\(\int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx\) [1188]

Optimal result

Integrand size = 35, antiderivative size = 162 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=-\frac {(A+3 C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {(3 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 a d}-\frac {(A+C) \sin (c+d x)}{d (a+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}+\frac {(3 A+5 C) \sin (c+d x)}{3 a d \sqrt {\sec (c+d x)}} \] Output:

-(A+3*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) 
^(3/2)+1/3*(3*A+5*C)*sin(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 = 6.89 (sec) , antiderivative size = 439, normalized size of antiderivative = 2.71 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (2 \sqrt {2} A e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+6 \sqrt {2} C e^{-i d x} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \csc (c) \left (-3 \sqrt {1+e^{2 i (c+d x)}}+e^{2 i d x} \left (-1+e^{2 i c}\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )\right )+12 A \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+20 C \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}+\frac {\csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \sec \left (\frac {1}{2} (c+d x)\right ) \left ((6 A+13 C) \cos \left (\frac {1}{2} (c-d x)\right )+C \left (5 \cos \left (\frac {1}{2} (3 c+d x)\right )+2 \sin (c) \sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{\sqrt {\sec (c+d x)}}\right )}{6 a d (1+\cos (c+d x))} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*((2*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) + (6*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*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) + 12*A*Sqrt[Cos[c + d*x]]*E 
llipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + 20*C*Sqrt[Cos[c + d*x]]*Elli 
pticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] + (Csc[c/2]*Sec[c/2]*Sec[(c + d*x 
)/2]*((6*A + 13*C)*Cos[(c - d*x)/2] + C*(5*Cos[(3*c + d*x)/2] + 2*Sin[c]*S 
in[(3*(c + d*x))/2])))/Sqrt[Sec[c + d*x]]))/(6*a*d*(1 + Cos[c + d*x]))
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.86, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3521

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(-(((A + C)*Cos[c + d*x]^(3/2)*Sin[c 
 + d*x])/(d*(a + a*Cos[c + d*x]))) - ((2*a*(A + 3*C)*EllipticE[(c + d*x)/2 
, 2])/d - 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)))/(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 = 4.52 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.62

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \left (3 A \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 A \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+9 C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )-8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} C +\left (6 A +18 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-3 A -7 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{3 a \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(262\)

Input:

int((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-1/3*((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)*(3*A* 
EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*A*EllipticE(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))+9*C*EllipticE(cos(1/2*d*x+ 
1/2*c),2^(1/2)))-8*sin(1/2*d*x+1/2*c)^6*C+(6*A+18*C)*sin(1/2*d*x+1/2*c)^4+ 
(-3*A-7*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 = 260, normalized size of antiderivative = 1.60 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+a \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {{\left (\sqrt {2} {\left (-3 i \, A - 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-3 i \, A - 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (\sqrt {2} {\left (3 i \, A + 5 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (3 i \, A + 5 i \, C\right )}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 3 \, {\left (\sqrt {2} {\left (i \, A + 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (i \, A + 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 \, {\left (\sqrt {2} {\left (-i \, A - 3 i \, C\right )} \cos \left (d x + c\right ) + \sqrt {2} {\left (-i \, A - 3 i \, C\right )}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + \frac {2 \, {\left (2 \, C \cos \left (d x + c\right )^{2} + {\left (3 \, A + 5 \, C\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorith 
m="fricas")
 

Output:

1/6*((sqrt(2)*(-3*I*A - 5*I*C)*cos(d*x + c) + sqrt(2)*(-3*I*A - 5*I*C))*we 
ierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + (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)) - 3*(sqrt(2)*(I*A + 3*I*C)*cos(d*x + c) + 
 sqrt(2)*(I*A + 3*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
cos(d*x + c) + I*sin(d*x + c))) - 3*(sqrt(2)*(-I*A - 3*I*C)*cos(d*x + c) + 
 sqrt(2)*(-I*A - 3*I*C))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, 
 cos(d*x + c) - I*sin(d*x + c))) + 2*(2*C*cos(d*x + c)^2 + (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]

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

integrate((A+C*cos(d*x+c)**2)/(a+a*cos(d*x+c))/sec(d*x+c)**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorith 
m="maxima")
 

Output:

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

Giac [F]

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

integrate((A+C*cos(d*x+c)^2)/(a+a*cos(d*x+c))/sec(d*x+c)^(1/2),x, algorith 
m="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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