\(\int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2} (e+f x^2)} \, dx\) [115]

Optimal result
Mathematica [C] (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 = 32, antiderivative size = 212 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )} \, dx=\frac {\sqrt {a} \sqrt {b} \sqrt {c+d x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{c (b e-a f) \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}}-\frac {a^{3/2} f \sqrt {c+d x^2} \operatorname {EllipticPi}\left (1-\frac {a f}{b e},\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),1-\frac {a d}{b c}\right )}{\sqrt {b} c e (b e-a f) \sqrt {a+b x^2} \sqrt {\frac {a \left (c+d x^2\right )}{c \left (a+b x^2\right )}}} \] Output:

a^(1/2)*b^(1/2)*(d*x^2+c)^(1/2)*InverseJacobiAM(arctan(b^(1/2)*x/a^(1/2)), 
(1-a*d/b/c)^(1/2))/c/(-a*f+b*e)/(b*x^2+a)^(1/2)/(a*(d*x^2+c)/c/(b*x^2+a))^ 
(1/2)-a^(3/2)*f*(d*x^2+c)^(1/2)*EllipticPi(b^(1/2)*x/a^(1/2)/(1+b*x^2/a)^( 
1/2),1-a*f/b/e,(1-a*d/b/c)^(1/2))/b^(1/2)/c/e/(-a*f+b*e)/(b*x^2+a)^(1/2)/( 
a*(d*x^2+c)/c/(b*x^2+a))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.70 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.48 \[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )} \, dx=-\frac {i \sqrt {1+\frac {b x^2}{a}} \sqrt {1+\frac {d x^2}{c}} \operatorname {EllipticPi}\left (\frac {a f}{b e},i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )}{\sqrt {\frac {b}{a}} e \sqrt {a+b x^2} \sqrt {c+d x^2}} \] Input:

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

Output:

((-I)*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticPi[(a*f)/(b*e), I*Ar 
cSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(Sqrt[b/a]*e*Sqrt[a + b*x^2]*Sqrt[c + d* 
x^2])
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.47, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {413, 413, 412}

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.

\(\displaystyle \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )} \, dx\)

\(\Big \downarrow \) 413

\(\displaystyle \frac {\sqrt {\frac {b x^2}{a}+1} \int \frac {1}{\sqrt {\frac {b x^2}{a}+1} \sqrt {d x^2+c} \left (f x^2+e\right )}dx}{\sqrt {a+b x^2}}\)

\(\Big \downarrow \) 413

\(\displaystyle \frac {\sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \int \frac {1}{\sqrt {\frac {b x^2}{a}+1} \sqrt {\frac {d x^2}{c}+1} \left (f x^2+e\right )}dx}{\sqrt {a+b x^2} \sqrt {c+d x^2}}\)

\(\Big \downarrow \) 412

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

Input:

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

Output:

(Sqrt[-a]*Sqrt[1 + (b*x^2)/a]*Sqrt[1 + (d*x^2)/c]*EllipticPi[(a*f)/(b*e), 
ArcSin[(Sqrt[b]*x)/Sqrt[-a]], (a*d)/(b*c)])/(Sqrt[b]*e*Sqrt[a + b*x^2]*Sqr 
t[c + d*x^2])
 

Defintions of rubi rules used

rule 412
Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x 
_)^2]), x_Symbol] :> Simp[(1/(a*Sqrt[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b* 
(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c, d, e, 
 f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && S 
implerSqrtQ[-f/e, -d/c])
 

rule 413
Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x 
_)^2]), x_Symbol] :> Simp[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*x^2]   Int[1/((a + 
 b*x^2)*Sqrt[1 + (d/c)*x^2]*Sqrt[e + f*x^2]), x], x] /; FreeQ[{a, b, c, d, 
e, f}, x] &&  !GtQ[c, 0]
 
Maple [A] (verified)

Time = 5.89 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.56

method result size
default \(\frac {\operatorname {EllipticPi}\left (x \sqrt {-\frac {b}{a}}, \frac {a f}{b e}, \frac {\sqrt {-\frac {d}{c}}}{\sqrt {-\frac {b}{a}}}\right ) \sqrt {\frac {x^{2} d +c}{c}}\, \sqrt {\frac {b \,x^{2}+a}{a}}\, \sqrt {x^{2} d +c}\, \sqrt {b \,x^{2}+a}}{e \sqrt {-\frac {b}{a}}\, \left (b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c \right )}\) \(118\)
elliptic \(\frac {\sqrt {\left (b \,x^{2}+a \right ) \left (x^{2} d +c \right )}\, \sqrt {1+\frac {b \,x^{2}}{a}}\, \sqrt {1+\frac {d \,x^{2}}{c}}\, \operatorname {EllipticPi}\left (x \sqrt {-\frac {b}{a}}, \frac {a f}{b e}, \frac {\sqrt {-\frac {d}{c}}}{\sqrt {-\frac {b}{a}}}\right )}{\sqrt {b \,x^{2}+a}\, \sqrt {x^{2} d +c}\, e \sqrt {-\frac {b}{a}}\, \sqrt {b d \,x^{4}+a d \,x^{2}+x^{2} b c +a c}}\) \(133\)

Input:

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

Output:

EllipticPi(x*(-b/a)^(1/2),a*f/b/e,(-1/c*d)^(1/2)/(-b/a)^(1/2))*((d*x^2+c)/ 
c)^(1/2)*((b*x^2+a)/a)^(1/2)*(d*x^2+c)^(1/2)*(b*x^2+a)^(1/2)/e/(-b/a)^(1/2 
)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
 

Fricas [F(-1)]

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(a + b*x**2)*sqrt(c + d*x**2)*(e + f*x**2)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {a+b x^2} \sqrt {c+d x^2} \left (e+f x^2\right )} \, dx=\int \frac {\sqrt {d \,x^{2}+c}\, \sqrt {b \,x^{2}+a}}{b d f \,x^{6}+a d f \,x^{4}+b c f \,x^{4}+b d e \,x^{4}+a c f \,x^{2}+a d e \,x^{2}+b c e \,x^{2}+a c e}d x \] Input:

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

Output:

int((sqrt(c + d*x**2)*sqrt(a + b*x**2))/(a*c*e + a*c*f*x**2 + a*d*e*x**2 + 
 a*d*f*x**4 + b*c*e*x**2 + b*c*f*x**4 + b*d*e*x**4 + b*d*f*x**6),x)