\(\int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx\) [329]

Optimal result

Integrand size = 25, antiderivative size = 185 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 (A b-a B) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 B \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}-\frac {2 (A b-a B) \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.82 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (-\left ((a+b) (-A b+a B) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )+\left (a^2-b^2\right ) B \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+b (-A b+a B) \sin (c+d x)\right )}{(a-b) b (a+b) d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 3233, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

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

Output:

((2*(A*b - a*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)*B*Sqrt[(a 
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(b*d*Sqr 
t[a + b*Cos[c + d*x]]))/(a^2 - b^2) - (2*(A*b - a*B)*Sin[c + d*x])/((a^2 - 
 b^2)*d*Sqrt[a + b*Cos[c + d*x]])
 

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 3233

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(431\) vs. \(2(184)=368\).

Time = 6.70 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.34

method	result	size
default	\(-\frac {\sqrt {-\left (-2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+a -b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )}{b \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}-\frac {2 \left (A b -B a \right ) \sqrt {-2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, b \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right ) \left (a^{2}-b^{2}\right )}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\)	\(432\)
parts	\(-\frac {2 A \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, b \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )\right )}{\left (a -b \right ) \left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}+\frac {2 B \left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b -\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}+b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-\frac {2 b \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{a -b}+\frac {a +b}{a -b}}\, b \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a \right )}{b \left (a -b \right ) \left (a +b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b}\, d}\)	\(592\)

Input:

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

Output:

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*B/b*(sin 
(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2* 
b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2 
*d*x+1/2*c),(-2*b/(a-b))^(1/2))-2*(A*b-B*a)/b/sin(1/2*d*x+1/2*c)^2/(2*sin( 
1/2*d*x+1/2*c)^2*b-a-b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2 
*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^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)*E 
llipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*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)*b*EllipticE(cos(1 
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2 
*c)^2*b+a+b)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 588, normalized size of antiderivative = 3.18 \[ \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

2/3*(sqrt(1/2)*(-2*I*B*a^3 - I*A*a^2*b + 3*I*B*a*b^2 + (-2*I*B*a^2*b - I*A 
*a*b^2 + 3*I*B*b^3)*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*si 
n(d*x + c) + 2*a)/b) + sqrt(1/2)*(2*I*B*a^3 + I*A*a^2*b - 3*I*B*a*b^2 + (2 
*I*B*a^2*b + I*A*a*b^2 - 3*I*B*b^3)*cos(d*x + c))*sqrt(b)*weierstrassPInve 
rse(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)*(-I*B*a^2*b + I*A*a*b^2 
 + (-I*B*a*b^2 + I*A*b^3)*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)*(I*B*a^2*b - I*A*a*b^2 + (I*B*a*b^2 
- I*A*b^3)*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*(B*a*b^2 - A*b^3)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/((a^ 
2*b^3 - b^5)*d*cos(d*x + c) + (a^3*b^2 - a*b^4)*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int((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)*b + a),x)