\(\int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\) [327]

Optimal result

Integrand size = 33, antiderivative size = 262 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (6 a A b-8 a^2 B-b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^3 d \sqrt {a+b \cos (c+d x)}}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 B \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^2 d} \] Output:

2/3*(6*A*a^2*b-3*A*b^3-8*B*a^3+5*B*a*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/(a^2-b^2)/d/((a+b*cos(d*x 
+c))/(a+b))^(1/2)-2/3*(6*A*a*b-8*B*a^2-B*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*a^2*(A*b-B*a)*sin(d*x+c)/b^2/(a^2-b^2)/d/(a+b*cos(d*x+c))^( 
1/2)+2/3*B*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b^2/d
 

Mathematica [A] (verified)

Time = 2.56 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (\frac {\sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (\left (6 a^2 A b-3 A b^3-8 a^3 B+5 a b^2 B\right ) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+(a-b) \left (-6 a A b+8 a^2 B+b^2 B\right ) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{a-b}+b \left (\frac {a \left (3 a A b-4 a^2 B+b^2 B\right )}{-a^2+b^2}+b B \cos (c+d x)\right ) \sin (c+d x)\right )}{3 b^3 d \sqrt {a+b \cos (c+d x)}} \] Input:

Integrate[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^(3/2) 
,x]
 

Output:

(2*((Sqrt[(a + b*Cos[c + d*x])/(a + b)]*((6*a^2*A*b - 3*A*b^3 - 8*a^3*B + 
5*a*b^2*B)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] + (a - b)*(-6*a*A*b + 8*a 
^2*B + b^2*B)*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/(a - b) + b*((a*(3*a 
*A*b - 4*a^2*B + b^2*B))/(-a^2 + b^2) + b*B*Cos[c + d*x])*Sin[c + d*x]))/( 
3*b^3*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.44 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.07, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3467, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3467

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {\frac {2 \int \frac {\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \cos (c+d x) b}{2 \sqrt {a+b \cos (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \cos (c+d x) b}{\sqrt {a+b \cos (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\left (-2 B a^2+3 A b a-b^2 B\right ) b^2+\left (-8 B a^3+6 A b a^2+5 b^2 B a-3 A b^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3231

\(\displaystyle \frac {\frac {\left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \sqrt {a+b \cos (c+d x)}dx-\left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \int \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3134

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}dx}{\sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {\frac {\frac {2 \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \int \frac {1}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3142

\(\displaystyle \frac {\frac {\frac {2 \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {2 \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin \left (c+d x+\frac {\pi }{2}\right )}{a+b}}}dx}{\sqrt {a+b \cos (c+d x)}}}{3 b}+\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {\frac {2 B \left (a^2-b^2\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}+\frac {\frac {2 \left (-8 a^3 B+6 a^2 A b+5 a b^2 B-3 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (-8 a^2 B+6 a A b-b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{d \sqrt {a+b \cos (c+d x)}}}{3 b}}{b^2 \left (a^2-b^2\right )}-\frac {2 a^2 (A b-a B) \sin (c+d x)}{b^2 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\)

Input:

Int[(Cos[c + d*x]^2*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^(3/2),x]
 

Output:

(-2*a^2*(A*b - a*B)*Sin[c + d*x])/(b^2*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d* 
x]]) + (((2*(6*a^2*A*b - 3*A*b^3 - 8*a^3*B + 5*a*b^2*B)*Sqrt[a + b*Cos[c + 
 d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[(a + b*Cos[c + d*x]) 
/(a + b)]) - (2*(a^2 - b^2)*(6*a*A*b - 8*a^2*B - b^2*B)*Sqrt[(a + b*Cos[c 
+ d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(d*Sqrt[a + b*Cos[ 
c + d*x]]))/(3*b) + (2*(a^2 - b^2)*B*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x] 
)/(3*d))/(b^2*(a^2 - b^2))
 

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 3467

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f 
_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ 
(B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 
*(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2))   Int[(c + d* 
Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c 
- 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* 
a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 
1)*(c^2 - d^2)*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] && LtQ[ 
n, -1]
 

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. \(1274\) vs. \(2(253)=506\).

Time = 15.99 (sec) , antiderivative size = 1275, normalized size of antiderivative = 4.87

method	result	size
parts	\(\text {Expression too large to display}\)	\(1275\)
default	\(\text {Expression too large to display}\)	\(1336\)

Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+cos(d*x+c)*b)^(3/2),x,method=_RETURNV 
ERBOSE)
 

Output:

-2*A*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2*b-2*(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(c 
os(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+2*b^2*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))+2*(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*(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-(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+(s 
in(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))*b^3)/b^2/(a-b)/(a+b) 
/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d-2/3*B*(4*cos(1 
/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4*a^2*b^2-4*cos(1/2*d*x+1/2*c)*sin(1/2*d* 
x+1/2*c)^4*b^4-8*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^3*b-2*cos(1/2*d 
*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*a^2*b^2+2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/ 
2*c)^2*a*b^3+2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b^4+8*(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^4-7*(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...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 767, normalized size of antiderivative = 2.93 \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x, algorith 
m="fricas")
 

Output:

2/9*(sqrt(1/2)*(-16*I*B*a^5 + 12*I*A*a^4*b + 16*I*B*a^3*b^2 - 15*I*A*a^2*b 
^3 + 3*I*B*a*b^4 + (-16*I*B*a^4*b + 12*I*A*a^3*b^2 + 16*I*B*a^2*b^3 - 15*I 
*A*a*b^4 + 3*I*B*b^5)*cos(d*x + c))*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^5 - 12*I*A*a^4*b - 16*I*B*a^3 
*b^2 + 15*I*A*a^2*b^3 - 3*I*B*a*b^4 + (16*I*B*a^4*b - 12*I*A*a^3*b^2 - 16* 
I*B*a^2*b^3 + 15*I*A*a*b^4 - 3*I*B*b^5)*cos(d*x + c))*sqrt(b)*weierstrassP 
Inverse(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^4*b + 6*I 
*A*a^3*b^2 + 5*I*B*a^2*b^3 - 3*I*A*a*b^4 + (-8*I*B*a^3*b^2 + 6*I*A*a^2*b^3 
 + 5*I*B*a*b^4 - 3*I*A*b^5)*cos(d*x + c))*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^4*b - 6*I*A*a^3*b^2 - 5*I 
*B*a^2*b^3 + 3*I*A*a*b^4 + (8*I*B*a^3*b^2 - 6*I*A*a^2*b^3 - 5*I*B*a*b^4 + 
3*I*A*b^5)*cos(d*x + c))*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*(4*B*a^3*b^2 - 3*A*a^2*b^3 - B*a*b^4 + (B*a^2*b^3 - B*b^5)*co 
s(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/((a^2*b^5 - b^7)*d*c...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**(3/2),x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x, algorith 
m="maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^2/(b*cos(d*x + c) + a)^(3/2), 
x)
 

Giac [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x, algorith 
m="giac")
 

Output:

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^2/(b*cos(d*x + c) + a)^(3/2), 
x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^(3/2),x)
 

Output:

int((cos(c + d*x)^2*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\cos \left (d x +c \right ) b +a}\, \cos \left (d x +c \right )^{2}}{\cos \left (d x +c \right ) b +a}d x \] Input:

int(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(3/2),x)
 

Output:

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