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

Optimal. Leaf size=82 \[ -\frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \]

[Out]

-1/12*ln(x^3+1)*2^(2/3)+1/4*ln(2^(1/3)-(-x^3+1)^(1/3))*2^(2/3)+1/6*arctan(1/3*(1+2^(2/3)*(-x^3+1)^(1/3))*3^(1/
2))*2^(2/3)*3^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {444, 55, 617, 204, 31} \[ -\frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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 {x^2}{\sqrt [3]{1-x^3} \left (1+x^3\right )} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{1-x} (1+x)} \, dx,x,x^3\right )\\ &=-\frac {\log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{2^{2/3}+\sqrt [3]{2} x+x^2} \, dx,x,\sqrt [3]{1-x^3}\right )-\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt [3]{2}-x} \, dx,x,\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\\ &=-\frac {\log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2^{2/3} \sqrt [3]{1-x^3}\right )}{\sqrt [3]{2}}\\ &=\frac {\tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\log \left (1+x^3\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2 \sqrt [3]{2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 73, normalized size = 0.89 \[ \frac {-\log \left (x^3+1\right )+3 \log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{6 \sqrt [3]{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 90, normalized size = 1.10 \[ \frac {1}{6} \, \sqrt {6} 2^{\frac {1}{6}} \arctan \left (\frac {1}{6} \cdot 2^{\frac {1}{6}} {\left (\sqrt {6} 2^{\frac {1}{3}} + 2 \, \sqrt {6} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.88, size = 87, normalized size = 1.06 \[ \frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 3.43, size = 655, normalized size = 7.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*ln((15*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^
2)*RootOf(_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-5*RootO
f(_Z^3-4)*x^3-6*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3+21*(-x^3+1)^(1/3)*RootOf(_Z^3-4)^2+42
*(-x^3+1)^(2/3)+35*RootOf(_Z^3-4)+42*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2))/(x+1)/(x^2-x+1))-1/
6*ln((-12*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z^3-4)^2
+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-12*RootOf(_Z^3-4)*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*Roo
tOf(_Z^3-4)+36*_Z^2)*x^3+21*(-x^3+1)^(1/3)*RootOf(_Z^3-4)^2+42*(-x^3+1)^(2/3)+28*RootOf(_Z^3-4)-42*RootOf(Root
Of(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2))/(x+1)/(x^2-x+1))*RootOf(_Z^3-4)-ln((-12*RootOf(RootOf(_Z^3-4)^2+6*_
Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^3*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootO
f(_Z^3-4)^2*x^3-12*RootOf(_Z^3-4)*x^3+18*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3+21*(-x^3+1)^
(1/3)*RootOf(_Z^3-4)^2+42*(-x^3+1)^(2/3)+28*RootOf(_Z^3-4)-42*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_
Z^2))/(x+1)/(x^2-x+1))*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)

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maxima [A]  time = 1.51, size = 86, normalized size = 1.05 \[ \frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{12} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{6} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.55, size = 100, normalized size = 1.22 \[ \frac {2^{2/3}\,\ln \left ({\left (1-x^3\right )}^{1/3}-2^{1/3}\right )}{6}+\frac {2^{2/3}\,\ln \left ({\left (1-x^3\right )}^{1/3}-\frac {2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{12}-\frac {2^{2/3}\,\ln \left ({\left (1-x^3\right )}^{1/3}-\frac {2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{4}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2^(2/3)*log((1 - x^3)^(1/3) - 2^(1/3)))/6 + (2^(2/3)*log((1 - x^3)^(1/3) - (2^(1/3)*(3^(1/2)*1i - 1)^2)/4)*(3
^(1/2)*1i - 1))/12 - (2^(2/3)*log((1 - x^3)^(1/3) - (2^(1/3)*(3^(1/2)*1i + 1)^2)/4)*(3^(1/2)*1i + 1))/12

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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