\(\int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [282]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [F(-1)]
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 35, antiderivative size = 209 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\frac {2 \sqrt {a+b} \cot (e+f x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (1+\sec (e+f x))}{a-b}}}{b d f}-\frac {2 c \operatorname {EllipticPi}\left (\frac {2 d}{c+d},\arcsin \left (\frac {\sqrt {1-\sec (e+f x)}}{\sqrt {2}}\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sec (e+f x)}{a+b}} \tan (e+f x)}{d (c+d) f \sqrt {a+b \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 3.99 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.79 \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=-\frac {4 \cos ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\frac {\cos (e+f x)}{1+\cos (e+f x)}} \sqrt {\frac {b+a \cos (e+f x)}{(a+b) (1+\cos (e+f x))}} \left ((c+d) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right ),\frac {a-b}{a+b}\right )-2 c \operatorname {EllipticPi}\left (\frac {c-d}{c+d},\arcsin \left (\tan \left (\frac {1}{2} (e+f x)\right )\right ),\frac {a-b}{a+b}\right )\right ) \sec (e+f x)}{(c-d) (c+d) f \sqrt {a+b \sec (e+f x)}} \] Input:

Integrate[Sec[e + f*x]^2/(Sqrt[a + b*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x 
]
 

Output:

(-4*Cos[(e + f*x)/2]^2*Sqrt[Cos[e + f*x]/(1 + Cos[e + f*x])]*Sqrt[(b + a*C 
os[e + f*x])/((a + b)*(1 + Cos[e + f*x]))]*((c + d)*EllipticF[ArcSin[Tan[( 
e + f*x)/2]], (a - b)/(a + b)] - 2*c*EllipticPi[(c - d)/(c + d), ArcSin[Ta 
n[(e + f*x)/2]], (a - b)/(a + b)])*Sec[e + f*x])/((c - d)*(c + d)*f*Sqrt[a 
 + b*Sec[e + f*x]])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4465, 3042, 4319, 4461}

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.

\(\displaystyle \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4465

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4319

\(\displaystyle \frac {2 \sqrt {a+b} \cot (e+f x) \sqrt {\frac {b (1-\sec (e+f x))}{a+b}} \sqrt {-\frac {b (\sec (e+f x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (e+f x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d f}-\frac {c \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (e+f x+\frac {\pi }{2}\right )} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{d}\)

\(\Big \downarrow \) 4461

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

Input:

Int[Sec[e + f*x]^2/(Sqrt[a + b*Sec[e + f*x]]*(c + d*Sec[e + f*x])),x]
 

Output:

(2*Sqrt[a + b]*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[e + f*x]]/Sqrt 
[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[e + f*x]))/(a + b)]*Sqrt[-((b* 
(1 + Sec[e + f*x]))/(a - b))])/(b*d*f) - (2*c*EllipticPi[(2*d)/(c + d), Ar 
cSin[Sqrt[1 - Sec[e + f*x]]/Sqrt[2]], (2*b)/(a + b)]*Sqrt[(a + b*Sec[e + f 
*x])/(a + b)]*Tan[e + f*x])/(d*(c + d)*f*Sqrt[a + b*Sec[e + f*x]]*Sqrt[-Ta 
n[e + f*x]^2])
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4319
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4461
Int[csc[(e_.) + (f_.)*(x_)]/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(cs 
c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))), x_Symbol] :> Simp[-2*(Cot[e + f*x]/(f 
*(c + d)*Sqrt[a + b*Csc[e + f*x]]*Sqrt[-Cot[e + f*x]^2]))*Sqrt[(a + b*Csc[e 
 + f*x])/(a + b)]*EllipticPi[2*(d/(c + d)), ArcSin[Sqrt[1 - Csc[e + f*x]]/S 
qrt[2]], 2*(b/(a + b))], 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 4465
Int[csc[(e_.) + (f_.)*(x_)]^2/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*( 
csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))), x_Symbol] :> Simp[1/d   Int[Csc[e + 
 f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[c/d   Int[Csc[e + f*x]/(Sqrt[ 
a + b*Csc[e + f*x]]*(c + d*Csc[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]
 
Maple [A] (verified)

Time = 7.15 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.98

method result size
default \(-\frac {2 \left (2 c \operatorname {EllipticPi}\left (\cot \left (f x +e \right )-\csc \left (f x +e \right ), \frac {c -d}{c +d}, \sqrt {\frac {a -b}{a +b}}\right )-\operatorname {EllipticF}\left (\cot \left (f x +e \right )-\csc \left (f x +e \right ), \sqrt {\frac {a -b}{a +b}}\right ) c -\operatorname {EllipticF}\left (\cot \left (f x +e \right )-\csc \left (f x +e \right ), \sqrt {\frac {a -b}{a +b}}\right ) d \right ) \left (\cos \left (f x +e \right )+1\right ) \sqrt {a +b \sec \left (f x +e \right )}\, \sqrt {\frac {\cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (a +b \right ) \left (\cos \left (f x +e \right )+1\right )}}}{f \left (c -d \right ) \left (c +d \right ) \left (b +a \cos \left (f x +e \right )\right )}\) \(205\)

Input:

int(sec(f*x+e)^2/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x,method=_RETURNV 
ERBOSE)
 

Output:

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(sec(f*x+e)^2/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorith 
m="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {\sec ^{2}{\left (e + f x \right )}}{\sqrt {a + b \sec {\left (e + f x \right )}} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \] Input:

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

Output:

Integral(sec(e + f*x)**2/(sqrt(a + b*sec(e + f*x))*(c + d*sec(e + f*x))), 
x)
 

Maxima [F]

\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int { \frac {\sec \left (f x + e\right )^{2}}{\sqrt {b \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}} \,d x } \] Input:

integrate(sec(f*x+e)^2/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorith 
m="maxima")
 

Output:

integrate(sec(f*x + e)^2/(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) + c)), 
x)
 

Giac [F]

\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int { \frac {\sec \left (f x + e\right )^{2}}{\sqrt {b \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}} \,d x } \] Input:

integrate(sec(f*x+e)^2/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, algorith 
m="giac")
 

Output:

integrate(sec(f*x + e)^2/(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) + c)), 
x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {1}{{\cos \left (e+f\,x\right )}^2\,\sqrt {a+\frac {b}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sec ^2(e+f x)}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {\sqrt {\sec \left (f x +e \right ) b +a}\, \sec \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{2} b d +\sec \left (f x +e \right ) a d +\sec \left (f x +e \right ) b c +a c}d x \] Input:

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

Output:

int((sqrt(sec(e + f*x)*b + a)*sec(e + f*x)**2)/(sec(e + f*x)**2*b*d + sec( 
e + f*x)*a*d + sec(e + f*x)*b*c + a*c),x)