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]

-1/6*(-x^2+1)^(2/3)/x^2-1/72*ln(x^2+3)*2^(1/3)+1/24*ln(2^(2/3)-(-x^2+1)^(1/3))*2^(1/3)+1/36*arctan(1/3*(1+(-2*
x^2+2)^(1/3))*3^(1/2))*3^(1/2)*2^(1/3)

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Rubi [A]  time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

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

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Mathematica [A]  time = 0.05, 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

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fricas [A]  time = 1.06, size = 115, normalized size = 1.19 \[ \frac {4 \cdot 4^{\frac {1}{6}} \sqrt {3} x^{2} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} {\left (4^{\frac {1}{3}} \sqrt {3} + 2 \, \sqrt {3} {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - 4^{\frac {2}{3}} x^{2} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + 2 \cdot 4^{\frac {2}{3}} x^{2} \log \left (-4^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - 24 \, {\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{144 \, x^{2}} \]

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

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giac [A]  time = 0.37, size = 100, normalized size = 1.03 \[ \frac {1}{72} \cdot 4^{\frac {2}{3}} \sqrt {3} \arctan \left (\frac {1}{12} \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (4^{\frac {1}{3}} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{144} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} + 4^{\frac {1}{3}} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} + {\left (-x^{2} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{72} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {1}{3}} - {\left (-x^{2} + 1\right )}^{\frac {1}{3}}\right ) - \frac {{\left (-x^{2} + 1\right )}^{\frac {2}{3}}}{6 \, x^{2}} \]

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]

1/72*4^(2/3)*sqrt(3)*arctan(1/12*4^(2/3)*sqrt(3)*(4^(1/3) + 2*(-x^2 + 1)^(1/3))) - 1/144*4^(2/3)*log(4^(2/3) +
4^(1/3)*(-x^2 + 1)^(1/3) + (-x^2 + 1)^(2/3)) + 1/72*4^(2/3)*log(4^(1/3) - (-x^2 + 1)^(1/3)) - 1/6*(-x^2 + 1)^
(2/3)/x^2

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maple [C]  time = 6.05, size = 754, normalized size = 7.77 \[ \text {result too large to display} \]

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]

1/6*(x^2-1)/x^2/(-x^2+1)^(1/3)-1/36*ln((64*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)^2*RootOf(_Z^3-
2)^2*x^2-8*x^2*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-168*(-x^2+1)^(1/3)*RootOf
(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)-42*RootOf(_Z^3-2)^2*(-x^2+1)^(1/3)-8*RootOf(Root
Of(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+RootOf(_Z^3-2)*x^2+42*(-x^2+1)^(2/3)+168*RootOf(RootOf(_Z^3-2)^2
+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-21*RootOf(_Z^3-2))/(x^2+3))*RootOf(_Z^3-2)-1/9*ln((64*RootOf(RootOf(_Z^3-2)^2+4*
_Z*RootOf(_Z^3-2)+16*_Z^2)^2*RootOf(_Z^3-2)^2*x^2-8*x^2*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_
Z^3-2)+16*_Z^2)-168*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)-42*Root
Of(_Z^3-2)^2*(-x^2+1)^(1/3)-8*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2+RootOf(_Z^3-2)*x^2+42*(
-x^2+1)^(2/3)+168*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)-21*RootOf(_Z^3-2))/(x^2+3))*RootOf(Root
Of(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)+1/36*RootOf(_Z^3-2)*ln((-96*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3
-2)+16*_Z^2)^2*RootOf(_Z^3-2)^2*x^2-8*x^2*RootOf(_Z^3-2)^3*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2
)-168*(-x^2+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*RootOf(_Z^3-2)-42*RootOf(_Z^3-2)^2*(
-x^2+1)^(1/3)-60*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)*x^2-5*RootOf(_Z^3-2)*x^2+42*(-x^2+1)^(2/
3)+252*RootOf(RootOf(_Z^3-2)^2+4*_Z*RootOf(_Z^3-2)+16*_Z^2)+21*RootOf(_Z^3-2))/(x^2+3))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{2} + 3\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{3}} x^{3}}\,{d x} \]

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)

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mupad [B]  time = 0.89, size = 120, normalized size = 1.24 \[ \frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{36}-\frac {2^{2/3}}{36}\right )}{36}-\frac {{\left (1-x^2\right )}^{2/3}}{6\,x^2}+\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{36}-\frac {2^{2/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{144}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{72}-\frac {2^{1/3}\,\ln \left (\frac {{\left (1-x^2\right )}^{1/3}}{36}-\frac {2^{2/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{144}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{72} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(1/3)*log((1 - x^2)^(1/3)/36 - 2^(2/3)/36))/36 - (1 - x^2)^(2/3)/(6*x^2) + (2^(1/3)*log((1 - x^2)^(1/3)/36
- (2^(2/3)*(3^(1/2)*1i - 1)^2)/144)*(3^(1/2)*1i - 1))/72 - (2^(1/3)*log((1 - x^2)^(1/3)/36 - (2^(2/3)*(3^(1/2)
*1i + 1)^2)/144)*(3^(1/2)*1i + 1))/72

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \sqrt [3]{- \left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 3\right )}\, dx \]

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)

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