\(\int \frac {a B+b B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx\) [439]

Optimal result

Integrand size = 38, antiderivative size = 110 \[ \int \frac {a B+b B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {a+b} B \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d} \] Output:

2*(a+b)^(1/2)*B*cot(d*x+c)*EllipticF((a+b*cos(d*x+c))^(1/2)/(a+b)^(1/2)/co 
s(d*x+c)^(1/2),(-(a+b)/(a-b))^(1/2))*(a*(1-sec(d*x+c))/(a+b))^(1/2)*(a*(1+ 
sec(d*x+c))/(a-b))^(1/2)/a/d
 

Mathematica [A] (verified)

Time = 2.04 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.55 \[ \int \frac {a B+b B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx=-\frac {4 (a+b) B \cos ^{\frac {3}{2}}(c+d x) \sqrt {-\frac {(a+b) \cot ^2\left (\frac {1}{2} (c+d x)\right )}{a-b}} \sqrt {\frac {(a+b \cos (c+d x)) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {-\frac {a+b \cos (c+d x)}{a (-1+\cos (c+d x))}}\right ),\frac {2 a}{a-b}\right )}{a d \sqrt {a+b \cos (c+d x)} \left (-\frac {(a+b) \cos (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{a}\right )^{3/2}} \] Input:

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

Output:

(-4*(a + b)*B*Cos[c + d*x]^(3/2)*Sqrt[-(((a + b)*Cot[(c + d*x)/2]^2)/(a - 
b))]*Sqrt[((a + b*Cos[c + d*x])*Csc[(c + d*x)/2]^2)/a]*Csc[c + d*x]*Ellipt 
icF[ArcSin[Sqrt[-((a + b*Cos[c + d*x])/(a*(-1 + Cos[c + d*x])))]], (2*a)/( 
a - b)])/(a*d*Sqrt[a + b*Cos[c + d*x]]*(-(((a + b)*Cos[c + d*x]*Csc[(c + d 
*x)/2]^2)/a))^(3/2))
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {2011, 3042, 3295}

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.

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

\(\Big \downarrow \) 2011

\(\displaystyle B \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

\(\displaystyle \frac {2 B \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}\)

Input:

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

Output:

(2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Cos[c + d*x]]/(S 
qrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sqrt[(a*(1 - Sec[c + 
d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/(a*d)
 

Defintions of rubi rules used

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 3295

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f 
_.)*(x_)]]), x_Symbol] :> Simp[-2*(Tan[e + f*x]/(a*f))*Rt[(a + b)/d, 2]*Sqr 
t[a*((1 - Csc[e + f*x])/(a + b))]*Sqrt[a*((1 + Csc[e + f*x])/(a - b))]*Elli 
pticF[ArcSin[Sqrt[a + b*Sin[e + f*x]]/Sqrt[d*Sin[e + f*x]]/Rt[(a + b)/d, 2] 
], -(a + b)/(a - b)], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] 
&& PosQ[(a + b)/d]
 

Maple [A] (verified)

Time = 12.84 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {2 B \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}}{d \sqrt {\cos \left (d x +c \right )}\, \sqrt {a +\cos \left (d x +c \right ) b}}\)	\(112\)
parts	\(\text {Expression too large to display}\)	\(988\)

Input:

int((B*a+b*B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+cos(d*x+c)*b)^(3/2),x,method= 
_RETURNVERBOSE)
 

Output:

-2*B/d*(cos(d*x+c)+1)/cos(d*x+c)^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)/( 
a+cos(d*x+c)*b)^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2) 
)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)
 

Fricas [F]

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

integrate((a*B+b*B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(3/2),x, 
algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {a B+b B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}} \, dx=B \int \frac {1}{\sqrt {a + b \cos {\left (c + d x \right )}} \sqrt {\cos {\left (c + d x \right )}}}\, dx \] Input:

integrate((a*B+b*B*cos(d*x+c))/cos(d*x+c)**(1/2)/(a+b*cos(d*x+c))**(3/2),x 
)
 

Output:

B*Integral(1/(sqrt(a + b*cos(c + d*x))*sqrt(cos(c + d*x))), x)
 

Maxima [F]

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

integrate((a*B+b*B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(3/2),x, 
algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((a*B+b*B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(3/2),x, 
algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

int((B*a + B*b*cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + b*cos(c + d*x))^(3/2 
)),x)
 

Output:

int((B*a + B*b*cos(c + d*x))/(cos(c + d*x)^(1/2)*(a + b*cos(c + d*x))^(3/2 
)), x)
 

Reduce [F]

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

int((a*B+b*B*cos(d*x+c))/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c))^(3/2),x)
 

Output:

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