\(\int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx\) [1189]

Optimal result

Integrand size = 35, antiderivative size = 199 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=\frac {3 (5 A+7 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 a d}-\frac {(3 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(A+C) \sin (c+d x)}{d (a+a \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)}+\frac {(5 A+7 C) \sin (c+d x)}{5 a d \sec ^{\frac {3}{2}}(c+d x)}-\frac {(3 A+5 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 5.72 (sec) , antiderivative size = 458, normalized size of antiderivative = 2.30 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (60 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+84 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+120 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+200 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+\frac {2 \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left ((60 A+83 C) \cos \left (\frac {1}{2} (c-d x)\right )+(30 A+43 C) \cos \left (\frac {1}{2} (3 c+d x)\right )+C \sin (c) \left (7 \sin \left (\frac {3}{2} (c+d x)\right )-3 \sin \left (\frac {5}{2} (c+d x)\right )\right )\right )}{\sqrt {\sec (c+d x)}}\right )}{60 a d (1+\cos (c+d x))} \] Input:

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

Output:

-1/60*(Cos[(c + d*x)/2]^2*((60*Sqrt[2]*A*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I 
)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-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))]))/E^(I*d*x) + (84*Sqrt[2]*C*Sqrt[E^(I*(c + d*x 
))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Csc[c]*(-3*Sqr 
t[1 + E^((2*I)*(c + d*x))] + E^((2*I)*d*x)*(-1 + E^((2*I)*c))*Hypergeometr 
ic2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))]))/E^(I*d*x) + 120*A*Sqrt[Cos[c 
+ d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 200*C*Sqrt[Cos[c + 
d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + (2*Csc[c/2]*Sec[c/2]* 
Sec[(c + d*x)/2]*((60*A + 83*C)*Cos[(c - d*x)/2] + (30*A + 43*C)*Cos[(3*c 
+ d*x)/2] + C*Sin[c]*(7*Sin[(3*(c + d*x))/2] - 3*Sin[(5*(c + d*x))/2])))/S 
qrt[Sec[c + d*x]]))/(a*d*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.85, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 4709, 3042, 3521, 27, 3042, 3227, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A+5 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A+7 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (-\frac {a (3 A+5 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )-a (5 A+7 C) \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{d (a \cos (c+d x)+a)}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A + C)*Cos[c + d*x]^(5/2)*Sin[c 
 + d*x])/(d*(a + a*Cos[c + d*x]))) - (a*(3*A + 5*C)*((2*EllipticF[(c + d*x 
)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) - a*(5*A + 7*C 
)*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x 
])/(5*d)))/(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 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[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 4709

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Simp[(c*Sec[a 
+ b*x])^m*(c*Cos[a + b*x])^m   Int[ActivateTrig[u]/(c*Cos[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 5.79 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (15 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+45 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+25 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+63 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-48 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+56 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} C +\left (30 A +30 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-15 A -23 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{15 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(276\)

Input:

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

Output:

1/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(15*A 
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+45*A*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))+63*C*EllipticE(cos(1/2* 
d*x+1/2*c),2^(1/2)))-48*C*sin(1/2*d*x+1/2*c)^8+56*sin(1/2*d*x+1/2*c)^6*C+( 
30*A+30*C)*sin(1/2*d*x+1/2*c)^4+(-15*A-23*C)*sin(1/2*d*x+1/2*c)^2)/a/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)/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 = 274, normalized size of antiderivative = 1.38 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {5 \, {\left (\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, {\left (\sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 \, {\left (\sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-5 i \, A - 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 9 \, {\left (\sqrt {2} {\left (5 i \, A + 7 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (5 i \, A + 7 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {2 \, {\left (6 \, C \cos \left (d x + c\right )^{3} - 4 \, C \cos \left (d x + c\right )^{2} - 5 \, {\left (3 \, A + 5 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

-1/30*(5*(sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A - 5*I*C) 
)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*(sqrt(2)*( 
3*I*A + 5*I*C)*cos(d*x + c) + sqrt(2)*(3*I*A + 5*I*C))*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) + sqrt(2)*(-5*I*A - 7*I*C))*weierstrassZeta(-4, 0, weierstrassPInv 
erse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 9*(sqrt(2)*(5*I*A + 7*I*C)*c 
os(d*x + c) + sqrt(2)*(5*I*A + 7*I*C))*weierstrassZeta(-4, 0, weierstrassP 
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - 2*(6*C*cos(d*x + c)^3 - 4 
*C*cos(d*x + c)^2 - 5*(3*A + 5*C)*cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x 
+ c)))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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