\(\int \frac {a+b x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx\) [47]

Optimal result

Integrand size = 30, antiderivative size = 191 \[ \int \frac {a+b x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx=\frac {b x \sqrt {3+f x^2}}{f \sqrt {2+d x^2}}-\frac {\sqrt {3} b \sqrt {3+f x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|1-\frac {2 f}{3 d}\right )}{\sqrt {d} f \sqrt {2+d x^2} \sqrt {\frac {3+f x^2}{2+d x^2}}}+\frac {a \sqrt {3+f x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {2}}\right ),1-\frac {2 f}{3 d}\right )}{\sqrt {3} \sqrt {d} \sqrt {2+d x^2} \sqrt {\frac {3+f x^2}{2+d x^2}}} \] Output:

b*x*(f*x^2+3)^(1/2)/f/(d*x^2+2)^(1/2)-3^(1/2)*b*(f*x^2+3)^(1/2)*EllipticE( 
d^(1/2)*x*2^(1/2)/(2*d*x^2+4)^(1/2),1/3*(9-6*f/d)^(1/2))/d^(1/2)/f/(d*x^2+ 
2)^(1/2)/((f*x^2+3)/(d*x^2+2))^(1/2)+1/3*a*(f*x^2+3)^(1/2)*InverseJacobiAM 
(arctan(1/2*d^(1/2)*x*2^(1/2)),1/3*(9-6*f/d)^(1/2))*3^(1/2)/d^(1/2)/(d*x^2 
+2)^(1/2)/((f*x^2+3)/(d*x^2+2))^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 3.41 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.42 \[ \int \frac {a+b x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx=-\frac {i \left (3 b E\left (i \text {arcsinh}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right )|\frac {2 f}{3 d}\right )+(-3 b+a f) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {d} x}{\sqrt {2}}\right ),\frac {2 f}{3 d}\right )\right )}{\sqrt {3} \sqrt {d} f} \] Input:

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

Output:

((-I)*(3*b*EllipticE[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(3*d)] + (-3*b 
+ a*f)*EllipticF[I*ArcSinh[(Sqrt[d]*x)/Sqrt[2]], (2*f)/(3*d)]))/(Sqrt[3]*S 
qrt[d]*f)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {406, 320, 388, 313}

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.

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

\(\Big \downarrow \) 406

\(\displaystyle a \int \frac {1}{\sqrt {d x^2+2} \sqrt {f x^2+3}}dx+b \int \frac {x^2}{\sqrt {d x^2+2} \sqrt {f x^2+3}}dx\)

\(\Big \downarrow \) 320

\(\displaystyle b \int \frac {x^2}{\sqrt {d x^2+2} \sqrt {f x^2+3}}dx+\frac {a \sqrt {d x^2+2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {f} x}{\sqrt {3}}\right ),1-\frac {3 d}{2 f}\right )}{\sqrt {2} \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}\)

\(\Big \downarrow \) 388

\(\displaystyle b \left (\frac {x \sqrt {d x^2+2}}{d \sqrt {f x^2+3}}-\frac {3 \int \frac {\sqrt {d x^2+2}}{\left (f x^2+3\right )^{3/2}}dx}{d}\right )+\frac {a \sqrt {d x^2+2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {f} x}{\sqrt {3}}\right ),1-\frac {3 d}{2 f}\right )}{\sqrt {2} \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}\)

\(\Big \downarrow \) 313

\(\displaystyle \frac {a \sqrt {d x^2+2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {f} x}{\sqrt {3}}\right ),1-\frac {3 d}{2 f}\right )}{\sqrt {2} \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}+b \left (\frac {x \sqrt {d x^2+2}}{d \sqrt {f x^2+3}}-\frac {\sqrt {2} \sqrt {d x^2+2} E\left (\arctan \left (\frac {\sqrt {f} x}{\sqrt {3}}\right )|1-\frac {3 d}{2 f}\right )}{d \sqrt {f} \sqrt {f x^2+3} \sqrt {\frac {d x^2+2}{f x^2+3}}}\right )\)

Input:

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

Output:

b*((x*Sqrt[2 + d*x^2])/(d*Sqrt[3 + f*x^2]) - (Sqrt[2]*Sqrt[2 + d*x^2]*Elli 
pticE[ArcTan[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(d*Sqrt[f]*Sqrt[(2 + 
d*x^2)/(3 + f*x^2)]*Sqrt[3 + f*x^2])) + (a*Sqrt[2 + d*x^2]*EllipticF[ArcTa 
n[(Sqrt[f]*x)/Sqrt[3]], 1 - (3*d)/(2*f)])/(Sqrt[2]*Sqrt[f]*Sqrt[(2 + d*x^2 
)/(3 + f*x^2)]*Sqrt[3 + f*x^2])
 

Defintions of rubi rules used

rule 313

Int[Sqrt[(a_) + (b_.)*(x_)^2]/((c_) + (d_.)*(x_)^2)^(3/2), x_Symbol] :> Sim 
p[(Sqrt[a + b*x^2]/(c*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*(c 
+ d*x^2)))]))*EllipticE[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; FreeQ 
[{a, b, c, d}, x] && PosQ[b/a] && PosQ[d/c]
 

rule 320

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S 
imp[(Sqrt[a + b*x^2]/(a*Rt[d/c, 2]*Sqrt[c + d*x^2]*Sqrt[c*((a + b*x^2)/(a*( 
c + d*x^2)))]))*EllipticF[ArcTan[Rt[d/c, 2]*x], 1 - b*(c/(a*d))], x] /; Fre 
eQ[{a, b, c, d}, x] && PosQ[d/c] && PosQ[b/a] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 388

Int[(x_)^2/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] 
 :> Simp[x*(Sqrt[a + b*x^2]/(b*Sqrt[c + d*x^2])), x] - Simp[c/b   Int[Sqrt[ 
a + b*x^2]/(c + d*x^2)^(3/2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && PosQ[b/a] && PosQ[d/c] &&  !SimplerSqrtQ[b/a, d/c]
 

rule 406

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[e   Int[(a + b*x^2)^p*(c + d*x^2)^q, x], x] + Sim 
p[f   Int[x^2*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, 
f, p, q}, x]
 

Maple [A] (verified)

Time = 4.23 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.55

method	result	size
default	\(\frac {\left (\operatorname {EllipticF}\left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right ) a d -2 \operatorname {EllipticF}\left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right ) b +2 \operatorname {EllipticE}\left (\frac {x \sqrt {3}\, \sqrt {-f}}{3}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {\frac {d}{f}}}{2}\right ) b \right ) \sqrt {2}}{2 d \sqrt {-f}}\)	\(105\)
elliptic	\(\frac {\sqrt {\left (f \,x^{2}+3\right ) \left (x^{2} d +2\right )}\, \left (\frac {a \sqrt {3 f \,x^{2}+9}\, \sqrt {2 x^{2} d +4}\, \operatorname {EllipticF}\left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )}{2 \sqrt {-3 f}\, \sqrt {d f \,x^{4}+3 x^{2} d +2 f \,x^{2}+6}}-\frac {b \sqrt {3 f \,x^{2}+9}\, \sqrt {2 x^{2} d +4}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {-3 f}}{3}, \frac {\sqrt {-4+\frac {6 d +4 f}{f}}}{2}\right )\right )}{\sqrt {-3 f}\, \sqrt {d f \,x^{4}+3 x^{2} d +2 f \,x^{2}+6}\, d}\right )}{\sqrt {f \,x^{2}+3}\, \sqrt {x^{2} d +2}}\)	\(225\)

Input:

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

Output:

1/2*(EllipticF(1/3*x*3^(1/2)*(-f)^(1/2),1/2*3^(1/2)*2^(1/2)*(1/f*d)^(1/2)) 
*a*d-2*EllipticF(1/3*x*3^(1/2)*(-f)^(1/2),1/2*3^(1/2)*2^(1/2)*(1/f*d)^(1/2 
))*b+2*EllipticE(1/3*x*3^(1/2)*(-f)^(1/2),1/2*3^(1/2)*2^(1/2)*(1/f*d)^(1/2 
))*b)*2^(1/2)/d/(-f)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.65 \[ \int \frac {a+b x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx=-\frac {9 \, \sqrt {3} \sqrt {d f} b x \sqrt {-\frac {1}{f}} E(\arcsin \left (\frac {\sqrt {3} \sqrt {-\frac {1}{f}}}{x}\right )\,|\,\frac {2 \, f}{3 \, d}) - \sqrt {3} {\left (a f^{2} + 9 \, b\right )} \sqrt {d f} x \sqrt {-\frac {1}{f}} F(\arcsin \left (\frac {\sqrt {3} \sqrt {-\frac {1}{f}}}{x}\right )\,|\,\frac {2 \, f}{3 \, d}) - 3 \, \sqrt {d x^{2} + 2} \sqrt {f x^{2} + 3} b f}{3 \, d f^{2} x} \] Input:

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

Output:

-1/3*(9*sqrt(3)*sqrt(d*f)*b*x*sqrt(-1/f)*elliptic_e(arcsin(sqrt(3)*sqrt(-1 
/f)/x), 2/3*f/d) - sqrt(3)*(a*f^2 + 9*b)*sqrt(d*f)*x*sqrt(-1/f)*elliptic_f 
(arcsin(sqrt(3)*sqrt(-1/f)/x), 2/3*f/d) - 3*sqrt(d*x^2 + 2)*sqrt(f*x^2 + 3 
)*b*f)/(d*f^2*x)
 

Sympy [F]

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

integrate((b*x**2+a)/(d*x**2+2)**(1/2)/(f*x**2+3)**(1/2),x)
 

Output:

Integral((a + b*x**2)/(sqrt(d*x**2 + 2)*sqrt(f*x**2 + 3)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx=\int \frac {b\,x^2+a}{\sqrt {d\,x^2+2}\,\sqrt {f\,x^2+3}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {a+b x^2}{\sqrt {2+d x^2} \sqrt {3+f x^2}} \, dx=\left (\int \frac {\sqrt {f \,x^{2}+3}\, \sqrt {d \,x^{2}+2}\, x^{2}}{d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}d x \right ) b +\left (\int \frac {\sqrt {f \,x^{2}+3}\, \sqrt {d \,x^{2}+2}}{d f \,x^{4}+3 d \,x^{2}+2 f \,x^{2}+6}d x \right ) a \] Input:

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

Output:

int((sqrt(f*x**2 + 3)*sqrt(d*x**2 + 2)*x**2)/(d*f*x**4 + 3*d*x**2 + 2*f*x* 
*2 + 6),x)*b + int((sqrt(f*x**2 + 3)*sqrt(d*x**2 + 2))/(d*f*x**4 + 3*d*x** 
2 + 2*f*x**2 + 6),x)*a