3.11.51 \(\int \frac {1}{x^2 (-2+3 x^2) \sqrt [4]{-1+3 x^2}} \, dx\) [1051]

Optimal. Leaf size=246 \[ -\frac {\left (-1+3 x^2\right )^{3/4}}{2 x}+\frac {3 x \sqrt [4]{-1+3 x^2}}{2 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{2 x}+\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{4 x} \]

[Out]

-1/2*(3*x^2-1)^(3/4)/x-1/8*arctan(1/2*x*6^(1/2)/(3*x^2-1)^(1/4))*6^(1/2)-1/8*arctanh(1/2*x*6^(1/2)/(3*x^2-1)^(
1/4))*6^(1/2)+3/2*x*(3*x^2-1)^(1/4)/(1+(3*x^2-1)^(1/2))-1/2*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2*arc
tan((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)+1/4*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2*arctan((3*x^2-1)^(1/4)))*Elli
pticF(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 = 0.08, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {451, 331, 236, 311, 226, 1210, 407} \begin {gather*} -\frac {1}{4} \sqrt {\frac {3}{2}} \text {ArcTan}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) F\left (2 \text {ArcTan}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{4 x}-\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) E\left (2 \text {ArcTan}\left (\sqrt [4]{3 x^2-1}\right )|\frac {1}{2}\right )}{2 x}+\frac {3 \sqrt [4]{3 x^2-1} x}{2 \left (\sqrt {3 x^2-1}+1\right )}-\frac {\left (3 x^2-1\right )^{3/4}}{2 x}-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(-1 + 3*x^2)^(3/4)/x + (3*x*(-1 + 3*x^2)^(1/4))/(2*(1 + Sqrt[-1 + 3*x^2])) - (Sqrt[3/2]*ArcTan[(Sqrt[3/2]
*x)/(-1 + 3*x^2)^(1/4)])/4 - (Sqrt[3/2]*ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/4 - (Sqrt[3]*Sqrt[x^2/(1 +
Sqrt[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*x^2])*EllipticE[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(2*x) + (Sqrt[3]*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])/(4*x)

Rule 226

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

Rule 236

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 311

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 331

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

Rule 407

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

Rule 451

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx &=\int \left (-\frac {1}{2 x^2 \sqrt [4]{-1+3 x^2}}+\frac {3}{2 \left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {1}{x^2 \sqrt [4]{-1+3 x^2}} \, dx\right )+\frac {3}{2} \int \frac {1}{\left (-2+3 x^2\right ) \sqrt [4]{-1+3 x^2}} \, dx\\ &=-\frac {\left (-1+3 x^2\right )^{3/4}}{2 x}-\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {3}{4} \int \frac {1}{\sqrt [4]{-1+3 x^2}} \, dx\\ &=-\frac {\left (-1+3 x^2\right )^{3/4}}{2 x}-\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {\left (\sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{2 x}\\ &=-\frac {\left (-1+3 x^2\right )^{3/4}}{2 x}-\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {\left (\sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{2 x}-\frac {\left (\sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1-x^2}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{2 x}\\ &=-\frac {\left (-1+3 x^2\right )^{3/4}}{2 x}+\frac {3 x \sqrt [4]{-1+3 x^2}}{2 \left (1+\sqrt {-1+3 x^2}\right )}-\frac {1}{4} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{4} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {\sqrt {3} \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 )}{2 x}+\frac {\sqrt {3} \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 )}{4 x}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 10.03, size = 64, normalized size = 0.26 \begin {gather*} \frac {4-12 x^2-3 x^4 \sqrt [4]{1-3 x^2} F_1\left (\frac {3}{2};\frac {1}{4},1;\frac {5}{2};3 x^2,\frac {3 x^2}{2}\right )}{8 x \sqrt [4]{-1+3 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4 - 12*x^2 - 3*x^4*(1 - 3*x^2)^(1/4)*AppellF1[3/2, 1/4, 1, 5/2, 3*x^2, (3*x^2)/2])/(8*x*(-1 + 3*x^2)^(1/4))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1}{x^{2} \left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/x^2/(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral((3*x^2 - 1)^(3/4)/(9*x^6 - 9*x^4 + 2*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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