\(\int \frac {(a B+b B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx\) [632]

Optimal result

Integrand size = 38, antiderivative size = 130 \[ \int \frac {(a B+b B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {a+b} B \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a d \sqrt {\sec (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.80 \[ \int \frac {(a B+b B \cos (c+d x)) \sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 B \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right )}{d \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {a+b \cos (c+d x)} \sqrt {\sec (c+d x)}} \] Input:

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

Output:

(2*B*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[Arc 
Sin[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]]*Sqrt[Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 130, 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, 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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3295

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

Input:

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

Output:

(2*Sqrt[a + b]*B*Sqrt[Cos[c + d*x]]*Csc[c + d*x]*EllipticF[ArcSin[Sqrt[a + 
 b*Cos[c + d*x]]/(Sqrt[a + b]*Sqrt[Cos[c + d*x]])], -((a + b)/(a - b))]*Sq 
rt[(a*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[(a*(1 + Sec[c + d*x]))/(a - b)])/( 
a*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 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]
 

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 = 13.70 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.85

method	result	size
default	\(-\frac {2 B \sqrt {\sec \left (d x +c \right )}\, \operatorname {EllipticF}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) \sqrt {\frac {a +b \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \cos \left (d x +c \right )}{d \sqrt {a +b \cos \left (d x +c \right )}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\)	\(110\)
parts	\(\text {Expression too large to display}\)	\(984\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Giac [F]

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

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

Mupad [F(-1)]

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

int(((1/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(((1/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]

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

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

Output:

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