3.10.35 \(\int \frac {\sqrt {a+b x^2+c x^4}}{x^6} \, dx\) [935]

Optimal. Leaf size=397 \[ -\frac {\sqrt {a+b x^2+c x^4}}{5 x^5}-\frac {b \sqrt {a+b x^2+c x^4}}{15 a x^3}+\frac {2 \left (b^2-3 a c\right ) \sqrt {a+b x^2+c x^4}}{15 a^2 x}-\frac {2 \sqrt {c} \left (b^2-3 a c\right ) x \sqrt {a+b x^2+c x^4}}{15 a^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {2 \sqrt [4]{c} \left (b^2-3 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{15 a^{7/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{c} \left (2 b^2+\sqrt {a} b \sqrt {c}-6 a c\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{30 a^{7/4} \sqrt {a+b x^2+c x^4}} \]

[Out]

-1/5*(c*x^4+b*x^2+a)^(1/2)/x^5-1/15*b*(c*x^4+b*x^2+a)^(1/2)/a/x^3+2/15*(-3*a*c+b^2)*(c*x^4+b*x^2+a)^(1/2)/a^2/
x-2/15*(-3*a*c+b^2)*x*c^(1/2)*(c*x^4+b*x^2+a)^(1/2)/a^2/(a^(1/2)+x^2*c^(1/2))+2/15*c^(1/4)*(-3*a*c+b^2)*(cos(2
*arctan(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)/a^(7
/4)/(c*x^4+b*x^2+a)^(1/2)-1/30*c^(1/4)*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(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))*(2*b^2-
6*a*c+b*a^(1/2)*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(7/4)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]
time = 0.17, antiderivative size = 397, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1131, 1295, 1211, 1117, 1209} \begin {gather*} -\frac {\sqrt [4]{c} \left (\sqrt {a} b \sqrt {c}-6 a c+2 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}} F\left (2 \text {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 a^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {2 \sqrt [4]{c} \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 \text {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 a^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {2 \left (b^2-3 a c\right ) \sqrt {a+b x^2+c x^4}}{15 a^2 x}-\frac {2 \sqrt {c} x \left (b^2-3 a c\right ) \sqrt {a+b x^2+c x^4}}{15 a^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt {a+b x^2+c x^4}}{5 x^5}-\frac {b \sqrt {a+b x^2+c x^4}}{15 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2 + c*x^4]/x^6,x]

[Out]

-1/5*Sqrt[a + b*x^2 + c*x^4]/x^5 - (b*Sqrt[a + b*x^2 + c*x^4])/(15*a*x^3) + (2*(b^2 - 3*a*c)*Sqrt[a + b*x^2 +
c*x^4])/(15*a^2*x) - (2*Sqrt[c]*(b^2 - 3*a*c)*x*Sqrt[a + b*x^2 + c*x^4])/(15*a^2*(Sqrt[a] + Sqrt[c]*x^2)) + (2
*c^(1/4)*(b^2 - 3*a*c)*(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])/(15*a^(7/4)*Sqrt[a + b*x^2 + c*x^4]) - (c^(1/4)*(2
*b^2 + Sqrt[a]*b*Sqrt[c] - 6*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])/(30*a^(7/4)*Sqrt[a + b*x^2 + c*x^4])

Rule 1117

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 1131

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] - Dist[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] && LtQ[m, -1] && IntegerQ[2*p] &&
(IntegerQ[p] || IntegerQ[m])

Rule 1209

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> 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 1211

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[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] && PosQ[c/a]

Rule 1295

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[d*(f
*x)^(m + 1)*((a + b*x^2 + c*x^4)^(p + 1)/(a*f*(m + 1))), x] + Dist[1/(a*f^2*(m + 1)), Int[(f*x)^(m + 2)*(a + b
*x^2 + c*x^4)^p*Simp[a*e*(m + 1) - b*d*(m + 2*p + 3) - c*d*(m + 4*p + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d,
e, f, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && IntegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {\sqrt {a+b x^2+c x^4}}{x^6} \, dx &=-\frac {\sqrt {a+b x^2+c x^4}}{5 x^5}+\frac {1}{5} \int \frac {b+2 c x^2}{x^4 \sqrt {a+b x^2+c x^4}} \, dx\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{5 x^5}-\frac {b \sqrt {a+b x^2+c x^4}}{15 a x^3}-\frac {\int \frac {2 \left (b^2-3 a c\right )+b c x^2}{x^2 \sqrt {a+b x^2+c x^4}} \, dx}{15 a}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{5 x^5}-\frac {b \sqrt {a+b x^2+c x^4}}{15 a x^3}+\frac {2 \left (b^2-3 a c\right ) \sqrt {a+b x^2+c x^4}}{15 a^2 x}+\frac {\int \frac {-a b c-2 c \left (b^2-3 a c\right ) x^2}{\sqrt {a+b x^2+c x^4}} \, dx}{15 a^2}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{5 x^5}-\frac {b \sqrt {a+b x^2+c x^4}}{15 a x^3}+\frac {2 \left (b^2-3 a c\right ) \sqrt {a+b x^2+c x^4}}{15 a^2 x}+\frac {\left (2 \sqrt {c} \left (b^2-3 a c\right )\right ) \int \frac {1-\frac {\sqrt {c} x^2}{\sqrt {a}}}{\sqrt {a+b x^2+c x^4}} \, dx}{15 a^{3/2}}--\frac {\left (-\sqrt {a} b c^{3/2}-2 c \left (b^2-3 a c\right )\right ) \int \frac {1}{\sqrt {a+b x^2+c x^4}} \, dx}{15 a^{3/2} \sqrt {c}}\\ &=-\frac {\sqrt {a+b x^2+c x^4}}{5 x^5}-\frac {b \sqrt {a+b x^2+c x^4}}{15 a x^3}+\frac {2 \left (b^2-3 a c\right ) \sqrt {a+b x^2+c x^4}}{15 a^2 x}-\frac {2 \sqrt {c} \left (b^2-3 a c\right ) x \sqrt {a+b x^2+c x^4}}{15 a^2 \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {2 \sqrt [4]{c} \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 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{15 a^{7/4} \sqrt {a+b x^2+c x^4}}-\frac {\sqrt [4]{c} \left (2 b^2+\sqrt {a} b \sqrt {c}-6 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}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{30 a^{7/4} \sqrt {a+b x^2+c x^4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 10.88, size = 530, normalized size = 1.34 \begin {gather*} \frac {-2 \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} \left (3 a^3-2 b^2 x^6 \left (b+c x^2\right )+a^2 \left (4 b x^2+9 c x^4\right )+a \left (-b^2 x^4+7 b c x^6+6 c^2 x^8\right )\right )-i \left (b^2-3 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right ) x^5 \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 \sinh ^{-1}\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 ) x^5 \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}}} F\left (i \sinh ^{-1}\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 a^2 \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x^5 \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2 + c*x^4]/x^6,x]

[Out]

(-2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*(3*a^3 - 2*b^2*x^6*(b + c*x^2) + a^2*(4*b*x^2 + 9*c*x^4) + a*(-(b^2*x^4) +
7*b*c*x^6 + 6*c^2*x^8)) - I*(b^2 - 3*a*c)*(-b + Sqrt[b^2 - 4*a*c])*x^5*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*ArcS
inh[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])*x^5*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])])/(30*a^2*Sqrt[c/(b + Sq
rt[b^2 - 4*a*c])]*x^5*Sqrt[a + b*x^2 + c*x^4])

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Maple [A]
time = 0.05, size = 452, normalized size = 1.14

method	result	size
risch	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (6 c \,x^{4} a -2 b^{2} x^{4}+a b \,x^{2}+3 a^{2}\right )}{15 x^{5} a^{2}}-\frac {c \left (\frac {\left (6 a c -2 b^{2}\right ) a \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\EllipticF \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{2 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}+\frac {a b \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \EllipticF \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{4 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}\right )}{15 a^{2}}\)	\(428\)
default	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{5 x^{5}}-\frac {b \sqrt {c \,x^{4}+b \,x^{2}+a}}{15 a \,x^{3}}-\frac {2 \left (3 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{15 a^{2} x}-\frac {b c \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \EllipticF \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{60 a \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {c \left (3 a c -b^{2}\right ) \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\EllipticF \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{15 a \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(452\)
elliptic	\(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{5 x^{5}}-\frac {b \sqrt {c \,x^{4}+b \,x^{2}+a}}{15 a \,x^{3}}-\frac {2 \left (3 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}}{15 a^{2} x}-\frac {b c \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \EllipticF \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )}{60 a \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}-\frac {c \left (3 a c -b^{2}\right ) \sqrt {2}\, \sqrt {4-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \sqrt {4+\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}}\, \left (\EllipticF \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )-\EllipticE \left (\frac {x \sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}}{2}, \frac {\sqrt {-4+\frac {2 b \left (b +\sqrt {-4 a c +b^{2}}\right )}{a c}}}{2}\right )\right )}{15 a \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right )}\)	\(452\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+b*x^2+a)^(1/2)/x^6,x,method=_RETURNVERBOSE)

[Out]

-1/5*(c*x^4+b*x^2+a)^(1/2)/x^5-1/15*b*(c*x^4+b*x^2+a)^(1/2)/a/x^3-2/15*(3*a*c-b^2)/a^2*(c*x^4+b*x^2+a)^(1/2)/x
-1/60*b*c/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)*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))-1/15*c*(3*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)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^6,x, algorithm="maxima")

[Out]

integrate(sqrt(c*x^4 + b*x^2 + a)/x^6, x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^6,x, algorithm="fricas")

[Out]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b x^{2} + c x^{4}}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+b*x**2+a)**(1/2)/x**6,x)

[Out]

Integral(sqrt(a + b*x**2 + c*x**4)/x**6, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+b*x^2+a)^(1/2)/x^6,x, algorithm="giac")

[Out]

integrate(sqrt(c*x^4 + b*x^2 + a)/x^6, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\sqrt {c\,x^4+b\,x^2+a}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2 + c*x^4)^(1/2)/x^6,x)

[Out]

int((a + b*x^2 + c*x^4)^(1/2)/x^6, x)

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