\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx\) [1124]

Optimal result

Integrand size = 35, antiderivative size = 180 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=-\frac {(A-9 C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac {(A+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{6 a^3 d}-\frac {(A+C) \sqrt {\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {2 (2 A-3 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {(A-9 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:

-1/10*(A-9*C)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^3/d+1/6*(A+3*C)*Inve 
rseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/a^3/d-1/5*(A+C)*cos(d*x+c)^(1/2)*sin(d* 
x+c)/d/(a+a*cos(d*x+c))^3+2/15*(2*A-3*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/a/d/( 
a+a*cos(d*x+c))^2+1/10*(A-9*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d/(a^3+a^3*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 = 8.20 (sec) , antiderivative size = 1200, normalized size of antiderivative = 6.67 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(-4*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S 
in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*S 
ec[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]]]])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*S 
ec[c + d*x])^3) - (4*C*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*HypergeometricPFQ[{1/ 
4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*Sec[c + d*x]*(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])^3) + (Cos[c/2 + (d*x)/2]^6*(A + C*Sec[c + d*x] 
^2)*((8*(A - 9*C)*Csc[c])/(5*d) + (8*Sec[c/2]*Sec[c/2 + (d*x)/2]*(A*Sin[(d 
*x)/2] - 9*C*Sin[(d*x)/2]))/(5*d) + (16*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(2*A 
*Sin[(d*x)/2] - 3*C*Sin[(d*x)/2]))/(15*d) - (4*Sec[c/2]*Sec[c/2 + (d*x)/2] 
^5*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/(5*d) + (16*(2*A - 3*C)*Sec[c/2 + (d 
*x)/2]^2*Tan[c/2])/(15*d) - (4*(A + C)*Sec[c/2 + (d*x)/2]^4*Tan[c/2])/(5*d 
)))/(Sqrt[Cos[c + d*x]]*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x] 
)^3) + (2*A*Cos[c/2 + (d*x)/2]^6*Csc[c/2]*Sec[c/2]*Sec[c + d*x]*(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[T...
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.08, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4602, 3042, 3521, 27, 3042, 3457, 25, 3042, 3457, 27, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sec (c+d x)^2}{\cos (c+d x)^{3/2} (a \sec (c+d x)+a)^3}dx\)

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3457

\(\displaystyle -\frac {\frac {\int -\frac {(A+21 C) a^2+2 (2 A-3 C) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {\int \frac {(A+21 C) a^2+2 (2 A-3 C) \cos (c+d x) a^2}{\sqrt {\cos (c+d x)} (\cos (c+d x) a+a)}dx}{3 a^2}-\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \frac {(A+21 C) a^2+2 (2 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {4 a (2 A-3 C) \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2}}{10 a^2}-\frac {(A+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3}\)

\(\Big \downarrow \) 3457

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

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

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 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 3457

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

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. \(450\) vs. \(2(167)=334\).

Time = 4.98 (sec) , antiderivative size = 451, normalized size of antiderivative = 2.51

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 (12 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+10 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 )+6 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 )-108 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+30 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 )-54 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 )-2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+138 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-24 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-24 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+17 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 A -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}\)	\(451\)

Input:

int((A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x,method=_RETUR 
NVERBOSE)
 

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)*(12*A*cos(1/ 
2*d*x+1/2*c)^8+10*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))+6*A*co 
s(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))-108*C*cos(1/2*d*x+1/2*c)^8+3 
0*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-54*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)*Ellipti 
cE(cos(1/2*d*x+1/2*c),2^(1/2))-2*A*cos(1/2*d*x+1/2*c)^6+138*C*cos(1/2*d*x+ 
1/2*c)^6-24*A*cos(1/2*d*x+1/2*c)^4-24*C*cos(1/2*d*x+1/2*c)^4+17*A*cos(1/2* 
d*x+1/2*c)^2-3*C*cos(1/2*d*x+1/2*c)^2-3*A-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*c 
os(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 = 0.10 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.58 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algori 
thm="fricas")
 

Output:

1/60*(2*(3*(A - 9*C)*cos(d*x + c)^2 + 2*(7*A - 33*C)*cos(d*x + c) + 5*A - 
45*C)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + 
 c)^3 + 3*sqrt(2)*(I*A + 3*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A + 3*I*C)*c 
os(d*x + c) + sqrt(2)*(I*A + 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + 
c) + I*sin(d*x + c)) - 5*(sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c)^3 + 3*sqrt(2 
)*(-I*A - 3*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c) + 
sqrt(2)*(-I*A - 3*I*C))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d* 
x + c)) - 3*(sqrt(2)*(I*A - 9*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(I*A - 9*I*C 
)*cos(d*x + c)^2 + 3*sqrt(2)*(I*A - 9*I*C)*cos(d*x + c) + sqrt(2)*(I*A - 9 
*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I* 
sin(d*x + c))) - 3*(sqrt(2)*(-I*A + 9*I*C)*cos(d*x + c)^3 + 3*sqrt(2)*(-I* 
A + 9*I*C)*cos(d*x + c)^2 + 3*sqrt(2)*(-I*A + 9*I*C)*cos(d*x + c) + sqrt(2 
)*(-I*A + 9*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d* 
x + c) - I*sin(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. \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^3} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algori 
thm="maxima")
 

Output:

Timed out
 

Giac [F]

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

integrate((A+C*sec(d*x+c)^2)/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^3,x, algori 
thm="giac")
 

Output:

integrate((C*sec(d*x + c)^2 + A)/((a*sec(d*x + c) + a)^3*cos(d*x + c)^(3/2 
)), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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