3.1048 \(\int \frac {x^4}{(-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\)

Optimal. Leaf size=244 \[ \frac {2}{45} \left (3 x^2-1\right )^{3/4} x+\frac {8 \sqrt [4]{3 x^2-1} x}{15 \left (\sqrt {3 x^2-1}+1\right )}-\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {4 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}-\frac {8 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x} \]

[Out]

2/45*x*(3*x^2-1)^(3/4)-1/27*arctan(1/2*x*6^(1/2)/(3*x^2-1)^(1/4))*6^(1/2)-1/27*arctanh(1/2*x*6^(1/2)/(3*x^2-1)
^(1/4))*6^(1/2)+8/15*x*(3*x^2-1)^(1/4)/(1+(3*x^2-1)^(1/2))-8/45*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2
*arctan((3*x^2-1)^(1/4)))*EllipticE(sin(2*arctan((3*x^2-1)^(1/4))),1/2*2^(1/2))*(1+(3*x^2-1)^(1/2))*(x^2/(1+(3
*x^2-1)^(1/2))^2)^(1/2)/x*3^(1/2)+4/45*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2*arctan((3*x^2-1)^(1/4)))
*EllipticF(sin(2*arctan((3*x^2-1)^(1/4))),1/2*2^(1/2))*(1+(3*x^2-1)^(1/2))*(x^2/(1+(3*x^2-1)^(1/2))^2)^(1/2)/x
*3^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {440, 230, 305, 220, 1196, 321, 398} \[ \frac {2}{45} \left (3 x^2-1\right )^{3/4} x+\frac {8 \sqrt [4]{3 x^2-1} x}{15 \left (\sqrt {3 x^2-1}+1\right )}-\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {4 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}-\frac {8 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x} \]

Antiderivative was successfully verified.

[In]

Int[x^4/((-2 + 3*x^2)*(-1 + 3*x^2)^(1/4)),x]

[Out]

(2*x*(-1 + 3*x^2)^(3/4))/45 + (8*x*(-1 + 3*x^2)^(1/4))/(15*(1 + Sqrt[-1 + 3*x^2])) - (Sqrt[2/3]*ArcTan[(Sqrt[3
/2]*x)/(-1 + 3*x^2)^(1/4)])/9 - (Sqrt[2/3]*ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/9 - (8*Sqrt[x^2/(1 + Sqr
t[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*x^2])*EllipticE[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(15*Sqrt[3]*x) + (4*Sqr
t[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(15*Sqrt[
3]*x)

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 230

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(2*Sqrt[-((b*x^2)/a)])/(b*x), Subst[Int[x^2/Sqrt[1 - x^4/a
], x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[a]

Rule 305

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 398

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[-(b^2/a), 4]}, Simp[(b*Ar
cTan[(q*x)/(Sqrt[2]*(a + b*x^2)^(1/4))])/(2*Sqrt[2]*a*d*q), x] + Simp[(b*ArcTanh[(q*x)/(Sqrt[2]*(a + b*x^2)^(1
/4))])/(2*Sqrt[2]*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && NegQ[b^2/a]

Rule 440

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

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {x^4}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx &=\int \left (\frac {2}{9 \sqrt [4]{-1+3 x^2}}+\frac {x^2}{3 \sqrt [4]{-1+3 x^2}}+\frac {4}{9 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}}\right ) \, dx\\ &=\frac {2}{9} \int \frac {1}{\sqrt [4]{-1+3 x^2}} \, dx+\frac {1}{3} \int \frac {x^2}{\sqrt [4]{-1+3 x^2}} \, dx+\frac {4}{9} \int \frac {1}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx\\ &=\frac {2}{45} x \left (-1+3 x^2\right )^{3/4}-\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {2}{45} \int \frac {1}{\sqrt [4]{-1+3 x^2}} \, dx+\frac {\left (4 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{9 \sqrt {3} x}\\ &=\frac {2}{45} x \left (-1+3 x^2\right )^{3/4}-\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {\left (4 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{45 \sqrt {3} x}+\frac {\left (4 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{9 \sqrt {3} x}-\frac {\left (4 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{9 \sqrt {3} x}\\ &=\frac {2}{45} x \left (-1+3 x^2\right )^{3/4}+\frac {4 x \sqrt [4]{-1+3 x^2}}{9 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {4 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{9 \sqrt {3} x}+\frac {2 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{9 \sqrt {3} x}+\frac {\left (4 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{45 \sqrt {3} x}-\frac {\left (4 \sqrt {x^2}\right ) \operatorname {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{45 \sqrt {3} x}\\ &=\frac {2}{45} x \left (-1+3 x^2\right )^{3/4}+\frac {8 x \sqrt [4]{-1+3 x^2}}{15 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {8 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}+\frac {4 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{15 \sqrt {3} x}\\ \end {align*}

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Mathematica [C]  time = 0.21, size = 177, normalized size = 0.73 \[ \frac {2 x \left (-3 \sqrt [4]{1-3 x^2} x^2 F_1\left (\frac {3}{2};\frac {1}{4},1;\frac {5}{2};3 x^2,\frac {3 x^2}{2}\right )-\frac {4 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};3 x^2,\frac {3 x^2}{2}\right )}{\left (3 x^2-2\right ) \left (x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};3 x^2,\frac {3 x^2}{2}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};3 x^2,\frac {3 x^2}{2}\right )\right )+2 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};3 x^2,\frac {3 x^2}{2}\right )\right )}+3 x^2-1\right )}{45 \sqrt [4]{3 x^2-1}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^4/((-2 + 3*x^2)*(-1 + 3*x^2)^(1/4)),x]

[Out]

(2*x*(-1 + 3*x^2 - 3*x^2*(1 - 3*x^2)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, 3*x^2, (3*x^2)/2] - (4*AppellF1[1/2, 1/4
, 1, 3/2, 3*x^2, (3*x^2)/2])/((-2 + 3*x^2)*(2*AppellF1[1/2, 1/4, 1, 3/2, 3*x^2, (3*x^2)/2] + x^2*(2*AppellF1[3
/2, 1/4, 2, 5/2, 3*x^2, (3*x^2)/2] + AppellF1[3/2, 5/4, 1, 5/2, 3*x^2, (3*x^2)/2])))))/(45*(-1 + 3*x^2)^(1/4))

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fricas [F]  time = 10.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (3 \, x^{2} - 1\right )}^{\frac {3}{4}} x^{4}}{9 \, x^{4} - 9 \, x^{2} + 2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(3*x^2-2)/(3*x^2-1)^(1/4),x, algorithm="fricas")

[Out]

integral((3*x^2 - 1)^(3/4)*x^4/(9*x^4 - 9*x^2 + 2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} {\left (3 \, x^{2} - 2\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(3*x^2-2)/(3*x^2-1)^(1/4),x, algorithm="giac")

[Out]

integrate(x^4/((3*x^2 - 1)^(1/4)*(3*x^2 - 2)), x)

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maple [F]  time = 0.19, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(3*x^2-2)/(3*x^2-1)^(1/4),x)

[Out]

int(x^4/(3*x^2-2)/(3*x^2-1)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{{\left (3 \, x^{2} - 1\right )}^{\frac {1}{4}} {\left (3 \, x^{2} - 2\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(3*x^2-2)/(3*x^2-1)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^4/((3*x^2 - 1)^(1/4)*(3*x^2 - 2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^4}{{\left (3\,x^2-1\right )}^{1/4}\,\left (3\,x^2-2\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((3*x^2 - 1)^(1/4)*(3*x^2 - 2)),x)

[Out]

int(x^4/((3*x^2 - 1)^(1/4)*(3*x^2 - 2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (3 x^{2} - 2\right ) \sqrt [4]{3 x^{2} - 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(3*x**2-2)/(3*x**2-1)**(1/4),x)

[Out]

Integral(x**4/((3*x**2 - 2)*(3*x**2 - 1)**(1/4)), x)

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