\(\int \frac {\sqrt {a+b \sec (c+d x)}}{1+\cos (c+d x)} \, dx\) [65]

Optimal result

Integrand size = 25, antiderivative size = 92 \[ \int \frac {\sqrt {a+b \sec (c+d x)}}{1+\cos (c+d x)} \, dx=\frac {E\left (\arcsin \left (\frac {\tan (c+d x)}{1+\sec (c+d x)}\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {a+b \sec (c+d x)}}{d \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}} \] Output:

EllipticE(tan(d*x+c)/(1+sec(d*x+c)),((a-b)/(a+b))^(1/2))*(1/(1+sec(d*x+c)) 
)^(1/2)*(a+b*sec(d*x+c))^(1/2)/d/((a+b*sec(d*x+c))/(a+b)/(1+sec(d*x+c)))^( 
1/2)
 

Mathematica [A] (verified)

Time = 5.62 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {a+b \sec (c+d x)}}{1+\cos (c+d x)} \, dx=\frac {E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {a+b \sec (c+d x)}}{d \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}}} \] Input:

Integrate[Sqrt[a + b*Sec[c + d*x]]/(1 + Cos[c + d*x]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3308, 3042, 4456}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3308

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4456

\(\displaystyle \frac {\sqrt {\frac {1}{\sec (c+d x)+1}} \sqrt {a+b \sec (c+d x)} E\left (\arcsin \left (\frac {\tan (c+d x)}{\sec (c+d x)+1}\right )|\frac {a-b}{a+b}\right )}{d \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (\sec (c+d x)+1)}}}\)

Input:

Int[Sqrt[a + b*Sec[c + d*x]]/(1 + Cos[c + d*x]),x]
 

Output:

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

Defintions of rubi rules used

rule 3042

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

rule 3308

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_)*((a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)])^(m_.), x_Symbol] :> Int[(b + a*Csc[e + f*x])^m*((c + d*Csc[e + 
f*x])^n/Csc[e + f*x]^m), x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !Integer 
Q[n] && IntegerQ[m]
 

rule 4456

Int[(csc[(e_.) + (f_.)*(x_)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)])/(c 
sc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(-Sqrt[a + b*Csc[e 
+ f*x]])*(Sqrt[c/(c + d*Csc[e + f*x])]/(d*f*Sqrt[c*d*((a + b*Csc[e + f*x])/ 
((b*c + a*d)*(c + d*Csc[e + f*x])))]))*EllipticE[ArcSin[c*(Cot[e + f*x]/(c 
+ d*Csc[e + f*x]))], -(b*c - a*d)/(b*c + a*d)], x] /; FreeQ[{a, b, c, d, e, 
 f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && EqQ[c^2 - d^2, 0]
 

Maple [A] (verified)

Time = 4.07 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {\left (-a -b \right ) \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right ) a +b}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \operatorname {EllipticE}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right ) \sqrt {a +b \sec \left (d x +c \right )}}{d \left (\cos \left (d x +c \right ) a +b \right )}\)	\(120\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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