\(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx\) [1350]

Optimal result

Integrand size = 45, antiderivative size = 189 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {2 C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{a^{3/2} d}+\frac {(3 A+B-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a+a \cos (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \] Output:

2*C*arcsin(a^(1/2)*sin(d*x+c)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec 
(d*x+c)^(1/2)/a^(3/2)/d+1/4*(3*A+B-5*C)*arctan(1/2*a^(1/2)*sin(d*x+c)*2^(1 
/2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^( 
1/2)*2^(1/2)/a^(3/2)/d-1/2*(A-B+C)*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)/sec 
(d*x+c)^(1/2)
 

Mathematica [A] (warning: unable to verify)

Time = 4.19 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.90 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {\left ((3 A+B-5 C) \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+4 \sqrt {2} C \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}}}\right )\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {1+\sec (c+d x)}}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}-\frac {(A-B+C) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\sec (c+d x)}}\right )}{d (a (1+\cos (c+d x)))^{3/2}} \] Input:

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

Output:

(Cos[(c + d*x)/2]^2*((((3*A + B - 5*C)*ArcSin[Tan[(c + d*x)/2]] + 4*Sqrt[2 
]*C*ArcTan[Tan[(c + d*x)/2]/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]])*Sqrt[C 
os[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[1 + Sec[c + d*x]])/Sqrt[Sec[(c + d*x) 
/2]^2] - ((A - B + C)*Tan[(c + d*x)/2])/Sqrt[Sec[c + d*x]]))/(d*(a*(1 + Co 
s[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.93, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 4709, 3042, 3520, 27, 3042, 3461, 3042, 3253, 223, 3261, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4709

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3520

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (3 A+B-C)+4 a C \cos (c+d x)}{2 \sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {a (3 A+B-C)+4 a C \cos (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3461

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3253

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (3 A+B-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {8 C \int \frac {1}{\sqrt {1-\frac {a \sin ^2(c+d x)}{\cos (c+d x) a+a}}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 223

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {a (3 A+B-5 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx+\frac {8 \sqrt {a} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 3261

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {8 \sqrt {a} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {2 a^2 (3 A+B-5 C) \int \frac {1}{\frac {\sin (c+d x) \tan (c+d x) a^3}{\cos (c+d x) a+a}+2 a^2}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {\cos (c+d x) a+a}}\right )}{d}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\)

\(\Big \downarrow \) 218

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\frac {\sqrt {2} \sqrt {a} (3 A+B-5 C) \arctan \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a \cos (c+d x)+a}}\right )}{d}+\frac {8 \sqrt {a} C \arcsin \left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {a \cos (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {(A-B+C) \sin (c+d x) \sqrt {\cos (c+d x)}}{2 d (a \cos (c+d x)+a)^{3/2}}\right )\)

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((8*Sqrt[a]*C*ArcSin[(Sqrt[a]*Sin[c 
 + d*x])/Sqrt[a + a*Cos[c + d*x]]])/d + (Sqrt[2]*Sqrt[a]*(3*A + B - 5*C)*A 
rcTan[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c 
+ d*x]])])/d)/(4*a^2) - ((A - B + C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(2*d 
*(a + a*Cos[c + d*x])^(3/2)))
 

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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt 
[a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3253

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.) 
*(x_)]], x_Symbol] :> Simp[-2/f   Subst[Int[1/Sqrt[1 - x^2/a], x], x, b*(Co 
s[e + f*x]/Sqrt[a + b*Sin[e + f*x]])], x] /; FreeQ[{a, b, d, e, f}, x] && E 
qQ[a^2 - b^2, 0] && EqQ[d, a/b]
 

rule 3261

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

rule 3461

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

rule 3520

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[(a*A - b*B + 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))   Int[(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)) + B*(b*c*m + a 
*d*(n + 1)) - C*(a*c*m + b*d*(n + 1)) + (d*(a*A - b*B)*(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, B, 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] (warning: unable to verify)

Time = 2.53 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.50

method	result	size
default	\(-\frac {\sqrt {\sec \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (-A \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+B \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+4 C \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )-C \sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-3 A \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )-B \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )+5 C \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right )}{8 d \,a^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(284\)
parts	\(\frac {A \sqrt {\sec \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+3 \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right )}{8 d \,a^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}-\frac {B \sqrt {\sec \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-\arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right )}{8 d \,a^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}+\frac {C \sqrt {\sec \left (d x +c \right )}\, \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right ) \sqrt {2}\, \sqrt {\left (1+\cos \left (d x +c \right )\right ) a}\, \left (-4 \sqrt {2}\, \arctan \left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \tan \left (d x +c \right )\right )+\sqrt {2}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-5 \arcsin \left (-\csc \left (d x +c \right )+\cot \left (d x +c \right )\right )\right )}{8 d \,a^{2} \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(427\)

Input:

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(3/2 
),x,method=_RETURNVERBOSE)
 

Output:

-1/8/d/a^2*sec(d*x+c)^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*2^(1/2)*((1+ 
cos(d*x+c))*a)^(1/2)*(-A*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(csc(d* 
x+c)-cot(d*x+c))+B*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(csc(d*x+c)-c 
ot(d*x+c))+4*C*2^(1/2)*arctan((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*tan(d*x+c) 
)-C*2^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(csc(d*x+c)-cot(d*x+c))-3*A* 
arcsin(-csc(d*x+c)+cot(d*x+c))-B*arcsin(-csc(d*x+c)+cot(d*x+c))+5*C*arcsin 
(-csc(d*x+c)+cot(d*x+c)))/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 15.37 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.13 \[ \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)}}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\sqrt {2} {\left ({\left (3 \, A + B - 5 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (3 \, A + B - 5 \, C\right )} \cos \left (d x + c\right ) + 3 \, A + B - 5 \, C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (A - B + C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, {\left (C \cos \left (d x + c\right )^{2} + 2 \, C \cos \left (d x + c\right ) + C\right )} \sqrt {a} \arctan \left (\frac {\sqrt {a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a} \sin \left (d x + c\right )}\right )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \] Input:

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c) 
)^(3/2),x, algorithm="fricas")
 

Output:

-1/4*(sqrt(2)*((3*A + B - 5*C)*cos(d*x + c)^2 + 2*(3*A + B - 5*C)*cos(d*x 
+ c) + 3*A + B - 5*C)*sqrt(a)*arctan(sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt 
(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) + 2*sqrt(a*cos(d*x + c) + a)*(A - B 
 + C)*sqrt(cos(d*x + c))*sin(d*x + c) + 8*(C*cos(d*x + c)^2 + 2*C*cos(d*x 
+ c) + C)*sqrt(a)*arctan(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt 
(a)*sin(d*x + c))))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
 

Sympy [F]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(1/2)/(a+a*cos(d*x+ 
c))**(3/2),x)
 

Output:

Integral((A + B*cos(c + d*x) + C*cos(c + d*x)**2)*sqrt(sec(c + d*x))/(a*(c 
os(c + d*x) + 1))**(3/2), x)
 

Maxima [F]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c) 
)^(3/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-1)]

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

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(1/2)/(a+a*cos(d*x+c) 
)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*(int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1)*cos(c + d*x))/(co 
s(c + d*x)**2 + 2*cos(c + d*x) + 1),x)*b + int((sqrt(sec(c + d*x))*sqrt(co 
s(c + d*x) + 1)*cos(c + d*x)**2)/(cos(c + d*x)**2 + 2*cos(c + d*x) + 1),x) 
*c + int((sqrt(sec(c + d*x))*sqrt(cos(c + d*x) + 1))/(cos(c + d*x)**2 + 2* 
cos(c + d*x) + 1),x)*a))/a**2