3.10.64 \(\int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx\) [964]

3.10.64.1 Optimal result
3.10.64.2 Mathematica [C] (verified)
3.10.64.3 Rubi [A] (verified)
3.10.64.4 Maple [A] (verified)
3.10.64.5 Fricas [A] (verification not implemented)
3.10.64.6 Sympy [F]
3.10.64.7 Maxima [F]
3.10.64.8 Giac [F]
3.10.64.9 Mupad [F(-1)]

3.10.64.1 Optimal result

Integrand size = 20, antiderivative size = 267 \[ \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {x \sqrt {a+b x^2+c x^4}}{\sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{c^{3/4} \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 c^{3/4} \sqrt {a+b x^2+c x^4}} \]

output
x*(c*x^4+b*x^2+a)^(1/2)/c^(1/2)/(a^(1/2)+x^2*c^(1/2))-a^(1/4)*(cos(2*arcta 
n(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticE( 
sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2) 
+x^2*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(3/4)/(c*x 
^4+b*x^2+a)^(1/2)+1/2*a^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/c 
os(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^(1/4))) 
,1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+b*x^2+a)/( 
a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(3/4)/(c*x^4+b*x^2+a)^(1/2)
 
3.10.64.2 Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.09 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {i \left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {1+\frac {2 c x^2}{b-\sqrt {b^2-4 a c}}} \left (E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )\right )}{2 \sqrt {2} c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {a+b x^2+c x^4}} \]

input
Integrate[x^2/Sqrt[a + b*x^2 + c*x^4],x]
 
output
((I/2)*(-b + Sqrt[b^2 - 4*a*c])*Sqrt[(b + Sqrt[b^2 - 4*a*c] + 2*c*x^2)/(b 
+ Sqrt[b^2 - 4*a*c])]*Sqrt[1 + (2*c*x^2)/(b - Sqrt[b^2 - 4*a*c])]*(Ellipti 
cE[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4 
*a*c])/(b - Sqrt[b^2 - 4*a*c])] - EllipticF[I*ArcSinh[Sqrt[2]*Sqrt[c/(b + 
Sqrt[b^2 - 4*a*c])]*x], (b + Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c])])) 
/(Sqrt[2]*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4])
 
3.10.64.3 Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.01, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1459, 27, 1416, 1509}

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.

\(\displaystyle \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx\)

\(\Big \downarrow \) 1459

\(\displaystyle \frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {a} \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 c^{3/4} \sqrt {a+b x^2+c x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 c^{3/4} \sqrt {a+b x^2+c x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4}}{\sqrt {a}+\sqrt {c} x^2}}{\sqrt {c}}\)

input
Int[x^2/Sqrt[a + b*x^2 + c*x^4],x]
 
output
-((-((x*Sqrt[a + b*x^2 + c*x^4])/(Sqrt[a] + Sqrt[c]*x^2)) + (a^(1/4)*(Sqrt 
[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*Ell 
ipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(c^(1/ 
4)*Sqrt[a + b*x^2 + c*x^4]))/Sqrt[c]) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*S 
qrt[(a + b*x^2 + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*ArcTan[(c^( 
1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(2*c^(3/4)*Sqrt[a + b*x^2 
+ c*x^4])
 

3.10.64.3.1 Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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]
 

rule 1459
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = 
 Rt[c/a, 2]}, Simp[1/q   Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Simp[1/q 
 Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x]] /; FreeQ[{a, b, c}, x] && 
 NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 
3.10.64.4 Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.81

method result size
default \(-\frac {a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (F\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-E\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\) \(216\)
elliptic \(-\frac {a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (F\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-E\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\) \(216\)

input
int(x^2/(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 
output
-1/2*a*2^(1/2)/((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2)*(4-2*(-b+(-4*a*c+b^2)^(1/ 
2))/a*x^2)^(1/2)*(4+2*(b+(-4*a*c+b^2)^(1/2))/a*x^2)^(1/2)/(c*x^4+b*x^2+a)^ 
(1/2)/(b+(-4*a*c+b^2)^(1/2))*(EllipticF(1/2*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1 
/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^(1/2))/a/c)^(1/2))-EllipticE(1/2 
*x*2^(1/2)*((-b+(-4*a*c+b^2)^(1/2))/a)^(1/2),1/2*(-4+2*b*(b+(-4*a*c+b^2)^( 
1/2))/a/c)^(1/2)))
 
3.10.64.5 Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.02 \[ \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\frac {\sqrt {\frac {1}{2}} {\left (c x \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b x\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} E(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - \sqrt {\frac {1}{2}} {\left (c x \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b x\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} F(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) + 2 \, \sqrt {c x^{4} + b x^{2} + a} c}{2 \, c^{2} x} \]

input
integrate(x^2/(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
 
output
1/2*(sqrt(1/2)*(c*x*sqrt((b^2 - 4*a*c)/c^2) - b*x)*sqrt(c)*sqrt((c*sqrt((b 
^2 - 4*a*c)/c^2) - b)/c)*elliptic_e(arcsin(sqrt(1/2)*sqrt((c*sqrt((b^2 - 4 
*a*c)/c^2) - b)/c)/x), 1/2*(b*c*sqrt((b^2 - 4*a*c)/c^2) + b^2 - 2*a*c)/(a* 
c)) - sqrt(1/2)*(c*x*sqrt((b^2 - 4*a*c)/c^2) - b*x)*sqrt(c)*sqrt((c*sqrt(( 
b^2 - 4*a*c)/c^2) - b)/c)*elliptic_f(arcsin(sqrt(1/2)*sqrt((c*sqrt((b^2 - 
4*a*c)/c^2) - b)/c)/x), 1/2*(b*c*sqrt((b^2 - 4*a*c)/c^2) + b^2 - 2*a*c)/(a 
*c)) + 2*sqrt(c*x^4 + b*x^2 + a)*c)/(c^2*x)
 
3.10.64.6 Sympy [F]

\[ \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {x^{2}}{\sqrt {a + b x^{2} + c x^{4}}}\, dx \]

input
integrate(x**2/(c*x**4+b*x**2+a)**(1/2),x)
 
output
Integral(x**2/sqrt(a + b*x**2 + c*x**4), x)
 
3.10.64.7 Maxima [F]

\[ \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {c x^{4} + b x^{2} + a}} \,d x } \]

input
integrate(x^2/(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")
 
output
integrate(x^2/sqrt(c*x^4 + b*x^2 + a), x)
 
3.10.64.8 Giac [F]

\[ \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {x^{2}}{\sqrt {c x^{4} + b x^{2} + a}} \,d x } \]

input
integrate(x^2/(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
 
output
integrate(x^2/sqrt(c*x^4 + b*x^2 + a), x)
 
3.10.64.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a+b x^2+c x^4}} \, dx=\int \frac {x^2}{\sqrt {c\,x^4+b\,x^2+a}} \,d x \]

input
int(x^2/(a + b*x^2 + c*x^4)^(1/2),x)
 
output
int(x^2/(a + b*x^2 + c*x^4)^(1/2), x)