\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\) [320]

Optimal result

Integrand size = 33, antiderivative size = 246 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {2 \left (10 a A b-8 a^2 B-9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (10 a^2 A b+5 A b^3-8 a^3 B-7 a b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 A b-4 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 B \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d} \] Output:

-2/15*(10*A*a*b-8*B*a^2-9*B*b^2)*(a+b*cos(d*x+c))^(1/2)*EllipticE(sin(1/2* 
d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))/b^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2 
/15*(10*A*a^2*b+5*A*b^3-8*B*a^3-7*B*a*b^2)*((a+b*cos(d*x+c))/(a+b))^(1/2)* 
InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/b^3/d/(a+b*cos(d*x+ 
c))^(1/2)+2/15*(5*A*b-4*B*a)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b^2/d+2/5*B 
*cos(d*x+c)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 1.85 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.73 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 (5 A b+2 a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (-10 a A b+8 a^2 B+9 b^2 B\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )+2 b (a+b \cos (c+d x)) (5 A b-4 a B+3 b B \cos (c+d x)) \sin (c+d x)}{15 b^3 d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*A*b + 2*a*B)*EllipticF[(c + 
d*x)/2, (2*b)/(a + b)] + (-10*a*A*b + 8*a^2*B + 9*b^2*B)*((a + b)*Elliptic 
E[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])) 
+ 2*b*(a + b*Cos[c + d*x])*(5*A*b - 4*a*B + 3*b*B*Cos[c + d*x])*Sin[c + d* 
x])/(15*b^3*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.26 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3469, 27, 3042, 3502, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3469

\(\displaystyle \frac {2 \int \frac {(5 A b-4 a B) \cos ^2(c+d x)+3 b B \cos (c+d x)+2 a B}{2 \sqrt {a+b \cos (c+d x)}}dx}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(5 A b-4 a B) \cos ^2(c+d x)+3 b B \cos (c+d x)+2 a B}{\sqrt {a+b \cos (c+d x)}}dx}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {2 \int \frac {b (5 A b+2 a B)-\left (-8 B a^2+10 A b a-9 b^2 B\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {b (5 A b+2 a B)-\left (-8 B a^2+10 A b a-9 b^2 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {b (5 A b+2 a B)+\left (8 B a^2-10 A b a+9 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{b}-\frac {\left (-8 a^2 B+10 a A b-9 b^2 B\right ) \int \sqrt {a+b \cos (c+d x)}dx}{b}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (-8 a^2 B+10 a A b-9 b^2 B\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (-8 a^2 B+10 a A b-9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {\left (-8 a^2 B+10 a A b-9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}-\frac {2 \left (-8 a^2 B+10 a A b-9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^2 B+10 a A b-9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{b \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^2 B+10 a A b-9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\frac {\frac {2 \left (-8 a^3 B+10 a^2 A b-7 a b^2 B+5 A b^3\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-8 a^2 B+10 a A b-9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}}{3 b}+\frac {2 (5 A b-4 a B) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d}}{5 b}+\frac {2 B \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d}\)

Input:

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

Output:

(2*B*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b*d) + (((-2*( 
10*a*A*b - 8*a^2*B - 9*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x) 
/2, (2*b)/(a + b)])/(b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(10*a^2* 
A*b + 5*A*b^3 - 8*a^3*B - 7*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*El 
lipticF[(c + d*x)/2, (2*b)/(a + b)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/(3*b) 
 + (2*(5*A*b - 4*a*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*b*d))/(5*b 
)
 

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 3042

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

rule 3132

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

rule 3134

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

rule 3140

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

rule 3142

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

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 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 3502

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co 
s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m 
+ 2))   Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m 
 + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] 
 &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(992\) vs. \(2(235)=470\).

Time = 14.19 (sec) , antiderivative size = 993, normalized size of antiderivative = 4.04

method	result	size
default	\(\text {Expression too large to display}\)	\(993\)
parts	\(\text {Expression too large to display}\)	\(1120\)

Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-2/15*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*B*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6*b^3+(20*A*b^3-4*B*a*b^2+24*B*b^3)*s 
in(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A*a*b^2-10*A*b^3+8*B*a^2*b+2*B 
*a*b^2-6*B*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+10*A*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ellipt 
icF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+5*A*b^3*(sin(1/2*d*x+1/2* 
c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF( 
cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-10*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*( 
-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1 
/2*c),(-2*b/(a-b))^(1/2))*a^2*b+10*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a 
-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),( 
-2*b/(a-b))^(1/2))*a*b^2-8*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin( 
1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a- 
b))^(1/2))*a^3-7*B*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+ 
1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2 
))*b^2+8*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+( 
a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-8*B 
*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b) 
)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+9*B*(sin(1/ 
2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.00 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

-2/45*(sqrt(1/2)*(-16*I*B*a^3 + 20*I*A*a^2*b - 12*I*B*a*b^2 + 15*I*A*b^3)* 
sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^ 
2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) + sqrt(1/2)*( 
16*I*B*a^3 - 20*I*A*a^2*b + 12*I*B*a*b^2 - 15*I*A*b^3)*sqrt(b)*weierstrass 
PInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*co 
s(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) + 3*sqrt(1/2)*(-8*I*B*a^2*b + 10 
*I*A*a*b^2 - 9*I*B*b^3)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, - 
8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, - 
8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2 
*a)/b)) + 3*sqrt(1/2)*(8*I*B*a^2*b - 10*I*A*a*b^2 + 9*I*B*b^3)*sqrt(b)*wei 
erstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierst 
rassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3* 
b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b)) - 3*(3*B*b^3*cos(d*x + c) - 
 4*B*a*b^2 + 5*A*b^3)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/(b^4*d)
 

Sympy [F]

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

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

Output:

Integral((A + B*cos(c + d*x))*cos(c + d*x)**2/sqrt(a + b*cos(c + d*x)), x)
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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