3.1032 \(\int \frac{x^5}{\sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\)

Optimal. Leaf size=121 \[ -\frac{2}{189} \left (2-3 x^2\right )^{7/4}+\frac{4}{27} \left (2-3 x^2\right )^{3/4}+\frac{8}{27} \sqrt [4]{2} \tan ^{-1}\left (\frac{\sqrt{2}-\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac{8}{27} \sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt{2-3 x^2}+\sqrt{2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right ) \]

[Out]

(4*(2 - 3*x^2)^(3/4))/27 - (2*(2 - 3*x^2)^(7/4))/189 + (8*2^(1/4)*ArcTan[(Sqrt[2] - Sqrt[2 - 3*x^2])/(2^(3/4)*
(2 - 3*x^2)^(1/4))])/27 + (8*2^(1/4)*ArcTanh[(Sqrt[2] + Sqrt[2 - 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))])/27

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Rubi [A]  time = 0.0645026, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {440, 261, 266, 43, 439} \[ -\frac{2}{189} \left (2-3 x^2\right )^{7/4}+\frac{4}{27} \left (2-3 x^2\right )^{3/4}+\frac{8}{27} \sqrt [4]{2} \tan ^{-1}\left (\frac{\sqrt{2}-\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac{8}{27} \sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt{2-3 x^2}+\sqrt{2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^5/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]

[Out]

(4*(2 - 3*x^2)^(3/4))/27 - (2*(2 - 3*x^2)^(7/4))/189 + (8*2^(1/4)*ArcTan[(Sqrt[2] - Sqrt[2 - 3*x^2])/(2^(3/4)*
(2 - 3*x^2)^(1/4))])/27 + (8*2^(1/4)*ArcTanh[(Sqrt[2] + Sqrt[2 - 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))])/27

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 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 439

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

Rubi steps

\begin{align*} \int \frac{x^5}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx &=\int \left (-\frac{4 x}{9 \sqrt [4]{2-3 x^2}}-\frac{x^3}{3 \sqrt [4]{2-3 x^2}}+\frac{16 x}{9 \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )}\right ) \, dx\\ &=-\left (\frac{1}{3} \int \frac{x^3}{\sqrt [4]{2-3 x^2}} \, dx\right )-\frac{4}{9} \int \frac{x}{\sqrt [4]{2-3 x^2}} \, dx+\frac{16}{9} \int \frac{x}{\sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx\\ &=\frac{8}{81} \left (2-3 x^2\right )^{3/4}+\frac{8}{27} \sqrt [4]{2} \tan ^{-1}\left (\frac{\sqrt{2}-\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac{8}{27} \sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt{2}+\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{x}{\sqrt [4]{2-3 x}} \, dx,x,x^2\right )\\ &=\frac{8}{81} \left (2-3 x^2\right )^{3/4}+\frac{8}{27} \sqrt [4]{2} \tan ^{-1}\left (\frac{\sqrt{2}-\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac{8}{27} \sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt{2}+\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \left (\frac{2}{3 \sqrt [4]{2-3 x}}-\frac{1}{3} (2-3 x)^{3/4}\right ) \, dx,x,x^2\right )\\ &=\frac{4}{27} \left (2-3 x^2\right )^{3/4}-\frac{2}{189} \left (2-3 x^2\right )^{7/4}+\frac{8}{27} \sqrt [4]{2} \tan ^{-1}\left (\frac{\sqrt{2}-\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )+\frac{8}{27} \sqrt [4]{2} \tanh ^{-1}\left (\frac{\sqrt{2}+\sqrt{2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0215337, size = 42, normalized size = 0.35 \[ \frac{2}{567} \left (2-3 x^2\right )^{3/4} \left (9 \left (x^2+4\right )-56 \, _2F_1\left (\frac{3}{4},1;\frac{7}{4};\frac{3 x^2}{2}-1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/((2 - 3*x^2)^(1/4)*(4 - 3*x^2)),x]

[Out]

(2*(2 - 3*x^2)^(3/4)*(9*(4 + x^2) - 56*Hypergeometric2F1[3/4, 1, 7/4, -1 + (3*x^2)/2]))/567

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Maple [F]  time = 0.061, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{-3\,{x}^{2}+4}{\frac{1}{\sqrt [4]{-3\,{x}^{2}+2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(-3*x^2+2)^(1/4)/(-3*x^2+4),x)

[Out]

int(x^5/(-3*x^2+2)^(1/4)/(-3*x^2+4),x)

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Maxima [A]  time = 1.52376, size = 189, normalized size = 1.56 \begin{align*} -\frac{2}{189} \,{\left (-3 \, x^{2} + 2\right )}^{\frac{7}{4}} - \frac{8}{27} \cdot 2^{\frac{1}{4}} \arctan \left (\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} + 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) - \frac{8}{27} \cdot 2^{\frac{1}{4}} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} - 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) + \frac{4}{27} \cdot 2^{\frac{1}{4}} \log \left (2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) - \frac{4}{27} \cdot 2^{\frac{1}{4}} \log \left (-2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) + \frac{4}{27} \,{\left (-3 \, x^{2} + 2\right )}^{\frac{3}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="maxima")

[Out]

-2/189*(-3*x^2 + 2)^(7/4) - 8/27*2^(1/4)*arctan(1/2*2^(1/4)*(2^(3/4) + 2*(-3*x^2 + 2)^(1/4))) - 8/27*2^(1/4)*a
rctan(-1/2*2^(1/4)*(2^(3/4) - 2*(-3*x^2 + 2)^(1/4))) + 4/27*2^(1/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) +
sqrt(-3*x^2 + 2)) - 4/27*2^(1/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) + 4/27*(-3*x^2
+ 2)^(3/4)

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Fricas [B]  time = 1.34583, size = 784, normalized size = 6.48 \begin{align*} \frac{2}{63} \,{\left (x^{2} + 4\right )}{\left (-3 \, x^{2} + 2\right )}^{\frac{3}{4}} + \frac{8}{27} \cdot 8^{\frac{1}{4}} \sqrt{2} \arctan \left (\frac{1}{4} \cdot 8^{\frac{1}{4}} \sqrt{2} \sqrt{8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 4 \, \sqrt{2} + 4 \, \sqrt{-3 \, x^{2} + 2}} - \frac{1}{2} \cdot 8^{\frac{1}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} - 1\right ) + \frac{8}{27} \cdot 8^{\frac{1}{4}} \sqrt{2} \arctan \left (\frac{1}{8} \cdot 8^{\frac{1}{4}} \sqrt{2} \sqrt{-4 \cdot 8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 16 \, \sqrt{2} + 16 \, \sqrt{-3 \, x^{2} + 2}} - \frac{1}{2} \cdot 8^{\frac{1}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 1\right ) + \frac{2}{27} \cdot 8^{\frac{1}{4}} \sqrt{2} \log \left (4 \cdot 8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 16 \, \sqrt{2} + 16 \, \sqrt{-3 \, x^{2} + 2}\right ) - \frac{2}{27} \cdot 8^{\frac{1}{4}} \sqrt{2} \log \left (-4 \cdot 8^{\frac{3}{4}} \sqrt{2}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + 16 \, \sqrt{2} + 16 \, \sqrt{-3 \, x^{2} + 2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="fricas")

[Out]

2/63*(x^2 + 4)*(-3*x^2 + 2)^(3/4) + 8/27*8^(1/4)*sqrt(2)*arctan(1/4*8^(1/4)*sqrt(2)*sqrt(8^(3/4)*sqrt(2)*(-3*x
^2 + 2)^(1/4) + 4*sqrt(2) + 4*sqrt(-3*x^2 + 2)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) - 1) + 8/27*8^(1/4)*s
qrt(2)*arctan(1/8*8^(1/4)*sqrt(2)*sqrt(-4*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 16*sqrt(2) + 16*sqrt(-3*x^2 + 2
)) - 1/2*8^(1/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 1) + 2/27*8^(1/4)*sqrt(2)*log(4*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/
4) + 16*sqrt(2) + 16*sqrt(-3*x^2 + 2)) - 2/27*8^(1/4)*sqrt(2)*log(-4*8^(3/4)*sqrt(2)*(-3*x^2 + 2)^(1/4) + 16*s
qrt(2) + 16*sqrt(-3*x^2 + 2))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x^{5}}{3 x^{2} \sqrt [4]{2 - 3 x^{2}} - 4 \sqrt [4]{2 - 3 x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(-3*x**2+2)**(1/4)/(-3*x**2+4),x)

[Out]

-Integral(x**5/(3*x**2*(2 - 3*x**2)**(1/4) - 4*(2 - 3*x**2)**(1/4)), x)

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Giac [A]  time = 1.72053, size = 189, normalized size = 1.56 \begin{align*} -\frac{2}{189} \,{\left (-3 \, x^{2} + 2\right )}^{\frac{7}{4}} + \frac{1}{27} \cdot 8^{\frac{3}{4}} \log \left (2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) - \frac{1}{27} \cdot 8^{\frac{3}{4}} \log \left (-2^{\frac{3}{4}}{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}} + \sqrt{2} + \sqrt{-3 \, x^{2} + 2}\right ) - \frac{8}{27} \cdot 2^{\frac{1}{4}} \arctan \left (\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} + 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) - \frac{8}{27} \cdot 2^{\frac{1}{4}} \arctan \left (-\frac{1}{2} \cdot 2^{\frac{1}{4}}{\left (2^{\frac{3}{4}} - 2 \,{\left (-3 \, x^{2} + 2\right )}^{\frac{1}{4}}\right )}\right ) + \frac{4}{27} \,{\left (-3 \, x^{2} + 2\right )}^{\frac{3}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(-3*x^2+2)^(1/4)/(-3*x^2+4),x, algorithm="giac")

[Out]

-2/189*(-3*x^2 + 2)^(7/4) + 1/27*8^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 1/27*8
^(3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 8/27*2^(1/4)*arctan(1/2*2^(1/4)*(2^(3/4
) + 2*(-3*x^2 + 2)^(1/4))) - 8/27*2^(1/4)*arctan(-1/2*2^(1/4)*(2^(3/4) - 2*(-3*x^2 + 2)^(1/4))) + 4/27*(-3*x^2
+ 2)^(3/4)