\(\int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx\) [136]

Optimal result

Integrand size = 21, antiderivative size = 193 \[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\frac {2 \sqrt {\sqrt {c} f+\sqrt {-a} g} \sqrt {1-\frac {\sqrt {c} (f+g x)}{\sqrt {c} f-\sqrt {-a} g}} \sqrt {1-\frac {\sqrt {c} (f+g x)}{\sqrt {c} f+\sqrt {-a} g}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{c} \sqrt {f+g x}}{\sqrt {\sqrt {c} f+\sqrt {-a} g}}\right ),\frac {\sqrt {c} f+\sqrt {-a} g}{\sqrt {c} f-\sqrt {-a} g}\right )}{\sqrt [4]{c} g \sqrt {a+c x^2}} \] Output:

2*(c^(1/2)*f+(-a)^(1/2)*g)^(1/2)*(1-c^(1/2)*(g*x+f)/(c^(1/2)*f-(-a)^(1/2)* 
g))^(1/2)*(1-c^(1/2)*(g*x+f)/(c^(1/2)*f+(-a)^(1/2)*g))^(1/2)*EllipticF(c^( 
1/4)*(g*x+f)^(1/2)/(c^(1/2)*f+(-a)^(1/2)*g)^(1/2),((c^(1/2)*f+(-a)^(1/2)*g 
)/(c^(1/2)*f-(-a)^(1/2)*g))^(1/2))/c^(1/4)/g/(c*x^2+a)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 21.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\frac {2 i \sqrt {\frac {g \left (\frac {i \sqrt {a}}{\sqrt {c}}+x\right )}{f+g x}} \sqrt {-\frac {\frac {i \sqrt {a} g}{\sqrt {c}}-g x}{f+g x}} (f+g x) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}}}{\sqrt {f+g x}}\right ),\frac {\sqrt {c} f-i \sqrt {a} g}{\sqrt {c} f+i \sqrt {a} g}\right )}{g \sqrt {-f-\frac {i \sqrt {a} g}{\sqrt {c}}} \sqrt {a+c x^2}} \] Input:

Integrate[1/(Sqrt[f + g*x]*Sqrt[a + c*x^2]),x]
 

Output:

((2*I)*Sqrt[(g*((I*Sqrt[a])/Sqrt[c] + x))/(f + g*x)]*Sqrt[-(((I*Sqrt[a]*g) 
/Sqrt[c] - g*x)/(f + g*x))]*(f + g*x)*EllipticF[I*ArcSinh[Sqrt[-f - (I*Sqr 
t[a]*g)/Sqrt[c]]/Sqrt[f + g*x]], (Sqrt[c]*f - I*Sqrt[a]*g)/(Sqrt[c]*f + I* 
Sqrt[a]*g)])/(g*Sqrt[-f - (I*Sqrt[a]*g)/Sqrt[c]]*Sqrt[a + c*x^2])
 

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.21, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {510, 1416}

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.

\(\displaystyle \int \frac {1}{\sqrt {a+c x^2} \sqrt {f+g x}} \, dx\)

\(\Big \downarrow \) 510

\(\displaystyle \frac {2 \int \frac {1}{\sqrt {\frac {c f^2}{g^2}-\frac {2 c (f+g x) f}{g^2}+\frac {c (f+g x)^2}{g^2}+a}}d\sqrt {f+g x}}{g}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {\sqrt [4]{a g^2+c f^2} \left (\frac {\sqrt {c} (f+g x)}{\sqrt {a g^2+c f^2}}+1\right ) \sqrt {\frac {a+\frac {c f^2}{g^2}-\frac {2 c f (f+g x)}{g^2}+\frac {c (f+g x)^2}{g^2}}{\left (a+\frac {c f^2}{g^2}\right ) \left (\frac {\sqrt {c} (f+g x)}{\sqrt {a g^2+c f^2}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {f+g x}}{\sqrt [4]{c f^2+a g^2}}\right ),\frac {1}{2} \left (\frac {\sqrt {c} f}{\sqrt {c f^2+a g^2}}+1\right )\right )}{\sqrt [4]{c} g \sqrt {a+\frac {c f^2}{g^2}-\frac {2 c f (f+g x)}{g^2}+\frac {c (f+g x)^2}{g^2}}}\)

Input:

Int[1/(Sqrt[f + g*x]*Sqrt[a + c*x^2]),x]
 

Output:

((c*f^2 + a*g^2)^(1/4)*(1 + (Sqrt[c]*(f + g*x))/Sqrt[c*f^2 + a*g^2])*Sqrt[ 
(a + (c*f^2)/g^2 - (2*c*f*(f + g*x))/g^2 + (c*(f + g*x)^2)/g^2)/((a + (c*f 
^2)/g^2)*(1 + (Sqrt[c]*(f + g*x))/Sqrt[c*f^2 + a*g^2])^2)]*EllipticF[2*Arc 
Tan[(c^(1/4)*Sqrt[f + g*x])/(c*f^2 + a*g^2)^(1/4)], (1 + (Sqrt[c]*f)/Sqrt[ 
c*f^2 + a*g^2])/2])/(c^(1/4)*g*Sqrt[a + (c*f^2)/g^2 - (2*c*f*(f + g*x))/g^ 
2 + (c*(f + g*x)^2)/g^2])
 

Defintions of rubi rules used

rule 510

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2/d   Subst[Int[1/Sqrt[(b*c^2 + a*d^2)/d^2 - 2*b*c*(x^2/d^2) + b*(x^4/d^2 
)], x], x, Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && PosQ[b/a]
 

rule 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

Maple [A] (verified)

Time = 2.23 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.04

method	result	size
default	\(\frac {2 \left (c f -\sqrt {-a c}\, g \right ) \operatorname {EllipticF}\left (\sqrt {-\frac {\left (g x +f \right ) c}{\sqrt {-a c}\, g -c f}}, \sqrt {-\frac {\sqrt {-a c}\, g -c f}{\sqrt {-a c}\, g +c f}}\right ) \sqrt {\frac {\left (c x +\sqrt {-a c}\right ) g}{\sqrt {-a c}\, g -c f}}\, \sqrt {\frac {\left (-c x +\sqrt {-a c}\right ) g}{\sqrt {-a c}\, g +c f}}\, \sqrt {-\frac {\left (g x +f \right ) c}{\sqrt {-a c}\, g -c f}}\, \sqrt {c \,x^{2}+a}\, \sqrt {g x +f}}{c g \left (c g \,x^{3}+c f \,x^{2}+a g x +a f \right )}\)	\(200\)
elliptic	\(\frac {2 \sqrt {\left (g x +f \right ) \left (c \,x^{2}+a \right )}\, \left (\frac {f}{g}-\frac {\sqrt {-a c}}{c}\right ) \sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\, \sqrt {\frac {x -\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\, \sqrt {\frac {x +\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}+\frac {\sqrt {-a c}}{c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {x +\frac {f}{g}}{\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}, \sqrt {\frac {-\frac {f}{g}+\frac {\sqrt {-a c}}{c}}{-\frac {f}{g}-\frac {\sqrt {-a c}}{c}}}\right )}{\sqrt {g x +f}\, \sqrt {c \,x^{2}+a}\, \sqrt {c g \,x^{3}+c f \,x^{2}+a g x +a f}}\)	\(242\)

Input:

int(1/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

2*(c*f-(-a*c)^(1/2)*g)*EllipticF((-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2),( 
-((-a*c)^(1/2)*g-c*f)/((-a*c)^(1/2)*g+c*f))^(1/2))*((c*x+(-a*c)^(1/2))*g/( 
(-a*c)^(1/2)*g-c*f))^(1/2)*((-c*x+(-a*c)^(1/2))*g/((-a*c)^(1/2)*g+c*f))^(1 
/2)*(-(g*x+f)*c/((-a*c)^(1/2)*g-c*f))^(1/2)*(c*x^2+a)^(1/2)*(g*x+f)^(1/2)/ 
c/g/(c*g*x^3+c*f*x^2+a*g*x+a*f)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.34 \[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\frac {2 \, \sqrt {c g} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (c f^{2} - 3 \, a g^{2}\right )}}{3 \, c g^{2}}, -\frac {8 \, {\left (c f^{3} + 9 \, a f g^{2}\right )}}{27 \, c g^{3}}, \frac {3 \, g x + f}{3 \, g}\right )}{c g} \] Input:

integrate(1/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

2*sqrt(c*g)*weierstrassPInverse(4/3*(c*f^2 - 3*a*g^2)/(c*g^2), -8/27*(c*f^ 
3 + 9*a*f*g^2)/(c*g^3), 1/3*(3*g*x + f)/g)/(c*g)
 

Sympy [F]

\[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int \frac {1}{\sqrt {a + c x^{2}} \sqrt {f + g x}}\, dx \] Input:

integrate(1/(g*x+f)**(1/2)/(c*x**2+a)**(1/2),x)
 

Output:

Integral(1/(sqrt(a + c*x**2)*sqrt(f + g*x)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + a} \sqrt {g x + f}} \,d x } \] Input:

integrate(1/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate(1/(sqrt(c*x^2 + a)*sqrt(g*x + f)), x)
 

Giac [F]

\[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int { \frac {1}{\sqrt {c x^{2} + a} \sqrt {g x + f}} \,d x } \] Input:

integrate(1/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

integrate(1/(sqrt(c*x^2 + a)*sqrt(g*x + f)), x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int \frac {1}{\sqrt {f+g\,x}\,\sqrt {c\,x^2+a}} \,d x \] Input:

int(1/((f + g*x)^(1/2)*(a + c*x^2)^(1/2)),x)
 

Output:

int(1/((f + g*x)^(1/2)*(a + c*x^2)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\sqrt {f+g x} \sqrt {a+c x^2}} \, dx=\int \frac {\sqrt {g x +f}\, \sqrt {c \,x^{2}+a}}{c g \,x^{3}+c f \,x^{2}+a g x +a f}d x \] Input:

int(1/(g*x+f)^(1/2)/(c*x^2+a)^(1/2),x)
 

Output:

int((sqrt(f + g*x)*sqrt(a + c*x**2))/(a*f + a*g*x + c*f*x**2 + c*g*x**3),x 
)