\(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx\) [1486]

Optimal result

Integrand size = 43, antiderivative size = 207 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {2 (b B-a C) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{b^2 d}+\frac {2 \left (b^2 (3 A+C)-3 a (b B-a C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{3 b^3 d}-\frac {2 a \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^3 (a+b) d}+\frac {2 C \sin (c+d x)}{3 b d \sqrt {\sec (c+d x)}} \] Output:

2*(B*b-C*a)*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x 
+c)^(1/2)/b^2/d+2/3*(b^2*(3*A+C)-3*a*(B*b-C*a))*cos(d*x+c)^(1/2)*InverseJa 
cobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2)/b^3/d-2*a*(A*b^2-a*(B*b-C*a 
))*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*sec(d 
*x+c)^(1/2)/b^3/(a+b)/d+2/3*C*sin(d*x+c)/b/d/sec(d*x+c)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 5.80 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.58 \[ \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sqrt {\sec (c+d x)}} \, dx=\frac {2 \csc (c+d x) \left (-3 b^2 B+3 a b C+3 b^2 B \sec ^2(c+d x)-3 a b C \sec ^2(c+d x)+b^2 C \sin (c+d x) \tan (c+d x)-3 b (b B-a C) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+b (3 b B-3 a C+b C) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+3 A b^2 \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-3 a b B \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}+3 a^2 C \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )}{3 b^3 d \sec ^{\frac {3}{2}}(c+d x)} \] Input:

Integrate[(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)/((a + b*Cos[c + d*x])*Sq 
rt[Sec[c + d*x]]),x]
 

Output:

(2*Csc[c + d*x]*(-3*b^2*B + 3*a*b*C + 3*b^2*B*Sec[c + d*x]^2 - 3*a*b*C*Sec 
[c + d*x]^2 + b^2*C*Sin[c + d*x]*Tan[c + d*x] - 3*b*(b*B - a*C)*EllipticE[ 
ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 
 b*(3*b*B - 3*a*C + b*C)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Se 
c[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 3*A*b^2*EllipticPi[-(a/b), ArcSin[Sqrt 
[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 3*a*b*B*El 
lipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[- 
Tan[c + d*x]^2] + 3*a^2*C*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], - 
1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2]))/(3*b^3*d*Sec[c + d*x]^(3/2))
 

Rubi [A] (verified)

Time = 1.40 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.86, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.326, Rules used = {3042, 4709, 3042, 3528, 27, 3042, 3538, 25, 3042, 3119, 3481, 3042, 3120, 3284}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3528

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

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 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(B*c - A*d)/d   Int[(a + b* 
Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, 
 B, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
 

rule 3528

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

rule 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(980\) vs. \(2(194)=388\).

Time = 5.65 (sec) , antiderivative size = 981, normalized size of antiderivative = 4.74

method	result	size
default	\(\text {Expression too large to display}\)	\(981\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))/sec(d*x+c)^(1/2),x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-2/3*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*C*b^2*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a-4*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d 
*x+1/2*c)^4+3*A*(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))*a*b^2-3*A*(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))*b^3-3*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)* 
EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a*b^2-3*B*(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))*a^2*b+3*B*(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))*a*b^2-3*B*(sin(1/2*d*x+1/2 
*c)^2)^(1/2)*(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))*a*b^2+3*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(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))*b^3+3*B*(sin(1/2*d*x+1/2*c)^ 
2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2 
*b/(a-b),2^(1/2))*a^2*b-2*C*b^2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)*a+ 
2*C*b^3*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+3*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))*a^3-3*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))*a^2*b+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))...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)/((a + b*cos(c + d*x))*sq 
rt(sec(c + d*x))), x)
 

Maxima [F]

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

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

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)*sq 
rt(sec(d*x + c))), x)
 

Giac [F]

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

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

Output:

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)/((b*cos(d*x + c) + a)*sq 
rt(sec(d*x + c))), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*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)*b + sec(c + d*x)*a),x)*a 
 + int((sqrt(sec(c + d*x))*cos(c + d*x))/(cos(c + d*x)*sec(c + d*x)*b + se 
c(c + d*x)*a),x)*b + int((sqrt(sec(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x 
)*sec(c + d*x)*b + sec(c + d*x)*a),x)*c