\(\int \frac {\sqrt {\cos (c+d x)} (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\) [438]

Optimal result

Integrand size = 38, antiderivative size = 117 \[ \int \frac {\sqrt {\cos (c+d x)} (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \sqrt {a+b} B \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d} \] Output:

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

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt {\cos (c+d x)} (a B+b B \cos (c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 B \sqrt {\cos (c+d x)} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \left (\operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )\right )}{d \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(-2*B*Sqrt[Cos[c + d*x]]*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d 
*x]))]*(EllipticF[ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)] - 2*Elliptic 
Pi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]))/(d*Sqrt[Cos[c + d*x]/ 
(1 + Cos[c + d*x])]*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 117, 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, 3288}

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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

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

Input:

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

Output:

(-2*Sqrt[a + b]*B*Cot[c + d*x]*EllipticPi[(a + b)/b, ArcSin[Sqrt[a + b*Cos 
[c + d*x]]/(Sqrt[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)])/(b*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 3288

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

Maple [A] (verified)

Time = 15.70 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.18

method	result	size
default	\(\frac {2 B \sqrt {\cos \left (d x +c \right )}\, \left (\operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right )-2 \operatorname {EllipticPi}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), -1, \sqrt {-\frac {a -b}{a +b}}\right )\right ) \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}}{d \sqrt {a +\cos \left (d x +c \right ) b}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\)	\(138\)
parts	\(\text {Expression too large to display}\)	\(1196\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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