\(\int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx\) [520]

Optimal result

Integrand size = 21, antiderivative size = 50 \[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=-\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+\pi +d x),\frac {8}{7}\right )}{\sqrt {7} d} \] Output:

-8/7*InverseJacobiAM(1/2*d*x+1/2*Pi+1/2*c,2/7*14^(1/2))*7^(1/2)/d-6/7*Elli 
pticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2))*7^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.22 \[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\frac {2 \sqrt {-3+4 \cos (c+d x)} \left (-4 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),8\right )+3 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x),8\right )\right )}{d \sqrt {3-4 \cos (c+d x)}} \] Input:

Integrate[Sqrt[3 - 4*Cos[c + d*x]]*Sec[c + d*x],x]
 

Output:

(2*Sqrt[-3 + 4*Cos[c + d*x]]*(-4*EllipticF[(c + d*x)/2, 8] + 3*EllipticPi[ 
2, (c + d*x)/2, 8]))/(d*Sqrt[3 - 4*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 3282, 3042, 3141, 3285}

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.

\(\displaystyle \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3282

\(\displaystyle 3 \int \frac {\sec (c+d x)}{\sqrt {3-4 \cos (c+d x)}}dx-4 \int \frac {1}{\sqrt {3-4 \cos (c+d x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 3 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx-4 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 3141

\(\displaystyle 3 \int \frac {1}{\sqrt {3-4 \sin \left (c+d x+\frac {\pi }{2}\right )} \sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}\)

\(\Big \downarrow \) 3285

\(\displaystyle -\frac {8 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}-\frac {6 \operatorname {EllipticPi}\left (2,\frac {1}{2} (c+d x+\pi ),\frac {8}{7}\right )}{\sqrt {7} d}\)

Input:

Int[Sqrt[3 - 4*Cos[c + d*x]]*Sec[c + d*x],x]
 

Output:

(-8*EllipticF[(c + Pi + d*x)/2, 8/7])/(Sqrt[7]*d) - (6*EllipticPi[2, (c + 
Pi + d*x)/2, 8/7])/(Sqrt[7]*d)
 

Defintions of rubi rules used

rule 3042

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

rule 3141

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a - b]))*EllipticF[(1/2)*(c + Pi/2 + d*x), -2*(b/(a - b))], x] /; FreeQ 
[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a - b, 0]
 

rule 3282

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

rule 3285

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(158\) vs. \(2(50)=100\).

Time = 3.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 3.18

method	result	size
default	\(\frac {2 \sqrt {-\left (8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {56 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-7}\, \left (4 \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \frac {2 \sqrt {14}}{7}\right )+3 \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), 2, \frac {2 \sqrt {14}}{7}\right )\right )}{7 \sqrt {8 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7}\, d}\)	\(159\)

Input:

int((3-4*cos(d*x+c))^(1/2)*sec(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

2/7*(-(8*cos(1/2*d*x+1/2*c)^2-7)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*(56*sin(1/2*d*x+1/2*c)^2-7)^(1/2)*(4*EllipticF(cos(1/2*d*x 
+1/2*c),2/7*14^(1/2))+3*EllipticPi(cos(1/2*d*x+1/2*c),2,2/7*14^(1/2)))/(8* 
sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-8*co 
s(1/2*d*x+1/2*c)^2+7)^(1/2)/d
 

Fricas [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right ) \,d x } \] Input:

integrate((3-4*cos(d*x+c))^(1/2)*sec(d*x+c),x, algorithm="fricas")
 

Output:

integral(sqrt(-4*cos(d*x + c) + 3)*sec(d*x + c), x)
 

Sympy [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int \sqrt {3 - 4 \cos {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx \] Input:

integrate((3-4*cos(d*x+c))**(1/2)*sec(d*x+c),x)
 

Output:

Integral(sqrt(3 - 4*cos(c + d*x))*sec(c + d*x), x)
 

Maxima [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right ) \,d x } \] Input:

integrate((3-4*cos(d*x+c))^(1/2)*sec(d*x+c),x, algorithm="maxima")
 

Output:

integrate(sqrt(-4*cos(d*x + c) + 3)*sec(d*x + c), x)
 

Giac [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int { \sqrt {-4 \, \cos \left (d x + c\right ) + 3} \sec \left (d x + c\right ) \,d x } \] Input:

integrate((3-4*cos(d*x+c))^(1/2)*sec(d*x+c),x, algorithm="giac")
 

Output:

integrate(sqrt(-4*cos(d*x + c) + 3)*sec(d*x + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int \frac {\sqrt {3-4\,\cos \left (c+d\,x\right )}}{\cos \left (c+d\,x\right )} \,d x \] Input:

int((3 - 4*cos(c + d*x))^(1/2)/cos(c + d*x),x)
 

Output:

int((3 - 4*cos(c + d*x))^(1/2)/cos(c + d*x), x)
 

Reduce [F]

\[ \int \sqrt {3-4 \cos (c+d x)} \sec (c+d x) \, dx=\int \sqrt {-4 \cos \left (d x +c \right )+3}\, \sec \left (d x +c \right )d x \] Input:

int((3-4*cos(d*x+c))^(1/2)*sec(d*x+c),x)
 

Output:

int(sqrt( - 4*cos(c + d*x) + 3)*sec(c + d*x),x)