\(\int \frac {(a+b x^2+c x^4)^{3/2}}{x^2} \, dx\) [978]

Optimal result

Integrand size = 20, antiderivative size = 361 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^2} \, dx=\frac {\left (b^2+12 a c\right ) x \sqrt {a+b x^2+c x^4}}{5 \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {1}{5} x \left (7 b+6 c x^2\right ) \sqrt {a+b x^2+c x^4}-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{x}-\frac {\sqrt [4]{a} \left (b^2+12 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{5 c^{3/4} \sqrt {a+b x^2+c x^4}}+\frac {\sqrt [4]{a} \left (b^2+8 \sqrt {a} b \sqrt {c}+12 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{10 c^{3/4} \sqrt {a+b x^2+c x^4}} \] Output:

1/5*(12*a*c+b^2)*x*(c*x^4+b*x^2+a)^(1/2)/c^(1/2)/(a^(1/2)+c^(1/2)*x^2)+1/5 
*x*(6*c*x^2+7*b)*(c*x^4+b*x^2+a)^(1/2)-(c*x^4+b*x^2+a)^(3/2)/x-1/5*a^(1/4) 
*(12*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^(3/4)/(c*x^4+b*x^2+a)^(1/2)+1/10*a^(1/4)*(b^2+8*a^(1/2)*b*c^ 
(1/2)+12*a*c)*(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)*InverseJacobiAM(2*arctan(c^(1/4)*x/a^(1/4)),1/2*(2-b/a^(1/2)/c^( 
1/2))^(1/2))/c^(3/4)/(c*x^4+b*x^2+a)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 10.83 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.40 \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^2} \, dx=\frac {4 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \left (-5 a^2-3 a b x^2+2 b^2 x^4-4 a c x^4+3 b c x^6+c^2 x^8\right )+i \left (b^2+12 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right ) x \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-i \left (-b^3+4 a b c+b^2 \sqrt {b^2-4 a c}+12 a c \sqrt {b^2-4 a c}\right ) x \sqrt {\frac {b+\sqrt {b^2-4 a c}+2 c x^2}{b+\sqrt {b^2-4 a c}}} \sqrt {\frac {2 b-2 \sqrt {b^2-4 a c}+4 c x^2}{b-\sqrt {b^2-4 a c}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right ),\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )}{20 c \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x \sqrt {a+b x^2+c x^4}} \] Input:

Integrate[(a + b*x^2 + c*x^4)^(3/2)/x^2,x]
 

Output:

(4*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*(-5*a^2 - 3*a*b*x^2 + 2*b^2*x^4 - 4*a 
*c*x^4 + 3*b*c*x^6 + c^2*x^8) + I*(b^2 + 12*a*c)*(-b + Sqrt[b^2 - 4*a*c])* 
x*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*Ar 
cSinh[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] + 12* 
a*c*Sqrt[b^2 - 4*a*c])*x*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])]*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])])/(20*c*Sqrt[c/(b + Sqrt[b^ 
2 - 4*a*c])]*x*Sqrt[a + b*x^2 + c*x^4])
 

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 359, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1437, 1490, 27, 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.

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

\(\Big \downarrow \) 1437

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

\(\Big \downarrow \) 1490

\(\displaystyle 3 \left (\frac {\int \frac {c \left (\left (b^2+12 a c\right ) x^2+8 a b\right )}{\sqrt {c x^4+b x^2+a}}dx}{15 c}+\frac {1}{15} x \left (7 b+6 c x^2\right ) \sqrt {a+b x^2+c x^4}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \left (\frac {1}{15} \int \frac {\left (b^2+12 a c\right ) x^2+8 a b}{\sqrt {c x^4+b x^2+a}}dx+\frac {1}{15} x \left (7 b+6 c x^2\right ) \sqrt {a+b x^2+c x^4}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 1511

\(\displaystyle 3 \left (\frac {1}{15} \left (\frac {\sqrt {a} \left (8 \sqrt {a} b \sqrt {c}+12 a c+b^2\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}-\frac {\sqrt {a} \left (12 a c+b^2\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )+\frac {1}{15} x \left (7 b+6 c x^2\right ) \sqrt {a+b x^2+c x^4}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 27

\(\displaystyle 3 \left (\frac {1}{15} \left (\frac {\sqrt {a} \left (8 \sqrt {a} b \sqrt {c}+12 a c+b^2\right ) \int \frac {1}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}-\frac {\left (12 a c+b^2\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )+\frac {1}{15} x \left (7 b+6 c x^2\right ) \sqrt {a+b x^2+c x^4}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 1416

\(\displaystyle 3 \left (\frac {1}{15} \left (\frac {\sqrt [4]{a} \left (8 \sqrt {a} b \sqrt {c}+12 a c+b^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 c^{3/4} \sqrt {a+b x^2+c x^4}}-\frac {\left (12 a c+b^2\right ) \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+b x^2+a}}dx}{\sqrt {c}}\right )+\frac {1}{15} x \left (7 b+6 c x^2\right ) \sqrt {a+b x^2+c x^4}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{x}\)

\(\Big \downarrow \) 1509

\(\displaystyle 3 \left (\frac {1}{15} \left (\frac {\sqrt [4]{a} \left (8 \sqrt {a} b \sqrt {c}+12 a c+b^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{2 c^{3/4} \sqrt {a+b x^2+c x^4}}-\frac {\left (12 a c+b^2\right ) \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{\sqrt [4]{c} \sqrt {a+b x^2+c x^4}}-\frac {x \sqrt {a+b x^2+c x^4}}{\sqrt {a}+\sqrt {c} x^2}\right )}{\sqrt {c}}\right )+\frac {1}{15} x \left (7 b+6 c x^2\right ) \sqrt {a+b x^2+c x^4}\right )-\frac {\left (a+b x^2+c x^4\right )^{3/2}}{x}\)

Input:

Int[(a + b*x^2 + c*x^4)^(3/2)/x^2,x]
 

Output:

-((a + b*x^2 + c*x^4)^(3/2)/x) + 3*((x*(7*b + 6*c*x^2)*Sqrt[a + b*x^2 + c* 
x^4])/15 + (-(((b^2 + 12*a*c)*(-((x*Sqrt[a + b*x^2 + c*x^4])/(Sqrt[a] + Sq 
rt[c]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + b*x^2 + c*x^4)/(S 
qrt[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)*Sqrt[a + b*x^2 + c*x^4])))/Sqrt[c]) + (a^(1 
/4)*(b^2 + 8*Sqrt[a]*b*Sqrt[c] + 12*a*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^(3/4)*Sqrt[a + b*x^2 + c*x^4] 
))/15)
 

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 1437

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

rule 1490

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb 
ol] :> Simp[x*(2*b*e*p + c*d*(4*p + 3) + c*e*(4*p + 1)*x^2)*((a + b*x^2 + c 
*x^4)^p/(c*(4*p + 1)*(4*p + 3))), x] + Simp[2*(p/(c*(4*p + 1)*(4*p + 3))) 
 Int[Simp[2*a*c*d*(4*p + 3) - a*b*e + (2*a*c*e*(4*p + 1) + b*c*d*(4*p + 3) 
- b^2*e*(2*p + 1))*x^2, x]*(a + b*x^2 + c*x^4)^(p - 1), x], x] /; FreeQ[{a, 
 b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && 
 GtQ[p, 0] && FractionQ[p] && IntegerQ[2*p]
 

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 = 2.50 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.12

method	result	size
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (-c \,x^{4}-2 b \,x^{2}+5 a \right )}{5 x}-\frac {\left (12 a c +b^{2}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{10 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}+\frac {2 a b \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{5 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\)	\(405\)
default	\(-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{x}+\frac {c \,x^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}+\frac {2 b x \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}+\frac {2 a b \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{5 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\left (\frac {12 a c}{5}+\frac {b^{2}}{5}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(430\)
elliptic	\(-\frac {a \sqrt {c \,x^{4}+b \,x^{2}+a}}{x}+\frac {c \,x^{3} \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}+\frac {2 b x \sqrt {c \,x^{4}+b \,x^{2}+a}}{5}+\frac {2 a b \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{5 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {\left (\frac {12 a c}{5}+\frac {b^{2}}{5}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\operatorname {EllipticF}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\operatorname {EllipticE}\left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(430\)

Input:

int((c*x^4+b*x^2+a)^(3/2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

-1/5*(c*x^4+b*x^2+a)^(1/2)*(-c*x^4-2*b*x^2+5*a)/x-1/10*(12*a*c+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)))+2/5*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 [F]

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

integrate((c*x^4+b*x^2+a)^(3/2)/x^2,x, algorithm="fricas")
                                                                                    
                                                                                    
 

Output:

integral((c*x^4 + b*x^2 + a)^(3/2)/x^2, x)
 

Sympy [F]

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

integrate((c*x**4+b*x**2+a)**(3/2)/x**2,x)
 

Output:

Integral((a + b*x**2 + c*x**4)**(3/2)/x**2, x)
 

Maxima [F]

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

integrate((c*x^4+b*x^2+a)^(3/2)/x^2,x, algorithm="maxima")
 

Output:

integrate((c*x^4 + b*x^2 + a)^(3/2)/x^2, x)
 

Giac [F]

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

integrate((c*x^4+b*x^2+a)^(3/2)/x^2,x, algorithm="giac")
 

Output:

integrate((c*x^4 + b*x^2 + a)^(3/2)/x^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^2} \, dx=\int \frac {{\left (c\,x^4+b\,x^2+a\right )}^{3/2}}{x^2} \,d x \] Input:

int((a + b*x^2 + c*x^4)^(3/2)/x^2,x)
 

Output:

int((a + b*x^2 + c*x^4)^(3/2)/x^2, x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2+c x^4\right )^{3/2}}{x^2} \, dx=\frac {7 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a c +\sqrt {c \,x^{4}+b \,x^{2}+a}\, b^{2}+2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, b c \,x^{2}+\sqrt {c \,x^{4}+b \,x^{2}+a}\, c^{2} x^{4}+12 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{6}+b \,x^{4}+a \,x^{2}}d x \right ) a^{2} c x +\left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{6}+b \,x^{4}+a \,x^{2}}d x \right ) a \,b^{2} x +8 \left (\int \frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{c \,x^{4}+b \,x^{2}+a}d x \right ) a b c x}{5 c x} \] Input:

int((c*x^4+b*x^2+a)^(3/2)/x^2,x)
 

Output:

(7*sqrt(a + b*x**2 + c*x**4)*a*c + sqrt(a + b*x**2 + c*x**4)*b**2 + 2*sqrt 
(a + b*x**2 + c*x**4)*b*c*x**2 + sqrt(a + b*x**2 + c*x**4)*c**2*x**4 + 12* 
int(sqrt(a + b*x**2 + c*x**4)/(a*x**2 + b*x**4 + c*x**6),x)*a**2*c*x + int 
(sqrt(a + b*x**2 + c*x**4)/(a*x**2 + b*x**4 + c*x**6),x)*a*b**2*x + 8*int( 
sqrt(a + b*x**2 + c*x**4)/(a + b*x**2 + c*x**4),x)*a*b*c*x)/(5*c*x)