\(\int \frac {\sqrt {\sec (c+d x)}}{a+a \sec (c+d x)} \, dx\) [197]

Optimal result

Integrand size = 23, antiderivative size = 110 \[ \int \frac {\sqrt {\sec (c+d x)}}{a+a \sec (c+d x)} \, dx=-\frac {\sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {\sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a d}+\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{d (a+a \sec (c+d x))} \] Output:

-cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/2)/a 
/d+cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*sec(d*x+c)^(1/2 
)/a/d+sec(d*x+c)^(1/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 1.02 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.84 \[ \int \frac {\sqrt {\sec (c+d x)}}{a+a \sec (c+d x)} \, dx=-\frac {2 i e^{-i (c+d x)} \cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (-1-e^{2 i (c+d x)}+\left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{4},-e^{2 i (c+d x)}\right )+e^{i (c+d x)} \left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-e^{2 i (c+d x)}\right )\right ) \sec ^{\frac {3}{2}}(c+d x)}{a d \left (1+e^{i (c+d x)}\right ) (1+\sec (c+d x))} \] Input:

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

Output:

((-2*I)*Cos[(c + d*x)/2]^2*(-1 - E^((2*I)*(c + d*x)) + (1 + E^(I*(c + d*x) 
))*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2* 
I)*(c + d*x))] + E^(I*(c + d*x))*(1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2*I)*( 
c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))])*Sec[c + 
 d*x]^(3/2))/(a*d*E^(I*(c + d*x))*(1 + E^(I*(c + d*x)))*(1 + Sec[c + d*x]) 
)
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4307, 3042, 4274, 3042, 4258, 3042, 3119, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4307

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4274

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4258

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

-1/2*((2*a*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] 
)/d - (2*a*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]] 
)/d)/a^2 + (Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(d*(a + a*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 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4307

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((d*Csc[e + f*x])^(n - 1)/(a*f* 
(a + b*Csc[e + f*x]))), x] + Simp[d*((n - 1)/(a*b))   Int[(d*Csc[e + f*x])^ 
(n - 1)*(a - b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ 
[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.80

method	result	size
default	\(-\frac {\sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (\operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+\operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )+2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}{a \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(198\)

Input:

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

Output:

-((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(cos(1/2*d*x+1/2* 
c)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(Elliptic 
F(cos(1/2*d*x+1/2*c),2^(1/2))+EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+2*sin 
(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)/a/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2 
*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+ 
1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int \frac {\sqrt {\sec (c+d x)}}{a+a \sec (c+d x)} \, dx=\frac {{\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + {\left (-i \, \sqrt {2} \cos \left (d x + c\right ) - i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + {\left (i \, \sqrt {2} \cos \left (d x + c\right ) + i \, \sqrt {2}\right )} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \] Input:

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

Output:

1/2*((-I*sqrt(2)*cos(d*x + c) - I*sqrt(2))*weierstrassPInverse(-4, 0, cos( 
d*x + c) + I*sin(d*x + c)) + (I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstr 
assPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + (-I*sqrt(2)*cos(d*x + 
c) - I*sqrt(2))*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c))) + (I*sqrt(2)*cos(d*x + c) + I*sqrt(2))*weierstrass 
Zeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) + 2 
*sqrt(cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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