3.1020 \(\int \frac{x^5}{\sqrt [3]{1-x^2} (3+x^2)^2} \, dx\)

Optimal. Leaf size=116 \[ -\frac{3}{4} \left (1-x^2\right )^{2/3}-\frac{9 \left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}+\frac{21 \log \left (x^2+3\right )}{16\ 2^{2/3}}-\frac{63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}-\frac{21 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{8\ 2^{2/3}} \]

[Out]

(-3*(1 - x^2)^(2/3))/4 - (9*(1 - x^2)^(2/3))/(8*(3 + x^2)) - (21*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3
]])/(8*2^(2/3)) + (21*Log[3 + x^2])/(16*2^(2/3)) - (63*Log[2^(2/3) - (1 - x^2)^(1/3)])/(16*2^(2/3))

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Rubi [A]  time = 0.0774002, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {446, 89, 80, 55, 617, 204, 31} \[ -\frac{3}{4} \left (1-x^2\right )^{2/3}-\frac{9 \left (1-x^2\right )^{2/3}}{8 \left (x^2+3\right )}+\frac{21 \log \left (x^2+3\right )}{16\ 2^{2/3}}-\frac{63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}-\frac{21 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{8\ 2^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(1 - x^2)^(2/3))/4 - (9*(1 - x^2)^(2/3))/(8*(3 + x^2)) - (21*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3
]])/(8*2^(2/3)) + (21*Log[3 + x^2])/(16*2^(2/3)) - (63*Log[2^(2/3) - (1 - x^2)^(1/3)])/(16*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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{\sqrt [3]{1-x} (3+x)^2} \, dx,x,x^2\right )\\ &=-\frac{9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac{1}{8} \operatorname{Subst}\left (\int \frac{-9+4 x}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac{3}{4} \left (1-x^2\right )^{2/3}-\frac{9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}-\frac{21}{8} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac{3}{4} \left (1-x^2\right )^{2/3}-\frac{9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac{21 \log \left (3+x^2\right )}{16\ 2^{2/3}}-\frac{63}{16} \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{63 \operatorname{Subst}\left (\int \frac{1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ &=-\frac{3}{4} \left (1-x^2\right )^{2/3}-\frac{9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}+\frac{21 \log \left (3+x^2\right )}{16\ 2^{2/3}}-\frac{63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac{63 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{8\ 2^{2/3}}\\ &=-\frac{3}{4} \left (1-x^2\right )^{2/3}-\frac{9 \left (1-x^2\right )^{2/3}}{8 \left (3+x^2\right )}-\frac{21 \sqrt{3} \tan ^{-1}\left (\frac{1+\sqrt [3]{2-2 x^2}}{\sqrt{3}}\right )}{8\ 2^{2/3}}+\frac{21 \log \left (3+x^2\right )}{16\ 2^{2/3}}-\frac{63 \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.147833, size = 110, normalized size = 0.95 \[ \frac{3}{32} \left (-8 \left (1-x^2\right )^{2/3}-\frac{12 \left (1-x^2\right )^{2/3}}{x^2+3}+7 \sqrt [3]{2} \log \left (x^2+3\right )-21 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )-14 \sqrt [3]{2} \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-8*(1 - x^2)^(2/3) - (12*(1 - x^2)^(2/3))/(3 + x^2) - 14*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sq
rt[3]] + 7*2^(1/3)*Log[3 + x^2] - 21*2^(1/3)*Log[2^(2/3) - (1 - x^2)^(1/3)]))/32

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Maple [F]  time = 0.043, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{5}}{ \left ({x}^{2}+3 \right ) ^{2}}{\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)^2,x)

[Out]

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

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Maxima [A]  time = 1.46894, size = 155, normalized size = 1.34 \begin{align*} -\frac{21}{32} \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{21}{64} \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{21}{32} \cdot 4^{\frac{2}{3}} \log \left (-4^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) - \frac{3}{4} \,{\left (-x^{2} + 1\right )}^{\frac{2}{3}} - \frac{9 \,{\left (-x^{2} + 1\right )}^{\frac{2}{3}}}{8 \,{\left (x^{2} + 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)^2,x, algorithm="maxima")

[Out]

-21/32*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) + 21/64*4^(2/3)*log(4^(2/3)
+ 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) - 21/32*4^(2/3)*log(-4^(1/3) + (-x^2 + 1)^(1/3)) - 3/4*(-x^2 +
1)^(2/3) - 9/8*(-x^2 + 1)^(2/3)/(x^2 + 3)

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Fricas [A]  time = 1.54901, size = 478, normalized size = 4.12 \begin{align*} -\frac{3 \,{\left (28 \cdot 4^{\frac{1}{6}} \sqrt{3} \left (-1\right )^{\frac{1}{3}}{\left (x^{2} + 3\right )} \arctan \left (\frac{1}{6} \cdot 4^{\frac{1}{6}} \sqrt{3}{\left (2 \, \left (-1\right )^{\frac{1}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} - 4^{\frac{1}{3}}\right )}\right ) + 7 \cdot 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}}{\left (x^{2} + 3\right )} \log \left (4^{\frac{1}{3}} \left (-1\right )^{\frac{2}{3}}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} - 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right ) - 14 \cdot 4^{\frac{2}{3}} \left (-1\right )^{\frac{1}{3}}{\left (x^{2} + 3\right )} \log \left (-4^{\frac{1}{3}} \left (-1\right )^{\frac{2}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) + 8 \,{\left (2 \, x^{2} + 9\right )}{\left (-x^{2} + 1\right )}^{\frac{2}{3}}\right )}}{64 \,{\left (x^{2} + 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)^2,x, algorithm="fricas")

[Out]

-3/64*(28*4^(1/6)*sqrt(3)*(-1)^(1/3)*(x^2 + 3)*arctan(1/6*4^(1/6)*sqrt(3)*(2*(-1)^(1/3)*(-x^2 + 1)^(1/3) - 4^(
1/3))) + 7*4^(2/3)*(-1)^(1/3)*(x^2 + 3)*log(4^(1/3)*(-1)^(2/3)*(-x^2 + 1)^(1/3) - 4^(2/3)*(-1)^(1/3) + (-x^2 +
1)^(2/3)) - 14*4^(2/3)*(-1)^(1/3)*(x^2 + 3)*log(-4^(1/3)*(-1)^(2/3) + (-x^2 + 1)^(1/3)) + 8*(2*x^2 + 9)*(-x^2
+ 1)^(2/3))/(x^2 + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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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)^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError