\(\int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx\) [1118]

Optimal result

Integrand size = 35, antiderivative size = 151 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=-\frac {4 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a^2 d}+\frac {(A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 a^2 d}+\frac {4 C \sin (c+d x)}{a^2 d \sqrt {\cos (c+d x)}}+\frac {(A-5 C) \sin (c+d x)}{3 a^2 d \sqrt {\cos (c+d x)} (1+\cos (c+d x))}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2} \] Output:

-4*C*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/a^2/d+1/3*(A-5*C)*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2))/a^2/d+4*C*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)+1/3* 
(A-5*C)*sin(d*x+c)/a^2/d/cos(d*x+c)^(1/2)/(1+cos(d*x+c))-1/3*(A+C)*sin(d*x 
+c)/d/cos(d*x+c)^(1/2)/(a+a*cos(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 = 7.61 (sec) , antiderivative size = 836, normalized size of antiderivative = 5.54 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(-4*A*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, S 
in[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*Sec[d*x - ArcT 
an[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*Sec[c + d*x])^ 
2) + (20*C*Cos[c/2 + (d*x)/2]^4*Csc[c/2]*HypergeometricPFQ[{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]]] 
])/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x])*Sqrt[1 + Cot[c]^2]*(a + a*Sec[c + d 
*x])^2) + (Cos[c/2 + (d*x)/2]^4*Sqrt[Cos[c + d*x]]*(A + C*Sec[c + d*x]^2)* 
((16*C*Cot[c/2]*Sec[c])/d + (16*C*Sec[c/2]*Sec[c/2 + (d*x)/2]*Sin[(d*x)/2] 
)/d + (4*Sec[c/2]*Sec[c/2 + (d*x)/2]^3*(A*Sin[(d*x)/2] + C*Sin[(d*x)/2]))/ 
(3*d) + (16*C*Sec[c]*Sec[c + d*x]*Sin[d*x])/d + (4*(A + C)*Sec[c/2 + (d*x) 
/2]^2*Tan[c/2])/(3*d)))/((A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x 
])^2) + (8*C*Cos[c/2 + (d*x)/2]^4*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]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 
 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])...
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.03, 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, 3457, 3042, 3227, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4602

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (A+7 C)+3 a (A-C) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (\cos (c+d x) a+a)}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (A+7 C)+3 a (A-C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3457

\(\displaystyle \frac {\frac {\int \frac {12 C a^2+(A-5 C) \cos (c+d x) a^2}{\cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {12 C a^2+(A-5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\cos (c+d x)}}dx+12 a^2 C \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}}dx}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3116

\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\cos (c+d x)}dx\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\frac {a^2 (A-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\frac {2 a^2 (A-5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{d}+12 a^2 C \left (\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}\right )}{a^2}+\frac {2 (A-5 C) \sin (c+d x)}{d \sqrt {\cos (c+d x)} (\cos (c+d x)+1)}}{6 a^2}-\frac {(A+C) \sin (c+d x)}{3 d \sqrt {\cos (c+d x)} (a \cos (c+d x)+a)^2}\)

Input:

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

Output:

-1/3*((A + C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2) 
+ ((2*(A - 5*C)*Sin[c + d*x])/(d*Sqrt[Cos[c + d*x]]*(1 + Cos[c + d*x])) + 
((2*a^2*(A - 5*C)*EllipticF[(c + d*x)/2, 2])/d + 12*a^2*C*((-2*EllipticE[( 
c + d*x)/2, 2])/d + (2*Sin[c + d*x])/(d*Sqrt[Cos[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 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 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. \(449\) vs. \(2(142)=284\).

Time = 3.30 (sec) , antiderivative size = 450, normalized size of antiderivative = 2.98

method	result	size
default	\(-\frac {-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}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 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}\, \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+12 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 )-48 C \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 )^{6}+2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +43 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A +37 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{6 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}\)	\(450\)

Input:

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

Output:

-1/6*(-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)*(-2 
*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticF(cos(1/2*d*x 
+1/2*c),2^(1/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*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*(si 
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(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)*(A*EllipticF(cos(1/2*d*x+1/2*c),2^(1 
/2))-5*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+12*C*EllipticE(cos(1/2*d*x+ 
1/2*c),2^(1/2)))*cos(1/2*d*x+1/2*c)-48*C*(-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)^6+2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*(A+43*C)*sin(1/2*d*x+1/2*c)^4-(-2*sin(1/2*d*x+1/2*c) 
^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(A+37*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 = 0.10 (sec) , antiderivative size = 367, normalized size of antiderivative = 2.43 \[ \int \frac {A+C \sec ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^2} \, dx=\frac {2 \, {\left (12 \, C \cos \left (d x + c\right )^{2} + {\left (A + 19 \, C\right )} \cos \left (d x + c\right ) + 6 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + {\left (\sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )^{3} - 2 \, \sqrt {2} {\left (-i \, A + 5 i \, C\right )} \cos \left (d x + c\right )^{2} + \sqrt {2} {\left (i \, A - 5 i \, C\right )} \cos \left (d x + c\right )\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 12 \, {\left (i \, \sqrt {2} C \cos \left (d x + c\right )^{3} + 2 i \, \sqrt {2} C \cos \left (d x + c\right )^{2} + i \, \sqrt {2} C \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 12 \, {\left (-i \, \sqrt {2} C \cos \left (d x + c\right )^{3} - 2 i \, \sqrt {2} C \cos \left (d x + c\right )^{2} - i \, \sqrt {2} C \cos \left (d x + c\right )\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}} \] Input:

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

Output:

1/6*(2*(12*C*cos(d*x + c)^2 + (A + 19*C)*cos(d*x + c) + 6*C)*sqrt(cos(d*x 
+ c))*sin(d*x + c) + (sqrt(2)*(-I*A + 5*I*C)*cos(d*x + c)^3 - 2*sqrt(2)*(I 
*A - 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(-I*A + 5*I*C)*cos(d*x + c))*weierstr 
assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (sqrt(2)*(I*A - 5*I*C) 
*cos(d*x + c)^3 - 2*sqrt(2)*(-I*A + 5*I*C)*cos(d*x + c)^2 + sqrt(2)*(I*A - 
 5*I*C)*cos(d*x + c))*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x 
+ c)) - 12*(I*sqrt(2)*C*cos(d*x + c)^3 + 2*I*sqrt(2)*C*cos(d*x + c)^2 + I* 
sqrt(2)*C*cos(d*x + c))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
cos(d*x + c) + I*sin(d*x + c))) - 12*(-I*sqrt(2)*C*cos(d*x + c)^3 - 2*I*sq 
rt(2)*C*cos(d*x + c)^2 - I*sqrt(2)*C*cos(d*x + c))*weierstrassZeta(-4, 0, 
weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))))/(a^2*d*cos(d*x 
 + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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