\(\int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx\) [283]

Optimal result

Integrand size = 39, antiderivative size = 83 \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\frac {2 g \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \operatorname {EllipticPi}\left (\frac {2 c}{c+d},\frac {1}{2} (e+f x),\frac {2 a}{a+b}\right ) \sqrt {g \sec (e+f x)}}{(c+d) f \sqrt {a+b \sec (e+f x)}} \] Output:

2*g*((b+a*cos(f*x+e))/(a+b))^(1/2)*EllipticPi(sin(1/2*f*x+1/2*e),2*c/(c+d) 
,2^(1/2)*(a/(a+b))^(1/2))*(g*sec(f*x+e))^(1/2)/(c+d)/f/(a+b*sec(f*x+e))^(1 
/2)
 

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\frac {2 g \sqrt {\frac {b+a \cos (e+f x)}{a+b}} \operatorname {EllipticPi}\left (\frac {2 c}{c+d},\frac {1}{2} (e+f x),\frac {2 a}{a+b}\right ) \sqrt {g \sec (e+f x)}}{(c+d) f \sqrt {a+b \sec (e+f x)}} \] Input:

Integrate[(g*Sec[e + f*x])^(3/2)/(Sqrt[a + b*Sec[e + f*x]]*(c + d*Sec[e + 
f*x])),x]
 

Output:

(2*g*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[(2*c)/(c + d), (e + f*x 
)/2, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/((c + d)*f*Sqrt[a + b*Sec[e + f* 
x]])
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3042, 4463, 3042, 3286, 3042, 3284}

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.

\(\displaystyle \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (g \csc \left (e+f x+\frac {\pi }{2}\right )\right )^{3/2}}{\sqrt {a+b \csc \left (e+f x+\frac {\pi }{2}\right )} \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 4463

\(\displaystyle \frac {g \sqrt {g \sec (e+f x)} \sqrt {a \cos (e+f x)+b} \int \frac {1}{\sqrt {b+a \cos (e+f x)} (d+c \cos (e+f x))}dx}{\sqrt {a+b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {g \sqrt {g \sec (e+f x)} \sqrt {a \cos (e+f x)+b} \int \frac {1}{\sqrt {b+a \sin \left (e+f x+\frac {\pi }{2}\right )} \left (d+c \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{\sqrt {a+b \sec (e+f x)}}\)

\(\Big \downarrow \) 3286

\(\displaystyle \frac {g \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (e+f x)}{a+b}} (d+c \cos (e+f x))}dx}{\sqrt {a+b \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {g \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \sin \left (e+f x+\frac {\pi }{2}\right )}{a+b}} \left (d+c \sin \left (e+f x+\frac {\pi }{2}\right )\right )}dx}{\sqrt {a+b \sec (e+f x)}}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 g \sqrt {g \sec (e+f x)} \sqrt {\frac {a \cos (e+f x)+b}{a+b}} \operatorname {EllipticPi}\left (\frac {2 c}{c+d},\frac {1}{2} (e+f x),\frac {2 a}{a+b}\right )}{f (c+d) \sqrt {a+b \sec (e+f x)}}\)

Input:

Int[(g*Sec[e + f*x])^(3/2)/(Sqrt[a + b*Sec[e + f*x]]*(c + d*Sec[e + f*x])) 
,x]
 

Output:

(2*g*Sqrt[(b + a*Cos[e + f*x])/(a + b)]*EllipticPi[(2*c)/(c + d), (e + f*x 
)/2, (2*a)/(a + b)]*Sqrt[g*Sec[e + f*x]])/((c + d)*f*Sqrt[a + b*Sec[e + f* 
x]])
 

Defintions of rubi rules used

rule 3042

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

rule 3284

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]
 

rule 3286

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt 
[c + d*Sin[e + f*x]]   Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(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] &&  !GtQ[c + d, 0]
 

rule 4463

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(3/2)/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))), x_Symbol] :> Simp[g*Sqr 
t[g*Csc[e + f*x]]*(Sqrt[b + a*Sin[e + f*x]]/Sqrt[a + b*Csc[e + f*x]])   Int 
[1/(Sqrt[b + a*Sin[e + f*x]]*(d + c*Sin[e + f*x])), x], x] /; FreeQ[{a, b, 
c, d, e, f, g}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 5.74 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.66

method	result	size
default	\(\frac {2 i g \left (c \operatorname {EllipticF}\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), \sqrt {-\frac {a -b}{a +b}}\right )+d \operatorname {EllipticF}\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), \sqrt {-\frac {a -b}{a +b}}\right )-2 c \operatorname {EllipticPi}\left (i \left (-\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ), -\frac {c -d}{c +d}, i \sqrt {\frac {a -b}{a +b}}\right )\right ) \sqrt {g \sec \left (f x +e \right )}\, \sqrt {a +b \sec \left (f x +e \right )}\, \sqrt {\frac {b +a \cos \left (f x +e \right )}{\left (a +b \right ) \left (\cos \left (f x +e \right )+1\right )}}\, \cos \left (f x +e \right )}{f \left (c -d \right ) \left (c +d \right ) \left (b +a \cos \left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}}\)	\(221\)

Input:

int((g*sec(f*x+e))^(3/2)/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x,method= 
_RETURNVERBOSE)
 

Output:

2*I*g/f/(c-d)/(c+d)*(c*EllipticF(I*(-cot(f*x+e)+csc(f*x+e)),(-(a-b)/(a+b)) 
^(1/2))+d*EllipticF(I*(-cot(f*x+e)+csc(f*x+e)),(-(a-b)/(a+b))^(1/2))-2*c*E 
llipticPi(I*(-cot(f*x+e)+csc(f*x+e)),-(c-d)/(c+d),I*((a-b)/(a+b))^(1/2)))* 
(g*sec(f*x+e))^(1/2)*(a+b*sec(f*x+e))^(1/2)*(1/(a+b)*(b+a*cos(f*x+e))/(cos 
(f*x+e)+1))^(1/2)*cos(f*x+e)/(b+a*cos(f*x+e))/(1/(cos(f*x+e)+1))^(1/2)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\text {Timed out} \] Input:

integrate((g*sec(f*x+e))^(3/2)/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, 
algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {\left (g \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\sqrt {a + b \sec {\left (e + f x \right )}} \left (c + d \sec {\left (e + f x \right )}\right )}\, dx \] Input:

integrate((g*sec(f*x+e))**(3/2)/(a+b*sec(f*x+e))**(1/2)/(c+d*sec(f*x+e)),x 
)
 

Output:

Integral((g*sec(e + f*x))**(3/2)/(sqrt(a + b*sec(e + f*x))*(c + d*sec(e + 
f*x))), x)
 

Maxima [F]

\[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int { \frac {\left (g \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {b \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}} \,d x } \] Input:

integrate((g*sec(f*x+e))^(3/2)/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, 
algorithm="maxima")
 

Output:

integrate((g*sec(f*x + e))^(3/2)/(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) 
 + c)), x)
 

Giac [F]

\[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int { \frac {\left (g \sec \left (f x + e\right )\right )^{\frac {3}{2}}}{\sqrt {b \sec \left (f x + e\right ) + a} {\left (d \sec \left (f x + e\right ) + c\right )}} \,d x } \] Input:

integrate((g*sec(f*x+e))^(3/2)/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x, 
algorithm="giac")
 

Output:

integrate((g*sec(f*x + e))^(3/2)/(sqrt(b*sec(f*x + e) + a)*(d*sec(f*x + e) 
 + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\int \frac {{\left (\frac {g}{\cos \left (e+f\,x\right )}\right )}^{3/2}}{\sqrt {a+\frac {b}{\cos \left (e+f\,x\right )}}\,\left (c+\frac {d}{\cos \left (e+f\,x\right )}\right )} \,d x \] Input:

int((g/cos(e + f*x))^(3/2)/((a + b/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x) 
)),x)
 

Output:

int((g/cos(e + f*x))^(3/2)/((a + b/cos(e + f*x))^(1/2)*(c + d/cos(e + f*x) 
)), x)
 

Reduce [F]

\[ \int \frac {(g \sec (e+f x))^{3/2}}{\sqrt {a+b \sec (e+f x)} (c+d \sec (e+f x))} \, dx=\sqrt {g}\, \left (\int \frac {\sqrt {\sec \left (f x +e \right )}\, \sqrt {\sec \left (f x +e \right ) b +a}\, \sec \left (f x +e \right )}{\sec \left (f x +e \right )^{2} b d +\sec \left (f x +e \right ) a d +\sec \left (f x +e \right ) b c +a c}d x \right ) g \] Input:

int((g*sec(f*x+e))^(3/2)/(a+b*sec(f*x+e))^(1/2)/(c+d*sec(f*x+e)),x)
 

Output:

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