3.7.26 \(\int \frac {x^2}{(1-x^3)^{2/3} (1+x^3)} \, dx\) [626]

Optimal. Leaf size=83 \[ -\frac {\tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}-\frac {\log \left (1+x^3\right )}{6\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 83, 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 = {455, 59, 631, 210, 31} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {2^{2/3} \sqrt [3]{1-x^3}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}-\frac {\log \left (x^3+1\right )}{6\ 2^{2/3}}+\frac {\log \left (\sqrt [3]{2}-\sqrt [3]{1-x^3}\right )}{2\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

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

Rule 59

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[3/(2*b*q^2), 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 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 455

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 631

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

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Mathematica [A]
time = 0.07, size = 102, normalized size = 1.23 \begin {gather*} -\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {1+2^{2/3} \sqrt [3]{1-x^3}}{\sqrt {3}}\right )-2 \log \left (-2+2^{2/3} \sqrt [3]{1-x^3}\right )+\log \left (2+2^{2/3} \sqrt [3]{1-x^3}+\sqrt [3]{2} \left (1-x^3\right )^{2/3}\right )}{6\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 2.44, size = 531, normalized size = 6.40

method	result	size
trager	\(\text {Expression too large to display}\)	\(531\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*ln(-(6*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^
2)*RootOf(_Z^3-2)^4*x^3-144*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3*x^3-x^3*Ro
otOf(_Z^3-2)^2+24*x^3*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)+7*RootOf(_Z^3-2)^2-1
68*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)+42*(-x^3+1)^(1/3)*RootOf(_Z^3-2)+252*(-
x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)-42*(-x^3+1)^(2/3))/(x+1)/(x^2-x+1))+1/6*Root
Of(_Z^3-2)*ln(-(180*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)^2*RootOf(_Z^3-2)^3*x^3+6*RootOf(RootO
f(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*RootOf(_Z^3-2)^4*x^3+90*x^3*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-
2)+36*_Z^2)*RootOf(_Z^3-2)+3*x^3*RootOf(_Z^3-2)^2-210*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)*Roo
tOf(_Z^3-2)+252*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-2)^2+6*_Z*RootOf(_Z^3-2)+36*_Z^2)-7*RootOf(_Z^3-2)^2+42*(-x^
3+1)^(1/3)*RootOf(_Z^3-2)-42*(-x^3+1)^(2/3))/(x+1)/(x^2-x+1))

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Maxima [A]
time = 0.54, size = 86, normalized size = 1.04 \begin {gather*} -\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{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 {1}{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 {1}{3}} \log \left (-2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.51, size = 98, normalized size = 1.18 \begin {gather*} -\frac {1}{6} \cdot 4^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {1}{6} \cdot 4^{\frac {1}{6}} {\left (4^{\frac {2}{3}} \sqrt {3} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 4^{\frac {1}{3}} \sqrt {3}\right )}\right ) - \frac {1}{24} \cdot 4^{\frac {2}{3}} \log \left (4^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 2 \cdot 4^{\frac {1}{3}}\right ) + \frac {1}{12} \cdot 4^{\frac {2}{3}} \log \left (-4^{\frac {2}{3}} + 2 \, {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*4^(1/6)*sqrt(3)*arctan(1/6*4^(1/6)*(4^(2/3)*sqrt(3)*(-x^3 + 1)^(1/3) + 4^(1/3)*sqrt(3))) - 1/24*4^(2/3)*l
og(4^(2/3)*(-x^3 + 1)^(1/3) + 2*(-x^3 + 1)^(2/3) + 2*4^(1/3)) + 1/12*4^(2/3)*log(-4^(2/3) + 2*(-x^3 + 1)^(1/3)
)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (- \left (x - 1\right ) \left (x^{2} + x + 1\right )\right )^{\frac {2}{3}} \left (x + 1\right ) \left (x^{2} - x + 1\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.29, size = 87, normalized size = 1.05 \begin {gather*} -\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{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 {1}{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 {1}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-x^{3} + 1\right )}^{\frac {1}{3}} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 5.05, size = 102, normalized size = 1.23 \begin {gather*} \frac {2^{1/3}\,\ln \left (3\,2^{1/3}-3\,{\left (1-x^3\right )}^{1/3}\right )}{6}+\frac {2^{1/3}\,\ln \left (3\,{\left (1-x^3\right )}^{1/3}-\frac {3\,2^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{12}-\frac {2^{1/3}\,\ln \left (\frac {3\,2^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{2}+3\,{\left (1-x^3\right )}^{1/3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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