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

Optimal. Leaf size=313 \[ -\frac {2 b x \sqrt {a+b x^2+c x^4}}{3 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \left (\sqrt {a} \sqrt {c}+2 b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{6 c^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {2 \sqrt [4]{a} b \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{3 c^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {x \sqrt {a+b x^2+c x^4}}{3 c} \]

[Out]

1/3*x*(c*x^4+b*x^2+a)^(1/2)/c-2/3*b*x*(c*x^4+b*x^2+a)^(1/2)/c^(3/2)/(a^(1/2)+x^2*c^(1/2))+2/3*a^(1/4)*b*(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)/c^(7
/4)/(c*x^4+b*x^2+a)^(1/2)-1/6*a^(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+a^(
1/2)*c^(1/2))*((c*x^4+b*x^2+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/c^(7/4)/(c*x^4+b*x^2+a)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 313, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1122, 1197, 1103, 1195} \[ -\frac {2 b x \sqrt {a+b x^2+c x^4}}{3 c^{3/2} \left (\sqrt {a}+\sqrt {c} x^2\right )}-\frac {\sqrt [4]{a} \left (\sqrt {a} \sqrt {c}+2 b\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{6 c^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {2 \sqrt [4]{a} b \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+b x^2+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{4} \left (2-\frac {b}{\sqrt {a} \sqrt {c}}\right )\right )}{3 c^{7/4} \sqrt {a+b x^2+c x^4}}+\frac {x \sqrt {a+b x^2+c x^4}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a + b*x^2 + c*x^4],x]

[Out]

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

Rule 1103

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)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 1122

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

Rule 1195

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)]*EllipticE[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(q*Sqrt[a + b*x^2 + c*x^4]), 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 1197

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]

Rubi steps

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

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Mathematica [C]  time = 0.86, size = 444, normalized size = 1.42 \[ \frac {i \left (b \sqrt {b^2-4 a c}+a c-b^2\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} F\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )-i b \left (\sqrt {b^2-4 a c}-b\right ) \sqrt {\frac {\sqrt {b^2-4 a c}+b+2 c x^2}{\sqrt {b^2-4 a c}+b}} \sqrt {\frac {-2 \sqrt {b^2-4 a c}+2 b+4 c x^2}{b-\sqrt {b^2-4 a c}}} E\left (i \sinh ^{-1}\left (\sqrt {2} \sqrt {\frac {c}{b+\sqrt {b^2-4 a c}}} x\right )|\frac {b+\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}}\right )+2 c x \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \left (a+b x^2+c x^4\right )}{6 c^2 \sqrt {\frac {c}{\sqrt {b^2-4 a c}+b}} \sqrt {a+b x^2+c x^4}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*x*(a + b*x^2 + c*x^4) - I*b*(-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^2 + 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])]*Sq
rt[(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])])/(6*c^2*Sqrt[c/(b + Sqrt[b^2 - 4*a*c])]*S
qrt[a + b*x^2 + c*x^4])

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{4}}{\sqrt {c x^{4} + b x^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 388, normalized size = 1.24 \[ \frac {\sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \left (-\EllipticE \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )+\EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )\right ) a b}{3 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, \left (b +\sqrt {-4 a c +b^{2}}\right ) c}-\frac {\sqrt {2}\, \sqrt {-\frac {2 \left (-b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, \sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) x^{2}}{a}+4}\, a \EllipticF \left (\frac {\sqrt {2}\, \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, x}{2}, \frac {\sqrt {\frac {2 \left (b +\sqrt {-4 a c +b^{2}}\right ) b}{a c}-4}}{2}\right )}{12 \sqrt {\frac {-b +\sqrt {-4 a c +b^{2}}}{a}}\, \sqrt {c \,x^{4}+b \,x^{2}+a}\, c}+\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}\, x}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*(c*x^4+b*x^2+a)^(1/2)/c-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 \[ \int \frac {x^{4}}{\sqrt {c x^{4} + b x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^2 + c*x^4)^(1/2),x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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