3.1012 \(\int \frac{1}{x^3 \sqrt [3]{1-x^2} (3+x^2)} \, dx\)

Optimal. Leaf size=97 \[ -\frac{\left (1-x^2\right )^{2/3}}{6 x^2}-\frac{\log \left (x^2+3\right )}{36\ 2^{2/3}}+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{12\ 2^{2/3}}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{6\ 2^{2/3} \sqrt{3}} \]

[Out]

-(1 - x^2)^(2/3)/(6*x^2) + ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]]/(6*2^(2/3)*Sqrt[3]) - Log[3 + x^2]/(36*2^(2
/3)) + Log[2^(2/3) - (1 - x^2)^(1/3)]/(12*2^(2/3))

________________________________________________________________________________________

Rubi [A]  time = 0.0744409, antiderivative size = 97, 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, 103, 12, 55, 617, 204, 31} \[ -\frac{\left (1-x^2\right )^{2/3}}{6 x^2}-\frac{\log \left (x^2+3\right )}{36\ 2^{2/3}}+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{12\ 2^{2/3}}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )}{6\ 2^{2/3} \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - x^2)^(2/3)/(6*x^2) + ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]]/(6*2^(2/3)*Sqrt[3]) - Log[3 + x^2]/(36*2^(2
/3)) + Log[2^(2/3) - (1 - x^2)^(1/3)]/(12*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 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{1}{x^3 \sqrt [3]{1-x^2} \left (3+x^2\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} x^2 (3+x)} \, dx,x,x^2\right )\\ &=-\frac{\left (1-x^2\right )^{2/3}}{6 x^2}-\frac{1}{6} \operatorname{Subst}\left (\int -\frac{1}{3 \sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac{\left (1-x^2\right )^{2/3}}{6 x^2}+\frac{1}{18} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-x} (3+x)} \, dx,x,x^2\right )\\ &=-\frac{\left (1-x^2\right )^{2/3}}{6 x^2}-\frac{\log \left (3+x^2\right )}{36\ 2^{2/3}}+\frac{1}{12} \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 )}{12\ 2^{2/3}}\\ &=-\frac{\left (1-x^2\right )^{2/3}}{6 x^2}-\frac{\log \left (3+x^2\right )}{36\ 2^{2/3}}+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{12\ 2^{2/3}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+\sqrt [3]{2-2 x^2}\right )}{6\ 2^{2/3}}\\ &=-\frac{\left (1-x^2\right )^{2/3}}{6 x^2}+\frac{\tan ^{-1}\left (\frac{1+\sqrt [3]{2-2 x^2}}{\sqrt{3}}\right )}{6\ 2^{2/3} \sqrt{3}}-\frac{\log \left (3+x^2\right )}{36\ 2^{2/3}}+\frac{\log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )}{12\ 2^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0447154, size = 93, normalized size = 0.96 \[ \frac{1}{72} \left (-\frac{12 \left (1-x^2\right )^{2/3}}{x^2}-\sqrt [3]{2} \log \left (x^2+3\right )+3 \sqrt [3]{2} \log \left (2^{2/3}-\sqrt [3]{1-x^2}\right )+2 \sqrt [3]{2} \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{2-2 x^2}+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*(1 - x^2)^(2/3))/x^2 + 2*2^(1/3)*Sqrt[3]*ArcTan[(1 + (2 - 2*x^2)^(1/3))/Sqrt[3]] - 2^(1/3)*Log[3 + x^2]
+ 3*2^(1/3)*Log[2^(2/3) - (1 - x^2)^(1/3)])/72

________________________________________________________________________________________

Maple [F]  time = 0.044, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ({x}^{2}+3 \right ) }{\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),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.52997, size = 332, normalized size = 3.42 \begin{align*} \frac{4 \cdot 4^{\frac{1}{6}} \sqrt{3} x^{2} \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 ) - 4^{\frac{2}{3}} x^{2} \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 ) + 2 \cdot 4^{\frac{2}{3}} x^{2} \log \left (-4^{\frac{1}{3}} +{\left (-x^{2} + 1\right )}^{\frac{1}{3}}\right ) - 24 \,{\left (-x^{2} + 1\right )}^{\frac{2}{3}}}{144 \, x^{2}} \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),x, algorithm="fricas")

[Out]

1/144*(4*4^(1/6)*sqrt(3)*x^2*arctan(1/6*4^(1/6)*(4^(1/3)*sqrt(3) + 2*sqrt(3)*(-x^2 + 1)^(1/3))) - 4^(2/3)*x^2*
log(4^(2/3) + 4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 2*4^(2/3)*x^2*log(-4^(1/3) + (-x^2 + 1)^(1/3)) -
24*(-x^2 + 1)^(2/3))/x^2

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError