3.1024 \(\int \frac{1}{x^3 \sqrt [3]{1-x^2} (3+x^2)^2} \, dx\)

Optimal. Leaf size=183 \[ -\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (x^2+3\right )}-\frac{5 \left (1-x^2\right )^{2/3}}{72 \left (x^2+3\right )}-\frac{\log \left (x^2+3\right )}{48\ 2^{2/3}}-\frac{1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{8\ 2^{2/3} \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^2}+1}{\sqrt{3}}\right )}{18 \sqrt{3}}+\frac{\log (x)}{54} \]

[Out]

(-5*(1 - x^2)^(2/3))/(72*(3 + x^2)) - (1 - x^2)^(2/3)/(6*x^2*(3 + x^2)) + ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[
3]]/(8*2^(2/3)*Sqrt[3]) - ArcTan[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]]/(18*Sqrt[3]) + Log[x]/54 - Log[3 + x^2]/(48*
2^(2/3)) - Log[1 - (1 - x^2)^(1/3)]/36 + Log[2^(2/3) - (1 - x^2)^(1/3)]/(16*2^(2/3))

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Rubi [A]  time = 0.130042, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.409, Rules used = {446, 103, 151, 156, 55, 618, 204, 31, 617} \[ -\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (x^2+3\right )}-\frac{5 \left (1-x^2\right )^{2/3}}{72 \left (x^2+3\right )}-\frac{\log \left (x^2+3\right )}{48\ 2^{2/3}}-\frac{1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{8\ 2^{2/3} \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^2}+1}{\sqrt{3}}\right )}{18 \sqrt{3}}+\frac{\log (x)}{54} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(1 - x^2)^(2/3))/(72*(3 + x^2)) - (1 - x^2)^(2/3)/(6*x^2*(3 + x^2)) + ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[
3]]/(8*2^(2/3)*Sqrt[3]) - ArcTan[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]]/(18*Sqrt[3]) + Log[x]/54 - Log[3 + x^2]/(48*
2^(2/3)) - Log[1 - (1 - x^2)^(1/3)]/36 + 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 103

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

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; 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]

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]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )^2} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} x^2 (3+x)^2} \, dx,x,x^2\right )\\ &=-\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1-\frac{4 x}{3}}{\sqrt [3]{1-x} x (3+x)^2} \, dx,x,x^2\right )\\ &=-\frac{5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}-\frac{1}{72} \operatorname{Subst}\left (\int \frac{4-\frac{5 x}{3}}{\sqrt [3]{1-x} x (3+x)} \, dx,x,x^2\right )\\ &=-\frac{5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}-\frac{1}{54} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} x} \, dx,x,x^2\right )+\frac{1}{24} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac{5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}+\frac{\log (x)}{54}-\frac{\log \left (3+x^2\right )}{48\ 2^{2/3}}+\frac{1}{36} \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\sqrt [3]{1-x^2}\right )-\frac{1}{36} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\sqrt [3]{1-x^2}\right )+\frac{1}{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{\operatorname{Subst}\left (\int \frac{1}{2^{2/3}-x} \, dx,x,\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ &=-\frac{5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}+\frac{\log (x)}{54}-\frac{\log \left (3+x^2\right )}{48\ 2^{2/3}}-\frac{1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}+\frac{1}{18} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1-x^2}\right )-\frac{\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{8\ 2^{2/3}}\\ &=-\frac{5 \left (1-x^2\right )^{2/3}}{72 \left (3+x^2\right )}-\frac{\left (1-x^2\right )^{2/3}}{6 x^2 \left (3+x^2\right )}+\frac{\tan ^{-1}\left (\frac{1+\sqrt [3]{2-2 x^2}}{\sqrt{3}}\right )}{8\ 2^{2/3} \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{1+2 \sqrt [3]{1-x^2}}{\sqrt{3}}\right )}{18 \sqrt{3}}+\frac{\log (x)}{54}-\frac{\log \left (3+x^2\right )}{48\ 2^{2/3}}-\frac{1}{36} \log \left (1-\sqrt [3]{1-x^2}\right )+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{16\ 2^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.149302, size = 171, normalized size = 0.93 \[ \frac{1}{864} \left (-\frac{144 \left (1-x^2\right )^{2/3}}{x^2 \left (x^2+3\right )}-\frac{60 \left (1-x^2\right )^{2/3}}{x^2+3}-9 \sqrt [3]{2} \log \left (x^2+3\right )-24 \log \left (1-\sqrt [3]{1-x^2}\right )+27 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )+18 \sqrt [3]{2} \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )-16 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{1-x^2}+1}{\sqrt{3}}\right )+16 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-60*(1 - x^2)^(2/3))/(3 + x^2) - (144*(1 - x^2)^(2/3))/(x^2*(3 + x^2)) + 18*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 -
2*x^2)^(1/3))/Sqrt[3]] - 16*Sqrt[3]*ArcTan[(1 + 2*(1 - x^2)^(1/3))/Sqrt[3]] + 16*Log[x] - 9*2^(1/3)*Log[3 + x
^2] - 24*Log[1 - (1 - x^2)^(1/3)] + 27*2^(1/3)*Log[2^(2/3) - (1 - x^2)^(1/3)])/864

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \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(1/x^3/(-x^2+1)^(1/3)/(x^2+3)^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.59575, size = 663, normalized size = 3.62 \begin{align*} \frac{36 \cdot 4^{\frac{1}{6}} \sqrt{3}{\left (x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac{1}{6} \cdot 4^{\frac{1}{6}}{\left (4^{\frac{1}{3}} \sqrt{3} + 2 \, \sqrt{3}{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right )}\right ) - 9 \cdot 4^{\frac{2}{3}}{\left (x^{4} + 3 \, x^{2}\right )} \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 ) + 18 \cdot 4^{\frac{2}{3}}{\left (x^{4} + 3 \, x^{2}\right )} \log \left (-4^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) - 32 \, \sqrt{3}{\left (x^{4} + 3 \, x^{2}\right )} \arctan \left (\frac{2}{3} \, \sqrt{3}{\left (-x^{2} + 1\right )}^{\frac{1}{3}} + \frac{1}{3} \, \sqrt{3}\right ) + 16 \,{\left (x^{4} + 3 \, x^{2}\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac{2}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}} + 1\right ) - 32 \,{\left (x^{4} + 3 \, x^{2}\right )} \log \left ({\left (-x^{2} + 1\right )}^{\frac{1}{3}} - 1\right ) - 24 \,{\left (5 \, x^{2} + 12\right )}{\left (-x^{2} + 1\right )}^{\frac{2}{3}}}{1728 \,{\left (x^{4} + 3 \, x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1728*(36*4^(1/6)*sqrt(3)*(x^4 + 3*x^2)*arctan(1/6*4^(1/6)*(4^(1/3)*sqrt(3) + 2*sqrt(3)*(-x^2 + 1)^(1/3))) -
9*4^(2/3)*(x^4 + 3*x^2)*log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 18*4^(2/3)*(x^4 + 3*x^2)*
log(-4^(1/3) + (-x^2 + 1)^(1/3)) - 32*sqrt(3)*(x^4 + 3*x^2)*arctan(2/3*sqrt(3)*(-x^2 + 1)^(1/3) + 1/3*sqrt(3))
+ 16*(x^4 + 3*x^2)*log((-x^2 + 1)^(2/3) + (-x^2 + 1)^(1/3) + 1) - 32*(x^4 + 3*x^2)*log((-x^2 + 1)^(1/3) - 1)
- 24*(5*x^2 + 12)*(-x^2 + 1)^(2/3))/(x^4 + 3*x^2)

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

[Out]

Exception raised: NotImplementedError