\(\int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx\) [633]

Optimal result

Integrand size = 38, antiderivative size = 137 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 \sqrt {a+b} B \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{b},\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{b d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 0.16 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07 \[ \int \frac {a B+b B \cos (c+d x)}{(a+b \cos (c+d x))^{3/2} \sqrt {\sec (c+d x)}} \, dx=-\frac {2 B \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \left (\operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )-2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )\right ) \sqrt {1+\sec (c+d x)}}{d \sqrt {\frac {1}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(-2*B*Sqrt[Cos[c + d*x]/(1 + 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*EllipticPi[-1, ArcSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)])*Sqr 
t[1 + Sec[c + d*x]])/(d*Sqrt[(1 + Cos[c + d*x])^(-1)]*Sqrt[a + b*Cos[c + d 
*x]])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {2011, 3042, 4710, 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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3288

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

Input:

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

Output:

(-2*Sqrt[a + b]*B*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticPi[(a + b)/b, Ar 
cSin[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*Sqrt[Sec[c + d*x]])
 

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]
 

rule 4710

Int[(csc[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Csc[a 
+ b*x])^m*(c*Sin[a + b*x])^m   Int[ActivateTrig[u]/(c*Sin[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u, x]
 

Maple [A] (verified)

Time = 11.74 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.02

method	result	size
default	\(-\frac {2 B \sqrt {\frac {a +b \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \left (-\operatorname {EllipticF}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right )+2 \operatorname {EllipticPi}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), -1, \sqrt {-\frac {a -b}{a +b}}\right )\right )}{d \sqrt {a +b \cos \left (d x +c \right )}\, \sqrt {\sec \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(140\)
parts	\(\text {Expression too large to display}\)	\(1144\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Reduce [F]

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

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

Output:

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