\(\int \frac {\sqrt {\cos (c+d x)} (B \cos (c+d x)+C \cos ^2(c+d x))}{a+b \cos (c+d x)} \, dx\) [878]

Optimal result

Integrand size = 42, antiderivative size = 137 \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {2 (b B-a C) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 d}-\frac {2 \left (3 a b B-3 a^2 C-b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^3 d}+\frac {2 a^2 (b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^3 (a+b) d}+\frac {2 C \sqrt {\cos (c+d x)} \sin (c+d x)}{3 b d} \] Output:

2*(B*b-C*a)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/d-2/3*(3*B*a*b-3*C*a 
^2-C*b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^3/d+2*a^2*(B*b-C*a)*Ell 
ipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/b^3/(a+b)/d+2/3*C*cos(d*x+c) 
^(1/2)*sin(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 1.70 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx=\frac {\frac {(3 b B-a C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+C \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )+2 C \sqrt {\cos (c+d x)} \sin (c+d x)+\frac {3 (b B-a C) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{3 b d} \] Input:

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

Output:

(((3*b*B - a*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + C*(2* 
EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2] 
)/(a + b)) + 2*C*Sqrt[Cos[c + d*x]]*Sin[c + d*x] + (3*(b*B - a*C)*(-2*a*b* 
EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[S 
qrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Co 
s[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/(3*b*d)
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.08, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3508, 3042, 3469, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3469

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

((6*(b*B - a*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((-2*(3*a*b*B - 3*a^2*C 
 - b^2*C)*EllipticF[(c + d*x)/2, 2])/(b*d) + (6*a^2*(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 3469

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(-b)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^( 
n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a + b*Sin[e + 
 f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*( 
m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c - b*d*(m 
+ n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin 
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] &&  !(IGt 
Q[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 3508

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[1/b^2   Int[(a + b*Sin[e + f*x])^(m 
+ 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ 
[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - 
a*b*B + a^2*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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(821\) vs. \(2(136)=272\).

Time = 6.60 (sec) , antiderivative size = 822, normalized size of antiderivative = 6.00

method	result	size
default	\(\text {Expression too large to display}\)	\(822\)

Input:

int(cos(d*x+c)^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2)/(a+b*cos(d*x+c)),x,meth 
od=_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*cos(1/2* 
d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a*b^2+4*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1 
/2*c)^4*b^3+3*B*a^2*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))-3*B*a*b^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)*EllipticF(cos(1/2*d*x+1/2*c) 
,2^(1/2))+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)*El 
lipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2*b+2*C*cos(1/2*d*x+1/2* 
c)*sin(1/2*d*x+1/2*c)^2*a*b^2-2*C*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2* 
b^3-3*a^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))+3*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))- 
C*a*b^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)*Elli 
pticF(cos(1/2*d*x+1/2*c),2^(1/2))+C*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si 
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-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)*EllipticE(cos(1/ 
2*d*x+1/2*c),2^(1/2))*a^2*b+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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2+3*C*(si...
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int((sqrt(cos(c + d*x))*cos(c + d*x))/(cos(c + d*x)*b + a),x)*b + int((sqr 
t(cos(c + d*x))*cos(c + d*x)**2)/(cos(c + d*x)*b + a),x)*c