\(\int \frac {\sqrt {\cos (c+d x)} (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [716]

Optimal result

Integrand size = 35, antiderivative size = 217 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {\left (A b^2+3 a^2 C-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}+\frac {a \left (A b^2-3 a^2 C+4 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {\left (A b^4-3 a^4 C+a^2 b^2 (A+5 C)\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^3 (a+b)^2 d}-\frac {\left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

(A*b^2+3*C*a^2-2*C*b^2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b^2/(a^2-b^2 
)/d+a*(A*b^2-3*C*a^2+4*C*b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/b^3/( 
a^2-b^2)/d-(A*b^4-3*a^4*C+a^2*b^2*(A+5*C))*EllipticPi(sin(1/2*d*x+1/2*c),2 
*b/(a+b),2^(1/2))/(a-b)/b^3/(a+b)^2/d-(A*b^2+C*a^2)*cos(d*x+c)^(1/2)*sin(d 
*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))
 

Mathematica [A] (warning: unable to verify)

Time = 4.05 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {\cos (c+d x)} \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\frac {4 \left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {2 \left (-A b^2+\left (a^2-2 b^2\right ) C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 a (A+C) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {2 \left (A b^2+3 a^2 C-2 b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(-a+b) (a+b)}}{4 b d} \] Input:

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

Output:

-1/4*((4*(A*b^2 + a^2*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(a 
+ b*Cos[c + d*x])) + ((2*(-(A*b^2) + (a^2 - 2*b^2)*C)*EllipticPi[(2*b)/(a 
+ b), (c + d*x)/2, 2])/(a + b) + (8*a*(A + C)*((a + b)*EllipticF[(c + d*x) 
/2, 2] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(a + b) + (2*(A*b^2 
 + 3*a^2*C - 2*b^2*C)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 
2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*Ell 
ipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt 
[Sin[c + d*x]^2]))/((-a + b)*(a + b)))/(b*d)
 

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.343, Rules used = {3042, 3527, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3527

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle -\frac {\frac {\int \frac {b \left (C a^2+A b^2\right )-a \left (-3 C a^2+A b^2+4 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

\(\displaystyle -\frac {\frac {\frac {\left (-3 a^4 C+a^2 b^2 (A+5 C)+A b^4\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 a \left (-3 a^2 C+A b^2+4 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}-\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\)

\(\Big \downarrow \) 3284

\(\displaystyle -\frac {\left (a^2 C+A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))}-\frac {\frac {\frac {2 \left (-3 a^4 C+a^2 b^2 (A+5 C)+A b^4\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}-\frac {2 a \left (-3 a^2 C+A b^2+4 b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}-\frac {2 \left (3 a^2 C+A b^2-2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}}{2 b \left (a^2-b^2\right )}\)

Input:

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

Output:

-1/2*((-2*(A*b^2 + 3*a^2*C - 2*b^2*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + ( 
(-2*a*(A*b^2 - 3*a^2*C + 4*b^2*C)*EllipticF[(c + d*x)/2, 2])/(b*d) + (2*(A 
*b^4 - 3*a^4*C + a^2*b^2*(A + 5*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 
 2])/(b*(a + b)*d))/b)/(b*(a^2 - b^2)) - ((A*b^2 + a^2*C)*Sqrt[Cos[c + d*x 
]]*Sin[c + d*x])/(b*(a^2 - b^2)*d*(a + b*Cos[c + d*x]))
 

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 3527

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

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. \(833\) vs. \(2(220)=440\).

Time = 7.79 (sec) , antiderivative size = 834, normalized size of antiderivative = 3.84

method	result	size
default	\(\text {Expression too large to display}\)	\(834\)

Input:

int(cos(d*x+c)^(1/2)*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^2,x,method=_RETUR 
NVERBOSE)
 

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*C/b^3/(-2*si 
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^ 
(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a*EllipticF(cos(1/2*d*x+1/2*c),2^(1/ 
2))+b*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))-4/b^2*(A*b^2+3*C*a^2)/(-2*a*b 
+2*b^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2* 
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/ 
2*c),-2*b/(a-b),2^(1/2))-2*a*(A*b^2+C*a^2)/b^3*(-b^2/a/(a^2-b^2)*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)/(2*b*cos(1/ 
2*d*x+1/2*c)^2+a-b)-1/2/(a+b)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2* 
d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*E 
llipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*b/a/(a^2-b^2)*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/a/(a^2-b 
^2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin( 
1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c), 
2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2* 
cos(1/2*d*x+1/2*c)^2)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2) 
^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2 
*a*b+2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1 
/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos...
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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