\(\int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx\) [330]

Optimal result

Integrand size = 14, antiderivative size = 106 \[ \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {4 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \left (\operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )-2 \operatorname {EllipticPi}\left (-1,\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \sec (c+d x)}{d \sqrt {a+b \sec (c+d x)}} \] Input:

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

Output:

(-4*Cos[(c + d*x)/2]^2*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*C 
os[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)])*Sec[c + d*x])/(d*Sqrt[a + b*Sec[c + d*x]])
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4271}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4271

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

Input:

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

Output:

(-2*Sqrt[a + b]*Cot[c + d*x]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Sec[c 
 + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b 
)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(a*d)
 

Defintions of rubi rules used

rule 3042

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

rule 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 3.96 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {2 \left (\operatorname {EllipticF}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), \sqrt {\frac {a -b}{a +b}}\right )-2 \operatorname {EllipticPi}\left (-\csc \left (d x +c \right )+\cot \left (d x +c \right ), -1, \sqrt {\frac {a -b}{a +b}}\right )\right ) \left (1+\cos \left (d x +c \right )\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (a +b \right ) \left (1+\cos \left (d x +c \right )\right )}}\, \sqrt {a +b \sec \left (d x +c \right )}}{d \left (b +a \cos \left (d x +c \right )\right )}\)	\(147\)

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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