\(\int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx\) [268]

Optimal result

Integrand size = 35, antiderivative size = 125 \[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=\frac {2 i \cot (e+f x) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {5} \sqrt {4-5 \sec (e+f x)}}{\sqrt {2+3 \sec (e+f x)}}\right ),\frac {1}{45}\right ) \sqrt {\frac {1-\sec (e+f x)}{2+3 \sec (e+f x)}} \sqrt {\frac {1+\sec (e+f x)}{2+3 \sec (e+f x)}} (2+3 \sec (e+f x))}{3 \sqrt {5} f} \] Output:

2/15*I*cot(f*x+e)*InverseJacobiAM(I*arcsinh(5^(1/2)*(4-5*sec(f*x+e))^(1/2) 
/(2+3*sec(f*x+e))^(1/2)),1/15*5^(1/2))*((1-sec(f*x+e))/(2+3*sec(f*x+e)))^( 
1/2)*((1+sec(f*x+e))/(2+3*sec(f*x+e)))^(1/2)*(2+3*sec(f*x+e))*5^(1/2)/f
 

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.41 \[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=-\frac {4 \sqrt {-\cot ^2\left (\frac {1}{2} (e+f x)\right )} \sqrt {-\left ((3+2 \cos (e+f x)) \csc ^2\left (\frac {1}{2} (e+f x)\right )\right )} \sqrt {-\left ((-5+4 \cos (e+f x)) \csc ^2\left (\frac {1}{2} (e+f x)\right )\right )} \csc (e+f x) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\frac {5}{22}} \sqrt {\frac {-5+4 \cos (e+f x)}{-1+\cos (e+f x)}}\right ),\frac {44}{45}\right ) \sec (e+f x) \sin ^4\left (\frac {1}{2} (e+f x)\right )}{3 \sqrt {5} f \sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \] Input:

Integrate[Sec[e + f*x]/(Sqrt[4 - 5*Sec[e + f*x]]*Sqrt[2 + 3*Sec[e + f*x]]) 
,x]
 

Output:

(-4*Sqrt[-Cot[(e + f*x)/2]^2]*Sqrt[-((3 + 2*Cos[e + f*x])*Csc[(e + f*x)/2] 
^2)]*Sqrt[-((-5 + 4*Cos[e + f*x])*Csc[(e + f*x)/2]^2)]*Csc[e + f*x]*Ellipt 
icF[ArcSin[Sqrt[5/22]*Sqrt[(-5 + 4*Cos[e + f*x])/(-1 + Cos[e + f*x])]], 44 
/45]*Sec[e + f*x]*Sin[(e + f*x)/2]^4)/(3*Sqrt[5]*f*Sqrt[4 - 5*Sec[e + f*x] 
]*Sqrt[2 + 3*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {3042, 4472}

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.

\(\displaystyle \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {3 \sec (e+f x)+2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {4-5 \csc \left (e+f x+\frac {\pi }{2}\right )} \sqrt {3 \csc \left (e+f x+\frac {\pi }{2}\right )+2}}dx\)

\(\Big \downarrow \) 4472

\(\displaystyle \frac {2 i \cot (e+f x) \sqrt {\frac {1-\sec (e+f x)}{3 \sec (e+f x)+2}} \sqrt {\frac {\sec (e+f x)+1}{3 \sec (e+f x)+2}} (3 \sec (e+f x)+2) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {5} \sqrt {4-5 \sec (e+f x)}}{\sqrt {3 \sec (e+f x)+2}}\right ),\frac {1}{45}\right )}{3 \sqrt {5} f}\)

Input:

Int[Sec[e + f*x]/(Sqrt[4 - 5*Sec[e + f*x]]*Sqrt[2 + 3*Sec[e + f*x]]),x]
 

Output:

(((2*I)/3)*Cot[e + f*x]*EllipticF[I*ArcSinh[(Sqrt[5]*Sqrt[4 - 5*Sec[e + f* 
x]])/Sqrt[2 + 3*Sec[e + f*x]]], 1/45]*Sqrt[(1 - Sec[e + f*x])/(2 + 3*Sec[e 
 + f*x])]*Sqrt[(1 + Sec[e + f*x])/(2 + 3*Sec[e + f*x])]*(2 + 3*Sec[e + f*x 
]))/(Sqrt[5]*f)
 

Defintions of rubi rules used

rule 3042

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

rule 4472

Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*Sqr 
t[csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)]), x_Symbol] :> Simp[-2*((c + d*Csc[ 
e + f*x])/(f*(b*c - a*d)*Rt[(c + d)/(a + b), 2]*Cot[e + f*x]))*Sqrt[(b*c - 
a*d)*((1 - Csc[e + f*x])/((a + b)*(c + d*Csc[e + f*x])))]*Sqrt[(-(b*c - a*d 
))*((1 + Csc[e + f*x])/((a - b)*(c + d*Csc[e + f*x])))]*EllipticF[ArcSin[Rt 
[(c + d)/(a + b), 2]*(Sqrt[a + b*Csc[e + f*x]]/Sqrt[c + d*Csc[e + f*x]])], 
(a + b)*((c - d)/((a - b)*(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]
 

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.14

method	result	size
default	\(\frac {i \operatorname {EllipticF}\left (3 i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), \frac {\sqrt {5}}{15}\right ) \sqrt {2+3 \sec \left (f x +e \right )}\, \sqrt {4-5 \sec \left (f x +e \right )}\, \sqrt {-\frac {2 \left (4 \cos \left (f x +e \right )-5\right )}{\cos \left (f x +e \right )+1}}\, \sqrt {10}\, \sqrt {\frac {2 \cos \left (f x +e \right )+3}{\cos \left (f x +e \right )+1}}\, \left (\cos \left (f x +e \right )^{2}+\cos \left (f x +e \right )\right )}{15 f \left (8 \cos \left (f x +e \right )^{2}+2 \cos \left (f x +e \right )-15\right )}\)	\(142\)

Input:

int(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x,method=_RET 
URNVERBOSE)
 

Output:

1/15*I/f*EllipticF(3*I*(cot(f*x+e)-csc(f*x+e)),1/15*5^(1/2))*(2+3*sec(f*x+ 
e))^(1/2)*(4-5*sec(f*x+e))^(1/2)*(-2*(4*cos(f*x+e)-5)/(cos(f*x+e)+1))^(1/2 
)*10^(1/2)*((2*cos(f*x+e)+3)/(cos(f*x+e)+1))^(1/2)/(8*cos(f*x+e)^2+2*cos(f 
*x+e)-15)*(cos(f*x+e)^2+cos(f*x+e))
 

Fricas [F]

\[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )}{\sqrt {3 \, \sec \left (f x + e\right ) + 2} \sqrt {-5 \, \sec \left (f x + e\right ) + 4}} \,d x } \] Input:

integrate(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x, algo 
rithm="fricas")
 

Output:

integral(-sqrt(3*sec(f*x + e) + 2)*sqrt(-5*sec(f*x + e) + 4)*sec(f*x + e)/ 
(15*sec(f*x + e)^2 - 2*sec(f*x + e) - 8), x)
 

Sympy [F]

\[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=\int \frac {\sec {\left (e + f x \right )}}{\sqrt {4 - 5 \sec {\left (e + f x \right )}} \sqrt {3 \sec {\left (e + f x \right )} + 2}}\, dx \] Input:

integrate(sec(f*x+e)/(4-5*sec(f*x+e))**(1/2)/(2+3*sec(f*x+e))**(1/2),x)
 

Output:

Integral(sec(e + f*x)/(sqrt(4 - 5*sec(e + f*x))*sqrt(3*sec(e + f*x) + 2)), 
 x)
 

Maxima [F]

\[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )}{\sqrt {3 \, \sec \left (f x + e\right ) + 2} \sqrt {-5 \, \sec \left (f x + e\right ) + 4}} \,d x } \] Input:

integrate(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x, algo 
rithm="maxima")
 

Output:

integrate(sec(f*x + e)/(sqrt(3*sec(f*x + e) + 2)*sqrt(-5*sec(f*x + e) + 4) 
), x)
 

Giac [F]

\[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=\int { \frac {\sec \left (f x + e\right )}{\sqrt {3 \, \sec \left (f x + e\right ) + 2} \sqrt {-5 \, \sec \left (f x + e\right ) + 4}} \,d x } \] Input:

integrate(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x, algo 
rithm="giac")
 

Output:

integrate(sec(f*x + e)/(sqrt(3*sec(f*x + e) + 2)*sqrt(-5*sec(f*x + e) + 4) 
), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=\int \frac {1}{\cos \left (e+f\,x\right )\,\sqrt {\frac {3}{\cos \left (e+f\,x\right )}+2}\,\sqrt {4-\frac {5}{\cos \left (e+f\,x\right )}}} \,d x \] Input:

int(1/(cos(e + f*x)*(3/cos(e + f*x) + 2)^(1/2)*(4 - 5/cos(e + f*x))^(1/2)) 
,x)
 

Output:

int(1/(cos(e + f*x)*(3/cos(e + f*x) + 2)^(1/2)*(4 - 5/cos(e + f*x))^(1/2)) 
, x)
 

Reduce [F]

\[ \int \frac {\sec (e+f x)}{\sqrt {4-5 \sec (e+f x)} \sqrt {2+3 \sec (e+f x)}} \, dx=-\left (\int \frac {\sqrt {3 \sec \left (f x +e \right )+2}\, \sqrt {-5 \sec \left (f x +e \right )+4}\, \sec \left (f x +e \right )}{15 \sec \left (f x +e \right )^{2}-2 \sec \left (f x +e \right )-8}d x \right ) \] Input:

int(sec(f*x+e)/(4-5*sec(f*x+e))^(1/2)/(2+3*sec(f*x+e))^(1/2),x)
 

Output:

 - int((sqrt(3*sec(e + f*x) + 2)*sqrt( - 5*sec(e + f*x) + 4)*sec(e + f*x)) 
/(15*sec(e + f*x)**2 - 2*sec(e + f*x) - 8),x)