\(\int x^2 \sqrt {a+b x^2+c x^4} \, dx\) [959]

Optimal result

Integrand size = 20, antiderivative size = 343 \[ \int x^2 \sqrt {a+b x^2+c x^4} \, dx=-\frac {2 \left (b^2-3 a c\right ) x \sqrt {a+b x^2+c x^4}}{15 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {x \left (b+3 c x^2\right ) \sqrt {a+b x^2+c x^4}}{15 c}+\frac {2 \sqrt [4]{a} \left (b^2-3 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{15 c^{7/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{a} \left (\sqrt {a} b+\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{30 c^{5/4} \sqrt {a+b x^2+c x^4}} \] Output:

-2/15*(-3*a*c+b^2)*x*(c*x^4+b*x^2+a)^(1/2)/c^(3/2)/(a^(1/2)+c^(1/2)*x^2)+1 
/15*x*(3*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c+2/15*a^(1/4)*(-3*a*c+b^2)*(a^(1/ 
2)+c^(1/2)*x^2)*((c*x^4+b*x^2+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*EllipticE( 
sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))/c^(7/4)/ 
(c*x^4+b*x^2+a)^(1/2)-1/30*a^(1/4)*(a^(1/2)*b+2*(-3*a*c+b^2)/c^(1/2))*(a^( 
1/2)+c^(1/2)*x^2)*((c*x^4+b*x^2+a)/(a^(1/2)+c^(1/2)*x^2)^2)^(1/2)*InverseJ 
acobiAM(2*arctan(c^(1/4)*x/a^(1/4)),1/2*(2-b/a^(1/2)/c^(1/2))^(1/2))/c^(5/ 
4)/(c*x^4+b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.89 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.40 \[ \int x^2 \sqrt {a+b x^2+c x^4} \, dx=\frac {2 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x \left (b+3 c x^2\right ) \left (a+b x^2+c x^4\right )-i \left (b^2-3 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+i \left (-b^3+4 a b c+b^2 \sqrt {b^2-4 a c}-3 a c \sqrt {b^2-4 a c}\right ) \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{30 c^2 \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \sqrt {a+b x^2+c x^4}} \] Input:

Integrate[x^2*Sqrt[a + b*x^2 + c*x^4],x]
 

Output:

(2*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(b + 3*c*x^2)*(a + b*x^2 + c*x^4) - 
 I*(b^2 - 3*a*c)*(-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[(2*b - 2*Sqrt[b^2 - 4*a*c] + 4*c*x^2) 
/(b - Sqrt[b^2 - 4*a*c])]*EllipticE[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])] + I*(-b^3 
 + 4*a*b*c + b^2*Sqrt[b^2 - 4*a*c] - 3*a*c*Sqrt[b^2 - 4*a*c])*Sqrt[(b + Sq 
rt[b^2 - 4*a*c] + 2*c*x^2)/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[(2*b - 2*Sqrt[b^2 
 - 4*a*c] + 4*c*x^2)/(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])])/(30*c^2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*Sqrt[a + b*x^2 + c*x^4 
])
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 341, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1436, 1511, 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.

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

\(\Big \downarrow \) 1436

\(\displaystyle \frac {x \left (b+3 c x^2\right ) \sqrt {a+b x^2+c x^4}}{15 c}-\frac {\int \frac {2 \left (b^2-3 a c\right ) x^2+a b}{\sqrt {c x^4+b x^2+a}}dx}{15 c}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {x \left (b+3 c x^2\right ) \sqrt {a+b x^2+c x^4}}{15 c}-\frac {\sqrt {a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx-\frac {2 \sqrt {a} \left (b^2-3 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}}{15 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x \left (b+3 c x^2\right ) \sqrt {a+b x^2+c x^4}}{15 c}-\frac {\sqrt {a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx-\frac {2 \left (b^2-3 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}}{15 c}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {x \left (b+3 c x^2\right ) \sqrt {a+b x^2+c x^4}}{15 c}-\frac {\frac {\sqrt [4]{a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {2 \left (b^2-3 a c\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}}{15 c}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {x \left (b+3 c x^2\right ) \sqrt {a+b x^2+c x^4}}{15 c}-\frac {\frac {\sqrt [4]{a} \left (\frac {2 \left (b^2-3 a c\right )}{\sqrt {c}}+\sqrt {a} b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 \sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {2 \left (b^2-3 a c\right ) \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4}}{\sqrt {a}+\sqrt {c} x^2}\right )}{\sqrt {c}}}{15 c}\)

Input:

Int[x^2*Sqrt[a + b*x^2 + c*x^4],x]
 

Output:

(x*(b + 3*c*x^2)*Sqrt[a + b*x^2 + c*x^4])/(15*c) - ((-2*(b^2 - 3*a*c)*(-(( 
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]*EllipticE 
[2*ArcTan[(c^(1/4)*x)/a^(1/4)], (2 - b/(Sqrt[a]*Sqrt[c]))/4])/(c^(1/4)*Sqr 
t[a + b*x^2 + c*x^4])))/Sqrt[c] + (a^(1/4)*(Sqrt[a]*b + (2*(b^2 - 3*a*c))/ 
Sqrt[c])*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(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^(1/4)*Sqrt[a + b*x^2 + c*x^4]))/(15*c)
 

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 1436

Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] 
:> Simp[d*(d*x)^(m - 1)*(a + b*x^2 + c*x^4)^p*((2*b*p + c*(m + 4*p - 1)*x^2 
)/(c*(m + 4*p + 1)*(m + 4*p - 1))), x] - Simp[2*p*(d^2/(c*(m + 4*p + 1)*(m 
+ 4*p - 1)))   Int[(d*x)^(m - 2)*(a + b*x^2 + c*x^4)^(p - 1)*Simp[a*b*(m - 
1) - (2*a*c*(m + 4*p - 1) - b^2*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, 
 b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && GtQ[m, 1] && IntegerQ[ 
2*p] && (IntegerQ[p] || IntegerQ[m])
 

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]
 

rule 1511

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

Maple [A] (verified)

Time = 3.02 (sec) , antiderivative size = 406, normalized size of antiderivative = 1.18

method	result	size
risch	\(\frac {x \left (3 c \,x^{2}+b \right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{15 c}-\frac {\frac {\left (6 a c -2 b^{2}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}+\frac {a b \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}}{15 c}\)	\(406\)
default	\(\frac {x^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}+\frac {b x \sqrt {c \,x^{4}+b \,x^{2}+a}}{15 c}-\frac {b a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{60 c \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\left (\frac {2 a}{5}-\frac {2 b^{2}}{15 c}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(417\)
elliptic	\(\frac {x^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}+\frac {b x \sqrt {c \,x^{4}+b \,x^{2}+a}}{15 c}-\frac {b a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{60 c \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\left (\frac {2 a}{5}-\frac {2 b^{2}}{15 c}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(417\)

Input:

int(x^2*(c*x^4+b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/15*x*(3*c*x^2+b)*(c*x^4+b*x^2+a)^(1/2)/c-1/15/c*(1/2*(6*a*c-2*b^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)))+1/4*a*b*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)*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)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.03 \[ \int x^2 \sqrt {a+b x^2+c x^4} \, dx=-\frac {2 \, \sqrt {\frac {1}{2}} {\left ({\left (b^{2} c - 3 \, a c^{2}\right )} x \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - {\left (b^{3} - 3 \, a b c\right )} x\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} E(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - \sqrt {\frac {1}{2}} {\left ({\left (2 \, b^{2} c - {\left (6 \, a + b\right )} c^{2}\right )} x \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - {\left (2 \, b^{3} - {\left (6 \, a b - b^{2}\right )} c\right )} x\right )} \sqrt {c} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}} F(\arcsin \left (\frac {\sqrt {\frac {1}{2}} \sqrt {\frac {c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} - b}{c}}}{x}\right )\,|\,\frac {b c \sqrt {\frac {b^{2} - 4 \, a c}{c^{2}}} + b^{2} - 2 \, a c}{2 \, a c}) - 2 \, {\left (3 \, c^{3} x^{4} + b c^{2} x^{2} - 2 \, b^{2} c + 6 \, a c^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{30 \, c^{3} x} \] Input:

integrate(x^2*(c*x^4+b*x^2+a)^(1/2),x, algorithm="fricas")
 

Output:

-1/30*(2*sqrt(1/2)*((b^2*c - 3*a*c^2)*x*sqrt((b^2 - 4*a*c)/c^2) - (b^3 - 3 
*a*b*c)*x)*sqrt(c)*sqrt((c*sqrt((b^2 - 4*a*c)/c^2) - b)/c)*elliptic_e(arcs 
in(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)*((2*b^2*c - (6*a + b)*c^ 
2)*x*sqrt((b^2 - 4*a*c)/c^2) - (2*b^3 - (6*a*b - b^2)*c)*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*(3*c^3*x^4 + b*c^2*x^2 - 2*b^2*c + 6*a*c^2)*sqrt(c*x^4 + 
b*x^2 + a))/(c^3*x)
                                                                                    
                                                                                    
 

Sympy [F]

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

Maxima [F]

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

integrate(x^2*(c*x^4+b*x^2+a)^(1/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^4 + b*x^2 + a)*x^2, x)
 

Giac [F]

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

integrate(x^2*(c*x^4+b*x^2+a)^(1/2),x, algorithm="giac")
 

Output:

integrate(sqrt(c*x^4 + b*x^2 + a)*x^2, x)
 

Mupad [F(-1)]

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

Reduce [F]

\[ \int x^2 \sqrt {a+b x^2+c x^4} \, dx=\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, b x +3 \sqrt {c \,x^{4}+b \,x^{2}+a}\, c \,x^{3}-\left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a b +6 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a c -2 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x^{2}}{c \,x^{4}+b \,x^{2}+a}d x \right ) b^{2}}{15 c} \] Input:

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

Output:

(sqrt(a + b*x**2 + c*x**4)*b*x + 3*sqrt(a + b*x**2 + c*x**4)*c*x**3 - int( 
sqrt(a + b*x**2 + c*x**4)/(a + b*x**2 + c*x**4),x)*a*b + 6*int((sqrt(a + b 
*x**2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*a*c - 2*int((sqrt(a + b*x** 
2 + c*x**4)*x**2)/(a + b*x**2 + c*x**4),x)*b**2)/(15*c)