\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx\) [1113]

Optimal result

Integrand size = 35, antiderivative size = 192 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=-\frac {3 (5 A+7 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 a d}-\frac {(3 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a d}+\frac {(5 A+7 C) \sin (c+d x)}{5 a d \cos ^{\frac {5}{2}}(c+d x)}-\frac {(3 A+5 C) \sin (c+d x)}{3 a d \cos ^{\frac {3}{2}}(c+d x)}+\frac {3 (5 A+7 C) \sin (c+d x)}{5 a d \sqrt {\cos (c+d x)}}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))} \] Output:

-3/5*(5*A+7*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a/d-1/3*(3*A+5*C)*Inv 
erseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a/d+1/5*(5*A+7*C)*sin(d*x+c)/a/d/cos(d 
*x+c)^(5/2)-1/3*(3*A+5*C)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)+3/5*(5*A+7*C)*si 
n(d*x+c)/a/d/cos(d*x+c)^(1/2)-(A+C)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*cos 
(d*x+c))
                                                                                    
                                                                                    
 

Mathematica [C] (warning: unable to verify)

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

Time = 8.15 (sec) , antiderivative size = 1144, normalized size of antiderivative = 5.96 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx =\text {Too large to display} \] Input:

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

Output:

(Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]^(3/2)*(A + C*Sec[c + d*x]^2)*((2*(10*A 
+ 16*C + 5*A*Cos[c] + 5*C*Cos[c])*Csc[c/2]*Sec[c/2]*Sec[c])/(5*d) + (4*Sec 
[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/d + (8*C*Sec[c 
]*Sec[c + d*x]^3*Sin[d*x])/(5*d) - (8*Sec[c]*Sec[c + d*x]*(5*C*Sin[c] - 15 
*A*Sin[d*x] - 24*C*Sin[d*x]))/(15*d) + (8*Sec[c]*Sec[c + d*x]^2*(3*C*Sin[c 
] - 5*C*Sin[d*x]))/(15*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + 
 d*x])) + (2*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*HypergeometricPF 
Q[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + 
d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt 
[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x 
- ArcTan[Cot[c]]]])/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*( 
a + a*Sec[c + d*x])) + (10*C*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Csc[c/2]*Hy 
pergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*( 
A + C*Sec[c + d*x]^2)*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[ 
Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqr 
t[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt 
[1 + Cot[c]^2]*(a + a*Sec[c + d*x])) + (3*A*Cos[c/2 + (d*x)/2]^2*Cos[c + d 
*x]*Csc[c/2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*((HypergeometricPFQ[{-1/2, -1 
/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c]) 
/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]...
 

Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.90, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4602, 3042, 3521, 27, 3042, 3227, 3042, 3116, 3042, 3116, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

\(\displaystyle \int \frac {A \cos ^2(c+d x)+C}{\cos ^{\frac {7}{2}}(c+d x) (a \cos (c+d x)+a)}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A \sin \left (c+d x+\frac {\pi }{2}\right )^2+C}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2} \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )}dx\)

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle \frac {a (5 A+7 C) \int \frac {1}{\cos ^{\frac {7}{2}}(c+d x)}dx-a (3 A+5 C) \int \frac {1}{\cos ^{\frac {5}{2}}(c+d x)}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (5 A+7 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{7/2}}dx-a (3 A+5 C) \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {a (5 A+7 C) \left (\frac {3}{5} \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (5 A+7 C) \left (\frac {3}{5} \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx+\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}\right )-a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3116

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \frac {a (5 A+7 C) \left (\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {a (5 A+7 C) \left (\frac {2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {3}{5} \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )\right )-a (3 A+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x)}{3 d \cos ^{\frac {3}{2}}(c+d x)}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x)}{d \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)}\)

Input:

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

Output:

-(((A + C)*Sin[c + d*x])/(d*Cos[c + d*x]^(5/2)*(a + a*Cos[c + d*x]))) + (- 
(a*(3*A + 5*C)*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sin[c + d*x])/(3* 
d*Cos[c + d*x]^(3/2)))) + a*(5*A + 7*C)*((2*Sin[c + d*x])/(5*d*Cos[c + d*x 
]^(5/2)) + (3*((-2*EllipticE[(c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt 
[Cos[c + d*x]])))/5))/(2*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 3116

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && 
 IntegerQ[2*n]
 

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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3521

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n 
 + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1))   I 
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) 
- b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* 
c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 
 - d^2, 0] && LtQ[m, -2^(-1)]
 

rule 4602

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x 
_)])^(m_.)*((A_.) + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[d^( 
m + 2)   Int[(b + a*Cos[e + f*x])^m*(d*Cos[e + f*x])^(n - m - 2)*(C + A*Cos 
[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] &&  !IntegerQ[n] 
 && IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(775\) vs. \(2(177)=354\).

Time = 5.22 (sec) , antiderivative size = 776, normalized size of antiderivative = 4.04

method	result	size
default	\(\text {Expression too large to display}\)	\(776\)

Input:

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)/a*((-A-C)*(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)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^( 
1/2)))-2*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)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(2*A+2*C)/sin(1/2*d*x+1/ 
2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 
*c)^2)^(1/2)*(2*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)^(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))+2/5*C/sin(1/2*d*x+1/2*c)^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x 
+1/2*c)^4+6*sin(1/2*d*x+1/2*c)^2-1)*(24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1 
/2*c)-12*(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)*sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2 
*c)*sin(1/2*d*x+1/2*c)^4+12*(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)*sin(1/2*d*x+1/2*c)^2 
+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-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*x+1/2*c)^2)^(1/2)-2*C*(-1/6*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)/(cos(1/2*d*x+1/ 
2*c)^2-1/2)^2+1/3*(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(c...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.73 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))} \, dx=\frac {2 \, {\left (9 \, {\left (5 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (15 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 6 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, {\left (\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, {\left (\sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 9 \, {\left (\sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \cos \left (d x + c\right )^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 \, {\left (\sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{4} + \sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \cos \left (d x + c\right )^{3}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{30 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c)),x, algorith 
m="fricas")
 

Output:

1/30*(2*(9*(5*A + 7*C)*cos(d*x + c)^3 + 2*(15*A + 19*C)*cos(d*x + c)^2 - 4 
*C*cos(d*x + c) + 6*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(-3*I* 
A - 5*I*C)*cos(d*x + c)^4 + sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c)^3)*weier 
strassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 5*(sqrt(2)*(3*I*A + 
 5*I*C)*cos(d*x + c)^4 + sqrt(2)*(3*I*A + 5*I*C)*cos(d*x + c)^3)*weierstra 
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 9*(sqrt(2)*(5*I*A + 7*I 
*C)*cos(d*x + c)^4 + sqrt(2)*(5*I*A + 7*I*C)*cos(d*x + c)^3)*weierstrassZe 
ta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*( 
sqrt(2)*(-5*I*A - 7*I*C)*cos(d*x + c)^4 + sqrt(2)*(-5*I*A - 7*I*C)*cos(d*x 
 + c)^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - 
I*sin(d*x + c))))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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