\(\int \cos (c+d x) \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [298]

Optimal result

Integrand size = 31, antiderivative size = 231 \[ \int \cos (c+d x) \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 \left (5 a A b-2 a^2 B+9 b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 A b-2 a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 A b-2 a B) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d} \] Output:

2/15*(5*A*a*b-2*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^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/1 
5*(a^2-b^2)*(5*A*b-2*B*a)*((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^2/d/(a+b*cos(d*x+c))^(1/2)+2/15*(5 
*A*b-2*B*a)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b/d+2/5*B*(a+b*cos(d*x+c))^( 
3/2)*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 3.56 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.77 \[ \int \cos (c+d x) \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 (5 A b+7 a B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (5 a A b-2 a^2 B+9 b^2 B\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )+2 b (a+b \cos (c+d x)) (5 A b+a B+3 b B \cos (c+d x)) \sin (c+d x)}{15 b^2 d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*A*b + 7*a*B)*EllipticF[(c + 
d*x)/2, (2*b)/(a + b)] + (5*a*A*b - 2*a^2*B + 9*b^2*B)*((a + b)*EllipticE[ 
(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 + a*B + 3*b*B*Cos[c + d*x])*Sin[c + d*x])/ 
(15*b^2*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.03, number of steps used = 17, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.548, Rules used = {3042, 3447, 3042, 3502, 27, 3042, 3232, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3447

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3232

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

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

Output:

(2*B*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b*d) + (((2*(5*a*A*b - 2* 
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*(a^2 - b^2)*(5*A*b - 
2*a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a 
+ b)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/3 + (2*(5*A*b - 2*a*B)*Sqrt[a + b*C 
os[c + d*x]]*Sin[c + d*x])/(3*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 3232

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

rule 3447

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a 
 + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), 
 x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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. \(2(220)=440\).

Time = 15.57 (sec) , antiderivative size = 993, normalized size of antiderivative = 4.30

method	result	size
default	\(\text {Expression too large to display}\)	\(993\)
parts	\(\text {Expression too large to display}\)	\(1119\)

Input:

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

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+16*B*a*b^2+24*B*b^3)* 
sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A*a*b^2-10*A*b^3-2*B*a^2*b-8* 
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)-5*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))+5*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-5*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+2*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^3-2*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^2-2*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+2*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))^(1/2)...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.13 \[ \int \cos (c+d x) \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx =\text {Too large to display} \] Input:

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

Output:

-2/45*(sqrt(1/2)*(4*I*B*a^3 - 10*I*A*a^2*b + 3*I*B*a*b^2 + 15*I*A*b^3)*sqr 
t(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)*(-4* 
I*B*a^3 + 10*I*A*a^2*b - 3*I*B*a*b^2 - 15*I*A*b^3)*sqrt(b)*weierstrassPInv 
erse(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)*(2*I*B*a^2*b - 5*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)*(-2*I*B*a^2*b + 5*I*A*a*b^2 + 9*I*B*b^3)*sqrt(b)*weierstra 
ssZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPI 
nverse(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) + B*a*b 
^2 + 5*A*b^3)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/(b^3*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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