\(\int \frac {\sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx\) [759]

Optimal result

Integrand size = 25, antiderivative size = 307 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 b \sqrt {\cos (c+d x)} \csc (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right )|-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a^2 \sqrt {a+b} d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \csc (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right ) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (1+\sec (c+d x))}{a-b}}}{a \sqrt {a+b} d \sqrt {\sec (c+d x)}}-\frac {2 b \sqrt {\sec (c+d x)} \sin (c+d x)}{\left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 6.40 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {\sec (c+d x)}}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {\sec (c+d x)} \left (-2 b (a+b) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}}+2 a (a+b) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {a+b \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}}+b (-a+b) \cos (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(2*Sqrt[Sec[c + d*x]]*(-2*b*(a + b)*Cos[(c + d*x)/2]^2*Sqrt[(a + b*Cos[c + 
 d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], ( 
-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] + 2*a*(a + b)*Cos[(c + d*x) 
/2]^2*Sqrt[(a + b*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[Ar 
cSin[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)] + 
b*(-a + b)*Cos[c + d*x]*Tan[(c + d*x)/2]))/(a*(a^2 - b^2)*d*Sqrt[a + b*Cos 
[c + d*x]])
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 311, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 4710, 3042, 3279, 3042, 3477, 3042, 3295, 3473}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4710

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\)

\(\Big \downarrow \) 3279

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {b+a \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 b \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {\int \frac {b+a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 b \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3477

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {b \int \frac {\cos (c+d x)+1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \cos (c+d x)}}dx+(a-b) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}dx}{a^2-b^2}-\frac {2 b \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {(a-b) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+b \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2-b^2}-\frac {2 b \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3295

\(\displaystyle \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (\frac {b \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )+1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \sqrt {\frac {a (1-\sec (c+d x))}{a+b}} \sqrt {\frac {a (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \cos (c+d x)}}{\sqrt {a+b} \sqrt {\cos (c+d x)}}\right ),-\frac {a+b}{a-b}\right )}{a d}}{a^2-b^2}-\frac {2 b \sin (c+d x)}{d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \cos (c+d x)}}\right )\)

\(\Big \downarrow \) 3473

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

Input:

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

Output:

Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]]*(((2*(a - b)*b*Sqrt[a + b]*Cot[c + d 
*x]*EllipticE[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)])/(a^2*d) + (2*(a - b)*Sqrt[a + b]*Cot[c + d*x]* 
EllipticF[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 + S 
ec[c + d*x]))/(a - b)])/(a*d))/(a^2 - b^2) - (2*b*Sin[c + d*x])/((a^2 - b^ 
2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Cos[c + d*x]]))
 

Defintions of rubi rules used

rule 3042

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

rule 3279

Int[1/(Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]*((a_) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(3/2)), x_Symbol] :> Simp[2*b*(Cos[e + f*x]/(f*(a^2 - b^2)*Sqrt[a + 
b*Sin[e + f*x]]*Sqrt[d*Sin[e + f*x]])), x] + Simp[d/(a^2 - b^2)   Int[(b + 
a*Sin[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*(d*Sin[e + f*x])^(3/2)), x], x] / 
; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
 

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 3473

Int[((A_) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((b_.)*sin[(e_.) + (f_.)*(x_)]) 
^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*A* 
(c - d)*(Tan[e + f*x]/(f*b*c^2))*Rt[(c + d)/b, 2]*Sqrt[c*((1 + Csc[e + f*x] 
)/(c - d))]*Sqrt[c*((1 - Csc[e + f*x])/(c + d))]*EllipticE[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, A, B}, x] && NeQ[c^2 - d^2, 0] && EqQ[A, B] && 
PosQ[(c + d)/b]
 

rule 3477

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ 
.)*(x_)])^(3/2)*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> S 
imp[(A - B)/(a - b)   Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f* 
x]]), x], x] - Simp[(A*b - a*B)/(a - b)   Int[(1 + Sin[e + f*x])/((a + b*Si 
n[e + f*x])^(3/2)*Sqrt[c + d*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e 
, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && NeQ[A, 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 = 17.74 (sec) , antiderivative size = 498, normalized size of antiderivative = 1.62

method	result	size
default	\(\frac {2 \sqrt {\sec \left (d x +c \right )}\, \sqrt {a +\cos \left (d x +c \right ) b}\, \left (\left (\csc \left (d x +c \right )^{3} \left (1-\cos \left (d x +c \right )\right )^{3}+\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ) a b +\left (-\csc \left (d x +c \right )^{3} \left (1-\cos \left (d x +c \right )\right )^{3}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) b^{2}-2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}\, \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) a^{2}-2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}\, \operatorname {EllipticF}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) a b +2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}\, \operatorname {EllipticE}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) a b +2 \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {a +\cos \left (d x +c \right ) b}{\left (\cos \left (d x +c \right )+1\right ) \left (a +b \right )}}\, \operatorname {EllipticE}\left (\cot \left (d x +c \right )-\csc \left (d x +c \right ), \sqrt {-\frac {a -b}{a +b}}\right ) b^{2}\right )}{d \left (\csc \left (d x +c \right )^{2} a \left (1-\cos \left (d x +c \right )\right )^{2}-\csc \left (d x +c \right )^{2} b \left (1-\cos \left (d x +c \right )\right )^{2}+a +b \right ) a \left (a -b \right ) \left (a +b \right )}\)	\(498\)

Input:

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

Output:

2/d*sec(d*x+c)^(1/2)*(a+cos(d*x+c)*b)^(1/2)*((csc(d*x+c)^3*(1-cos(d*x+c))^ 
3+cot(d*x+c)-csc(d*x+c))*a*b+(-csc(d*x+c)^3*(1-cos(d*x+c))^3-cot(d*x+c)+cs 
c(d*x+c))*b^2-2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d 
*x+c)+1)/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2) 
)*a^2-2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1) 
/(a+b))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b+2* 
(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^ 
(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*a*b+2*(cos(d*x 
+c)/(cos(d*x+c)+1))^(1/2)*((a+cos(d*x+c)*b)/(cos(d*x+c)+1)/(a+b))^(1/2)*El 
lipticE(cot(d*x+c)-csc(d*x+c),(-(a-b)/(a+b))^(1/2))*b^2)/(csc(d*x+c)^2*a*( 
1-cos(d*x+c))^2-csc(d*x+c)^2*b*(1-cos(d*x+c))^2+a+b)/a/(a-b)/(a+b)
 

Fricas [F]

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

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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