\(\int \frac {\sqrt {\cos (c+d x)} (B \cos (c+d x)+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^2} \, dx\) [886]

Optimal result

Integrand size = 42, antiderivative size = 224 \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=-\frac {\left (a b B-3 a^2 C+2 b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}+\frac {\left (a^2 b B-2 b^3 B-3 a^3 C+4 a b^2 C\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^3 \left (a^2-b^2\right ) d}-\frac {a \left (a^2 b B-3 b^3 B-3 a^3 C+5 a b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{(a-b) b^3 (a+b)^2 d}+\frac {a (b B-a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \] Output:

-(B*a*b-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+(B*a^2*b-2*B*b^3-3*C*a^3+4*C*a*b^2)*InverseJacobiAM(1/2*d*x+1/2*c,2^( 
1/2))/b^3/(a^2-b^2)/d-a*(B*a^2*b-3*B*b^3-3*C*a^3+5*C*a*b^2)*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*b-C*a)*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 = 3.11 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^2} \, dx=\frac {-\frac {4 a (-b B+a C) \sqrt {\cos (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac {\frac {2 \left (a b B+a^2 C-2 b^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {8 (-b B+a C) \left ((a+b) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{a+b}+\frac {2 \left (-a b B+3 a^2 C-2 b^2 C\right ) \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{(a-b) (a+b)}}{4 b d} \] Input:

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

Output:

((-4*a*(-(b*B) + a*C)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((a^2 - b^2)*(a + b 
*Cos[c + d*x])) + ((2*(a*b*B + a^2*C - 2*b^2*C)*EllipticPi[(2*b)/(a + b), 
(c + d*x)/2, 2])/(a + b) + (8*(-(b*B) + 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* 
B) + 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)*E 
llipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sq 
rt[Sin[c + d*x]^2]))/((a - b)*(a + b)))/(4*b*d)
 

Rubi [A] (verified)

Time = 1.47 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.98, 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, 3468, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3508

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3468

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3481

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

((-2*(a*b*B - 3*a^2*C + 2*b^2*C)*EllipticE[(c + d*x)/2, 2])/(b*d) + ((2*(a 
^2*b*B - 2*b^3*B - 3*a^3*C + 4*a*b^2*C)*EllipticF[(c + d*x)/2, 2])/(b*d) - 
 (2*a*(a^2*b*B - 3*b^3*B - 3*a^3*C + 5*a*b^2*C)*EllipticPi[(2*b)/(a + b), 
(c + d*x)/2, 2])/(b*(a + b)*d))/b)/(2*b*(a^2 - b^2)) + (a*(b*B - a*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 3468

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

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. \(848\) vs. \(2(227)=454\).

Time = 6.18 (sec) , antiderivative size = 849, normalized size of antiderivative = 3.79

method	result	size
default	\(\text {Expression too large to display}\)	\(849\)

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))^2,x,me 
thod=_RETURNVERBOSE)
                                                                                    
                                                                                    
 

Output:

-(-(1-2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^3/(-2*sin(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)*(B*b*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)) 
-2*C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-C*b*EllipticE(cos(1/2*d*x+1/2 
*c),2^(1/2)))+2*a^2*(B*b-C*a)/b^3*(-1/a*b^2/(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)*EllipticF(co 
s(1/2*d*x+1/2*c),2^(1/2))-1/2/a*b/(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/a*b/(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)*Elli 
pticPi(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(1/2*d*x+1/2*c 
),-2*b/(a-b),2^(1/2)))+4*a*(2*B*b-3*C*a)/b^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)^...
 

Fricas [F(-1)]

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

Output:

Timed out
 

Sympy [F(-1)]

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

Output:

Timed out
 

Maxima [F]

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

Giac [F]

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

Mupad [F(-1)]

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

Reduce [F]

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

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