3.1008 \(\int \frac{x^5}{\sqrt [3]{1-x^2} (3+x^2)} \, dx\)

Optimal. Leaf size=109 \[ \frac{3}{10} \left (1-x^2\right )^{5/3}+\frac{3}{2} \left (1-x^2\right )^{2/3}-\frac{9 \log \left (x^2+3\right )}{4\ 2^{2/3}}+\frac{27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}+\frac{9 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{2\ 2^{2/3}} \]

[Out]

(3*(1 - x^2)^(2/3))/2 + (3*(1 - x^2)^(5/3))/10 + (9*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3
)) - (9*Log[3 + x^2])/(4*2^(2/3)) + (27*Log[2^(2/3) - (1 - x^2)^(1/3)])/(4*2^(2/3))

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Rubi [A]  time = 0.0877867, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {446, 88, 55, 617, 204, 31} \[ \frac{3}{10} \left (1-x^2\right )^{5/3}+\frac{3}{2} \left (1-x^2\right )^{2/3}-\frac{9 \log \left (x^2+3\right )}{4\ 2^{2/3}}+\frac{27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}+\frac{9 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{2\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(1 - x^2)^(2/3))/2 + (3*(1 - x^2)^(5/3))/10 + (9*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]])/(2*2^(2/3
)) - (9*Log[3 + x^2])/(4*2^(2/3)) + (27*Log[2^(2/3) - (1 - x^2)^(1/3)])/(4*2^(2/3))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
 NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
 FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{x^5}{\sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{2}{\sqrt [3]{1-x}}-(1-x)^{2/3}+\frac{9}{\sqrt [3]{1-x} (3+x)}\right ) \, dx,x,x^2\right )\\ &=\frac{3}{2} \left (1-x^2\right )^{2/3}+\frac{3}{10} \left (1-x^2\right )^{5/3}+\frac{9}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=\frac{3}{2} \left (1-x^2\right )^{2/3}+\frac{3}{10} \left (1-x^2\right )^{5/3}-\frac{9 \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac{27}{4} \operatorname{Subst}\left (\int \frac{1}{2 \sqrt [3]{2}+2^{2/3} x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac{27 \operatorname{Subst}\left (\int \frac{1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}\\ &=\frac{3}{2} \left (1-x^2\right )^{2/3}+\frac{3}{10} \left (1-x^2\right )^{5/3}-\frac{9 \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac{27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}-\frac{27 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{2\ 2^{2/3}}\\ &=\frac{3}{2} \left (1-x^2\right )^{2/3}+\frac{3}{10} \left (1-x^2\right )^{5/3}+\frac{9 \sqrt{3} \tan ^{-1}\left (\frac{1+\sqrt [3]{2-2 x^2}}{\sqrt{3}}\right )}{2\ 2^{2/3}}-\frac{9 \log \left (3+x^2\right )}{4\ 2^{2/3}}+\frac{27 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{4\ 2^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0754938, size = 106, normalized size = 0.97 \[ -\frac{3}{40} \left (4 \left (1-x^2\right )^{2/3} x^2-24 \left (1-x^2\right )^{2/3}+15 \sqrt [3]{2} \log \left (x^2+3\right )-45 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )-30 \sqrt [3]{2} \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(-24*(1 - x^2)^(2/3) + 4*x^2*(1 - x^2)^(2/3) - 30*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]]
+ 15*2^(1/3)*Log[3 + x^2] - 45*2^(1/3)*Log[2^(2/3) - (1 - x^2)^(1/3)]))/40

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{{x}^{2}+3}{\frac{1}{\sqrt [3]{-{x}^{2}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.48205, size = 146, normalized size = 1.34 \begin{align*} \frac{9}{8} \cdot 4^{\frac{2}{3}} \sqrt{3} \arctan \left (\frac{1}{12} \cdot 4^{\frac{2}{3}} \sqrt{3}{\left (4^{\frac{1}{3}} + 2 \,{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right )}\right ) + \frac{3}{10} \,{\left (-x^{2} + 1\right )}^{\frac{5}{3}} - \frac{9}{16} \cdot 4^{\frac{2}{3}} \log \left (4^{\frac{2}{3}} + 4^{\frac{1}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right ) + \frac{9}{8} \cdot 4^{\frac{2}{3}} \log \left (-4^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) + \frac{3}{2} \,{\left (-x^{2} + 1\right )}^{\frac{2}{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

9/8*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 3/10*(-x^2 + 1)^(5/3) - 9/16
*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 9/8*4^(2/3)*log(-4^(1/3) + (-x^2 + 1)^(1
/3)) + 3/2*(-x^2 + 1)^(2/3)

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Fricas [A]  time = 1.51807, size = 319, normalized size = 2.93 \begin{align*} -\frac{3}{10} \,{\left (x^{2} - 6\right )}{\left (-x^{2} + 1\right )}^{\frac{2}{3}} + \frac{9}{4} \cdot 4^{\frac{1}{6}} \sqrt{3} \arctan \left (\frac{1}{6} \cdot 4^{\frac{1}{6}} \sqrt{3}{\left (4^{\frac{1}{3}} + 2 \,{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right )}\right ) - \frac{9}{16} \cdot 4^{\frac{2}{3}} \log \left (4^{\frac{2}{3}} + 4^{\frac{1}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right ) + \frac{9}{8} \cdot 4^{\frac{2}{3}} \log \left (-4^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/10*(x^2 - 6)*(-x^2 + 1)^(2/3) + 9/4*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3
))) - 9/16*4^(2/3)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 9/8*4^(2/3)*log(-4^(1/3) + (-x
^2 + 1)^(1/3))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{5}}{\sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError