3.1037 \(\int \frac {x^4}{\sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\)

Optimal. Leaf size=164 \[ \frac {2}{45} \left (2-3 x^2\right )^{3/4} x+\frac {4 \sqrt [4]{2} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}} \]

[Out]

2/45*x*(-3*x^2+2)^(3/4)+4/27*2^(1/4)*arctan(1/3*(2^(3/4)-2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))
*3^(1/2)+4/27*2^(1/4)*arctanh(1/3*(2^(3/4)+2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^(1/4)*3^(1/2))*3^(1/2)-16/45
*2^(1/4)*(cos(1/2*arcsin(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcsin(1/2*x*6^(1/2)))*EllipticE(sin(1/2*arcsin(1/2*
x*6^(1/2))),2^(1/2))*3^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {440, 228, 321, 397} \[ \frac {2}{45} \left (2-3 x^2\right )^{3/4} x+\frac {4 \sqrt [4]{2} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(2 - 3*x^2)^(3/4))/45 + (4*2^(1/4)*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4
))])/(9*Sqrt[3]) + (4*2^(1/4)*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(9*S
qrt[3]) - (16*2^(1/4)*EllipticE[ArcSin[Sqrt[3/2]*x]/2, 2])/(15*Sqrt[3])

Rule 228

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(1/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[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 397

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> With[{q = Rt[b^2/a, 4]}, -Simp[(b*ArcT
an[(b + q^2*Sqrt[a + b*x^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), x] - Simp[(b*ArcTanh[(b - q^2*Sqrt[a + b*x
^2])/(q^3*x*(a + b*x^2)^(1/4))])/(2*a*d*q), x]] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[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])

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx &=\int \left (-\frac {4}{9 \sqrt [4]{2-3 x^2}}-\frac {x^2}{3 \sqrt [4]{2-3 x^2}}+\frac {16}{9 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )}\right ) \, dx\\ &=-\left (\frac {1}{3} \int \frac {x^2}{\sqrt [4]{2-3 x^2}} \, dx\right )-\frac {4}{9} \int \frac {1}{\sqrt [4]{2-3 x^2}} \, dx+\frac {16}{9} \int \frac {1}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx\\ &=\frac {2}{45} x \left (2-3 x^2\right )^{3/4}+\frac {4 \sqrt [4]{2} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {8 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{9 \sqrt {3}}-\frac {4}{45} \int \frac {1}{\sqrt [4]{2-3 x^2}} \, dx\\ &=\frac {2}{45} x \left (2-3 x^2\right )^{3/4}+\frac {4 \sqrt [4]{2} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}+\frac {4 \sqrt [4]{2} \tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{9 \sqrt {3}}-\frac {16 \sqrt [4]{2} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{15 \sqrt {3}}\\ \end {align*}

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Mathematica [C]  time = 0.26, size = 184, normalized size = 1.12 \[ \frac {1}{45} x \left (3\ 2^{3/4} x^2 F_1\left (\frac {3}{2};\frac {1}{4},1;\frac {5}{2};\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+\frac {2 \left (\frac {32 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{\left (3 x^2-4\right ) \left (x^2 \left (2 F_1\left (\frac {3}{2};\frac {1}{4},2;\frac {5}{2};\frac {3 x^2}{2},\frac {3 x^2}{4}\right )+F_1\left (\frac {3}{2};\frac {5}{4},1;\frac {5}{2};\frac {3 x^2}{2},\frac {3 x^2}{4}\right )\right )+4 F_1\left (\frac {1}{2};\frac {1}{4},1;\frac {3}{2};\frac {3 x^2}{2},\frac {3 x^2}{4}\right )\right )}-3 x^2+2\right )}{\sqrt [4]{2-3 x^2}}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(3*2^(3/4)*x^2*AppellF1[3/2, 1/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4] + (2*(2 - 3*x^2 + (32*AppellF1[1/2, 1/4, 1,
3/2, (3*x^2)/2, (3*x^2)/4])/((-4 + 3*x^2)*(4*AppellF1[1/2, 1/4, 1, 3/2, (3*x^2)/2, (3*x^2)/4] + x^2*(2*Appell
F1[3/2, 1/4, 2, 5/2, (3*x^2)/2, (3*x^2)/4] + AppellF1[3/2, 5/4, 1, 5/2, (3*x^2)/2, (3*x^2)/4])))))/(2 - 3*x^2)
^(1/4)))/45

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fricas [F]  time = 8.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} x^{4}}{9 \, x^{4} - 18 \, x^{2} + 8}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((-3*x^2 + 2)^(3/4)*x^4/(9*x^4 - 18*x^2 + 8), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {x^{4}}{\left (-3 x^{2}+2\right )^{\frac {1}{4}} \left (-3 x^{2}+4\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {x^4}{{\left (2-3\,x^2\right )}^{1/4}\,\left (3\,x^2-4\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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