\(\int \frac {(g \sec (e+f x))^{3/2}}{(a+b \sec (e+f x)) \sqrt {c+d \sec (e+f x)}} \, dx\) [271]

Optimal result

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

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

Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00 \[ \int \frac {(g \sec (e+f x))^{3/2}}{(a+b \sec (e+f x)) \sqrt {c+d \sec (e+f x)}} \, dx=\frac {2 g \sqrt {\frac {d+c \cos (e+f x)}{c+d}} \operatorname {EllipticPi}\left (\frac {2 a}{a+b},\frac {1}{2} (e+f x),\frac {2 c}{c+d}\right ) \sqrt {g \sec (e+f x)}}{(a+b) f \sqrt {c+d \sec (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.73 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4463

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3286

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

(2*g*Sqrt[(d + c*Cos[e + f*x])/(c + d)]*EllipticPi[(2*a)/(a + b), (e + f*x 
)/2, (2*c)/(c + d)]*Sqrt[g*Sec[e + f*x]])/((a + b)*f*Sqrt[c + d*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 = 5.78 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.69

method	result	size
default	\(\frac {2 i g \left (2 a \operatorname {EllipticPi}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), -\frac {a -b}{a +b}, i \sqrt {\frac {c -d}{c +d}}\right )-a \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), \sqrt {-\frac {c -d}{c +d}}\right )-b \operatorname {EllipticF}\left (i \left (\cot \left (f x +e \right )-\csc \left (f x +e \right )\right ), \sqrt {-\frac {c -d}{c +d}}\right )\right ) \sqrt {g \sec \left (f x +e \right )}\, \sqrt {c +d \sec \left (f x +e \right )}\, \sqrt {\frac {d +c \cos \left (f x +e \right )}{\left (c +d \right ) \left (\cos \left (f x +e \right )+1\right )}}\, \cos \left (f x +e \right )}{f \left (a -b \right ) \left (a +b \right ) \left (d +c \cos \left (f x +e \right )\right ) \sqrt {\frac {1}{\cos \left (f x +e \right )+1}}}\)	\(223\)

Input:

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

Output:

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

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(g)*int((sqrt(sec(e + f*x))*sqrt(sec(e + f*x)*d + c)*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