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

Optimal. Leaf size=445 \[ \frac {\sqrt {\sqrt {4 a c+b^2}+b} \left (-b \sqrt {4 a c+b^2}+a c+b^2\right ) \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{3 \sqrt {2} a^2 \sqrt {c} \sqrt {a+b x^2-c x^4}}-\frac {b \left (b-\sqrt {4 a c+b^2}\right ) \sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} E\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{3 \sqrt {2} a^2 \sqrt {c} \sqrt {a+b x^2-c x^4}}+\frac {2 b \sqrt {a+b x^2-c x^4}}{3 a^2 x}-\frac {\sqrt {a+b x^2-c x^4}}{3 a x^3} \]

[Out]

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

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Rubi [A]  time = 0.47, antiderivative size = 445, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1123, 1281, 1202, 524, 424, 419} \[ \frac {\sqrt {\sqrt {4 a c+b^2}+b} \left (-b \sqrt {4 a c+b^2}+a c+b^2\right ) \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{3 \sqrt {2} a^2 \sqrt {c} \sqrt {a+b x^2-c x^4}}-\frac {b \left (b-\sqrt {4 a c+b^2}\right ) \sqrt {\sqrt {4 a c+b^2}+b} \sqrt {1-\frac {2 c x^2}{b-\sqrt {4 a c+b^2}}} \sqrt {1-\frac {2 c x^2}{\sqrt {4 a c+b^2}+b}} E\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{3 \sqrt {2} a^2 \sqrt {c} \sqrt {a+b x^2-c x^4}}+\frac {2 b \sqrt {a+b x^2-c x^4}}{3 a^2 x}-\frac {\sqrt {a+b x^2-c x^4}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a + b*x^2 - c*x^4]),x]

[Out]

-Sqrt[a + b*x^2 - c*x^4]/(3*a*x^3) + (2*b*Sqrt[a + b*x^2 - c*x^4])/(3*a^2*x) - (b*(b - Sqrt[b^2 + 4*a*c])*Sqrt
[b + Sqrt[b^2 + 4*a*c]]*Sqrt[1 - (2*c*x^2)/(b - Sqrt[b^2 + 4*a*c])]*Sqrt[1 - (2*c*x^2)/(b + Sqrt[b^2 + 4*a*c])
]*EllipticE[ArcSin[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 + 4*a*c]]], (b + Sqrt[b^2 + 4*a*c])/(b - Sqrt[b^2 + 4
*a*c])])/(3*Sqrt[2]*a^2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4]) + (Sqrt[b + Sqrt[b^2 + 4*a*c]]*(b^2 + a*c - b*Sqrt[b^
2 + 4*a*c])*Sqrt[1 - (2*c*x^2)/(b - Sqrt[b^2 + 4*a*c])]*Sqrt[1 - (2*c*x^2)/(b + Sqrt[b^2 + 4*a*c])]*EllipticF[
ArcSin[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 + 4*a*c]]], (b + Sqrt[b^2 + 4*a*c])/(b - Sqrt[b^2 + 4*a*c])])/(3*
Sqrt[2]*a^2*Sqrt[c]*Sqrt[a + b*x^2 - c*x^4])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 1123

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 + 1))/(a*d*(m + 1)), x] - Dist[1/(a*d^2*(m + 1)), Int[(d*x)^(m + 2)*(b*(m + 2*p + 3) + c*(m + 4*p +
5)*x^2)*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[m, -1] && In
tegerQ[2*p] && (IntegerQ[p] || IntegerQ[m])

Rule 1202

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

Rule 1281

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 {1}{x^4 \sqrt {a+b x^2-c x^4}} \, dx &=-\frac {\sqrt {a+b x^2-c x^4}}{3 a x^3}+\frac {\int \frac {-2 b+c x^2}{x^2 \sqrt {a+b x^2-c x^4}} \, dx}{3 a}\\ &=-\frac {\sqrt {a+b x^2-c x^4}}{3 a x^3}+\frac {2 b \sqrt {a+b x^2-c x^4}}{3 a^2 x}-\frac {\int \frac {-a c-2 b c x^2}{\sqrt {a+b x^2-c x^4}} \, dx}{3 a^2}\\ &=-\frac {\sqrt {a+b x^2-c x^4}}{3 a x^3}+\frac {2 b \sqrt {a+b x^2-c x^4}}{3 a^2 x}-\frac {\left (\sqrt {1-\frac {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}}}\right ) \int \frac {-a c-2 b c x^2}{\sqrt {1-\frac {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}}}} \, dx}{3 a^2 \sqrt {a+b x^2-c x^4}}\\ &=-\frac {\sqrt {a+b x^2-c x^4}}{3 a x^3}+\frac {2 b \sqrt {a+b x^2-c x^4}}{3 a^2 x}-\frac {\left (b \left (b-\sqrt {b^2+4 a c}\right ) \sqrt {1-\frac {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}}}\right ) \int \frac {\sqrt {1-\frac {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}}}} \, dx}{3 a^2 \sqrt {a+b x^2-c x^4}}+\frac {\left (\left (b^2+a c-b \sqrt {b^2+4 a c}\right ) \sqrt {1-\frac {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}}}\right ) \int \frac {1}{\sqrt {1-\frac {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}}}} \, dx}{3 a^2 \sqrt {a+b x^2-c x^4}}\\ &=-\frac {\sqrt {a+b x^2-c x^4}}{3 a x^3}+\frac {2 b \sqrt {a+b x^2-c x^4}}{3 a^2 x}-\frac {b \left (b-\sqrt {b^2+4 a c}\right ) \sqrt {b+\sqrt {b^2+4 a c}} \sqrt {1-\frac {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}}} E\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{3 \sqrt {2} a^2 \sqrt {c} \sqrt {a+b x^2-c x^4}}+\frac {\sqrt {b+\sqrt {b^2+4 a c}} \left (b^2+a c-b \sqrt {b^2+4 a c}\right ) \sqrt {1-\frac {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}}} F\left (\sin ^{-1}\left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2+4 a c}}}\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{3 \sqrt {2} a^2 \sqrt {c} \sqrt {a+b x^2-c x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.76, size = 472, normalized size = 1.06 \[ \frac {-2 \sqrt {-\frac {c}{\sqrt {4 a c+b^2}+b}} \left (a-2 b x^2\right ) \left (a+b x^2-c x^4\right )+i \sqrt {2} x^3 \left (b \sqrt {4 a c+b^2}-a c-b^2\right ) \sqrt {\frac {\sqrt {4 a c+b^2}+b-2 c x^2}{\sqrt {4 a c+b^2}+b}} \sqrt {\frac {\sqrt {4 a c+b^2}-b+2 c x^2}{\sqrt {4 a c+b^2}-b}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2+4 a c}}} x\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )-i \sqrt {2} b x^3 \left (\sqrt {4 a c+b^2}-b\right ) \sqrt {\frac {\sqrt {4 a c+b^2}+b-2 c x^2}{\sqrt {4 a c+b^2}+b}} \sqrt {\frac {\sqrt {4 a c+b^2}-b+2 c x^2}{\sqrt {4 a c+b^2}-b}} E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {-\frac {c}{b+\sqrt {b^2+4 a c}}} x\right )|\frac {b+\sqrt {b^2+4 a c}}{b-\sqrt {b^2+4 a c}}\right )}{6 a^2 x^3 \sqrt {-\frac {c}{\sqrt {4 a c+b^2}+b}} \sqrt {a+b x^2-c x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a + b*x^2 - c*x^4]),x]

[Out]

(-2*Sqrt[-(c/(b + Sqrt[b^2 + 4*a*c]))]*(a - 2*b*x^2)*(a + b*x^2 - c*x^4) - I*Sqrt[2]*b*(-b + Sqrt[b^2 + 4*a*c]
)*x^3*Sqrt[(b + Sqrt[b^2 + 4*a*c] - 2*c*x^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])]*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*Sqrt[2]*(-b^2 - a*c + b*Sqrt[b^2 + 4*a*c])*x^3*Sqrt[(b + Sqrt[b^2 + 4*a*c]
- 2*c*x^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])]*Ellipti
cF[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])])/
(6*a^2*Sqrt[-(c/(b + Sqrt[b^2 + 4*a*c]))]*x^3*Sqrt[a + b*x^2 - c*x^4])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c x^{4} + b x^{2} + a}}{c x^{8} - b x^{6} - a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-c*x^4 + b*x^2 + a)/(c*x^8 - b*x^6 - a*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c x^{4} + b x^{2} + a} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-c*x^4 + b*x^2 + a)*x^4), x)

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maple [A]  time = 0.02, size = 417, normalized size = 0.94 \[ -\frac {\sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {-\frac {2 \left (b +\sqrt {4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )+\EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {-\frac {2 \left (b +\sqrt {4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )\right ) b c}{3 \sqrt {\frac {-b +\sqrt {4 a c +b^{2}}}{a}}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {4 a c +b^{2}}\right ) a}+\frac {\sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {4 a c +b^{2}}\right ) x^{2}}{a}+4}\, c \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {-\frac {2 \left (b +\sqrt {4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )}{12 \sqrt {\frac {-b +\sqrt {4 a c +b^{2}}}{a}}\, \sqrt {-c \,x^{4}+b \,x^{2}+a}\, a}+\frac {2 \sqrt {-c \,x^{4}+b \,x^{2}+a}\, b}{3 a^{2} x}-\frac {\sqrt {-c \,x^{4}+b \,x^{2}+a}}{3 a \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(-c*x^4+b*x^2+a)^(1/2)/a/x^3+2/3*b*(-c*x^4+b*x^2+a)^(1/2)/a^2/x+1/12*c/a*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/
a)^(1/2)*(-2*(-b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)/(-c*x^4+b*x^2+a)^(1
/2)*EllipticF(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-1/3
*b*c/a*2^(1/2)/((-b+(4*a*c+b^2)^(1/2))/a)^(1/2)*(-2*(-b+(4*a*c+b^2)^(1/2))/a*x^2+4)^(1/2)*(2*(b+(4*a*c+b^2)^(1
/2))/a*x^2+4)^(1/2)/(-c*x^4+b*x^2+a)^(1/2)/(b+(4*a*c+b^2)^(1/2))*(EllipticF(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2)
)/a)^(1/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2))-EllipticE(1/2*2^(1/2)*((-b+(4*a*c+b^2)^(1/2))/a)^(1
/2)*x,1/2*(-2*(b+(4*a*c+b^2)^(1/2))/a*b/c-4)^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {-c x^{4} + b x^{2} + a} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(-c*x^4 + b*x^2 + a)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^4\,\sqrt {-c\,x^4+b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \sqrt {a + b x^{2} - c x^{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**4*sqrt(a + b*x**2 - c*x**4)), x)

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