 i \, A + 35 i \, B - 20 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 (56 i \, A - 35 i \, B + 20 i \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} {\left (56 i \, A - 35 i \, B + 20 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (56 i \, A - 35 i \, B + 20 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 )^{4} - 2 \, {\left (4 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} - {\left (94 \, A - 65 \, B + 30 \, C\right )} \cos \left (d x + c\right )^{2} - 25 \, {\left (3 \, A - 2 \, B + 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^{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((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c) 
)^2,x, algorithm="fricas")
 

Output:

-1/30*(25*(sqrt(2)*(-3*I*A + 2*I*B - I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I 
*A + 2*I*B - I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A + 2*I*B - I*C))*weierstra 
ssPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 25*(sqrt(2)*(3*I*A - 2* 
I*B + I*C)*cos(d*x + c)^2 + 2*sqrt(2)*(3*I*A - 2*I*B + I*C)*cos(d*x + c) + 
 sqrt(2)*(3*I*A - 2*I*B + I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c)) + 3*(sqrt(2)*(-56*I*A + 35*I*B - 20*I*C)*cos(d*x + c)^2 + 
2*sqrt(2)*(-56*I*A + 35*I*B - 20*I*C)*cos(d*x + c) + sqrt(2)*(-56*I*A + 35 
*I*B - 20*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c))) + 3*(sqrt(2)*(56*I*A - 35*I*B + 20*I*C)*cos(d*x + 
c)^2 + 2*sqrt(2)*(56*I*A - 35*I*B + 20*I*C)*cos(d*x + c) + sqrt(2)*(56*I*A 
 - 35*I*B + 20*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos 
(d*x + c) - I*sin(d*x + c))) - 2*(6*A*cos(d*x + c)^4 - 2*(4*A - 5*B)*cos(d 
*x + c)^3 - (94*A - 65*B + 30*C)*cos(d*x + c)^2 - 25*(3*A - 2*B + 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)
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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))^2,x)
 

Output:

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