3.2.16 \(\int \frac {\sqrt [3]{1-x^3}}{1+x^3} \, dx\) [116]

Optimal. Leaf size=272 \[ \frac {\sqrt [3]{2} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (2^{2/3}-\frac {1-x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}-\frac {\log \left (1+\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}+\frac {1}{3} \sqrt [3]{2} \log \left (1+\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}\right )-\frac {\log \left (2 \sqrt [3]{2}+\frac {(1-x)^2}{\left (1-x^3\right )^{2/3}}+\frac {2^{2/3} (1-x)}{\sqrt [3]{1-x^3}}\right )}{6\ 2^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {420, 493, 298, 31, 648, 631, 210, 642} \begin {gather*} \frac {\sqrt [3]{2} \text {ArcTan}\left (\frac {1-\frac {2 \sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\text {ArcTan}\left (\frac {\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1}{\sqrt {3}}\right )}{2^{2/3} \sqrt {3}}+\frac {\log \left (2^{2/3}-\frac {1-x}{\sqrt [3]{1-x^3}}\right )}{3\ 2^{2/3}}-\frac {\log \left (\frac {2^{2/3} (1-x)^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )}{3\ 2^{2/3}}+\frac {1}{3} \sqrt [3]{2} \log \left (\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{1-x^3}}+1\right )-\frac {\log \left (\frac {(1-x)^2}{\left (1-x^3\right )^{2/3}}+\frac {2^{2/3} (1-x)}{\sqrt [3]{1-x^3}}+2 \sqrt [3]{2}\right )}{6\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - x^3)^(1/3)/(1 + x^3),x]

[Out]

(2^(1/3)*ArcTan[(1 - (2*2^(1/3)*(1 - x))/(1 - x^3)^(1/3))/Sqrt[3]])/Sqrt[3] + ArcTan[(1 + (2^(1/3)*(1 - x))/(1
- x^3)^(1/3))/Sqrt[3]]/(2^(2/3)*Sqrt[3]) + Log[2^(2/3) - (1 - x)/(1 - x^3)^(1/3)]/(3*2^(2/3)) - Log[1 + (2^(2
/3)*(1 - x)^2)/(1 - x^3)^(2/3) - (2^(1/3)*(1 - x))/(1 - x^3)^(1/3)]/(3*2^(2/3)) + (2^(1/3)*Log[1 + (2^(1/3)*(1
- x))/(1 - x^3)^(1/3)])/3 - Log[2*2^(1/3) + (1 - x)^2/(1 - x^3)^(2/3) + (2^(2/3)*(1 - x))/(1 - x^3)^(1/3)]/(6
*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 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 298

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Dist[-(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 420

Int[((a_) + (b_.)*(x_)^3)^(1/3)/((c_) + (d_.)*(x_)^3), x_Symbol] :> With[{q = Rt[b/a, 3]}, Dist[9*(a/(c*q)), S
ubst[Int[x/((4 - a*x^3)*(1 + 2*a*x^3)), x], x, (1 + q*x)/(a + b*x^3)^(1/3)], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[b*c - a*d, 0] && EqQ[b*c + a*d, 0]

Rule 493

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[b/(b*c - a*d), I
nt[(e*x)^m/(a + b*x^n), x], x] - Dist[d/(b*c - a*d), Int[(e*x)^m/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e,
m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 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]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {\sqrt [3]{1-x^3}}{1+x^3} \, dx &=x F_1\left (\frac {1}{3};-\frac {1}{3},1;\frac {4}{3};x^3,-x^3\right )\\ \end {align*}

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Mathematica [A]
time = 2.14, size = 283, normalized size = 1.04 \begin {gather*} -\frac {2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{1-x^3}}{\sqrt [3]{2}-\sqrt [3]{2} x+\sqrt [3]{1-x^3}}\right )+4 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{1-x^3}}{-2 \sqrt [3]{2}+2 \sqrt [3]{2} x+\sqrt [3]{1-x^3}}\right )-4 \log \left (-\sqrt [3]{2}+\sqrt [3]{2} x-\sqrt [3]{1-x^3}\right )-2 \log \left (-\sqrt [3]{2}+\sqrt [3]{2} x+2 \sqrt [3]{1-x^3}\right )+2 \log \left (2^{2/3}-2\ 2^{2/3} x+2^{2/3} x^2+(-1+x) \sqrt [3]{2-2 x^3}+\left (1-x^3\right )^{2/3}\right )+\log \left (2^{2/3}-2\ 2^{2/3} x+2^{2/3} x^2-2 (-1+x) \sqrt [3]{2-2 x^3}+4 \left (1-x^3\right )^{2/3}\right )}{6\ 2^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(2*Sqrt[3]*ArcTan[(Sqrt[3]*(1 - x^3)^(1/3))/(2^(1/3) - 2^(1/3)*x + (1 - x^3)^(1/3))] + 4*Sqrt[3]*ArcTan[(
Sqrt[3]*(1 - x^3)^(1/3))/(-2*2^(1/3) + 2*2^(1/3)*x + (1 - x^3)^(1/3))] - 4*Log[-2^(1/3) + 2^(1/3)*x - (1 - x^3
)^(1/3)] - 2*Log[-2^(1/3) + 2^(1/3)*x + 2*(1 - x^3)^(1/3)] + 2*Log[2^(2/3) - 2*2^(2/3)*x + 2^(2/3)*x^2 + (-1 +
x)*(2 - 2*x^3)^(1/3) + (1 - x^3)^(2/3)] + Log[2^(2/3) - 2*2^(2/3)*x + 2^(2/3)*x^2 - 2*(-1 + x)*(2 - 2*x^3)^(1
/3) + 4*(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 = 4.35, size = 681, normalized size = 2.50

method	result	size
trager	\(\text {Expression too large to display}\)	\(681\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*RootOf(_Z^3-2)*ln(-(18*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)^2*RootOf(_Z^3-2)^2*x^3+12*RootO
f(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)^3*x^3-3*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3
-2)+9*_Z^2)*x^6-2*RootOf(_Z^3-2)*x^6+18*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)*(-x
^3+1)^(2/3)*x^2+6*(-x^3+1)^(1/3)*x^4+18*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*x^3+12*RootOf(_Z^3
-2)*x^3-6*x*(-x^3+1)^(1/3)-3*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)-2*RootOf(_Z^3-2))/(1+x)^2/(x^
2-x+1)^2)+1/2*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*ln((-18*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(
_Z^3-2)+9*_Z^2)^2*RootOf(_Z^3-2)^3*x^3-12*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)^4
*x^3-3*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)*x^6-2*RootOf(_Z^3-2)^2*x^6+18*RootOf
(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)^2*(-x^3+1)^(2/3)*x^2-18*RootOf(RootOf(_Z^3-2)^2+3
*_Z*RootOf(_Z^3-2)+9*_Z^2)*(-x^3+1)^(1/3)*x^4+6*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^
3-2)*x^3+4*RootOf(_Z^3-2)^2*x^3+12*(-x^3+1)^(2/3)*x^2+18*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*(
-x^3+1)^(1/3)*x-3*RootOf(RootOf(_Z^3-2)^2+3*_Z*RootOf(_Z^3-2)+9*_Z^2)*RootOf(_Z^3-2)-2*RootOf(_Z^3-2)^2)/(1+x)
^2/(x^2-x+1)^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 3.51, size = 341, normalized size = 1.25 \begin {gather*} \frac {1}{18} \, \sqrt {3} 2^{\frac {1}{3}} \arctan \left (-\frac {6 \, \sqrt {3} 2^{\frac {2}{3}} {\left (x^{16} - 33 \, x^{13} + 110 \, x^{10} - 110 \, x^{7} + 33 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - 24 \, \sqrt {3} 2^{\frac {1}{3}} {\left (x^{14} - 2 \, x^{11} - 6 \, x^{8} - 2 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - \sqrt {3} {\left (x^{18} + 42 \, x^{15} - 417 \, x^{12} + 812 \, x^{9} - 417 \, x^{6} + 42 \, x^{3} + 1\right )}}{3 \, {\left (x^{18} - 102 \, x^{15} + 447 \, x^{12} - 628 \, x^{9} + 447 \, x^{6} - 102 \, x^{3} + 1\right )}}\right ) + \frac {1}{18} \cdot 2^{\frac {1}{3}} \log \left (-\frac {12 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x^{2} + 2^{\frac {2}{3}} {\left (x^{6} + 2 \, x^{3} + 1\right )} - 6 \cdot 2^{\frac {1}{3}} {\left (x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} + 2 \, x^{3} + 1}\right ) - \frac {1}{36} \cdot 2^{\frac {1}{3}} \log \left (\frac {12 \cdot 2^{\frac {2}{3}} {\left (x^{8} - 4 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{12} - 32 \, x^{9} + 78 \, x^{6} - 32 \, x^{3} + 1\right )} + 6 \, {\left (x^{10} - 11 \, x^{7} + 11 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{12} + 4 \, x^{9} + 6 \, x^{6} + 4 \, x^{3} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*sqrt(3)*2^(1/3)*arctan(-1/3*(6*sqrt(3)*2^(2/3)*(x^16 - 33*x^13 + 110*x^10 - 110*x^7 + 33*x^4 - x)*(-x^3 +
1)^(1/3) - 24*sqrt(3)*2^(1/3)*(x^14 - 2*x^11 - 6*x^8 - 2*x^5 + x^2)*(-x^3 + 1)^(2/3) - sqrt(3)*(x^18 + 42*x^1
5 - 417*x^12 + 812*x^9 - 417*x^6 + 42*x^3 + 1))/(x^18 - 102*x^15 + 447*x^12 - 628*x^9 + 447*x^6 - 102*x^3 + 1)
) + 1/18*2^(1/3)*log(-(12*(-x^3 + 1)^(2/3)*x^2 + 2^(2/3)*(x^6 + 2*x^3 + 1) - 6*2^(1/3)*(x^4 - x)*(-x^3 + 1)^(1
/3))/(x^6 + 2*x^3 + 1)) - 1/36*2^(1/3)*log((12*2^(2/3)*(x^8 - 4*x^5 + x^2)*(-x^3 + 1)^(2/3) + 2^(1/3)*(x^12 -
32*x^9 + 78*x^6 - 32*x^3 + 1) + 6*(x^10 - 11*x^7 + 11*x^4 - x)*(-x^3 + 1)^(1/3))/(x^12 + 4*x^9 + 6*x^6 + 4*x^3
+ 1))

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - x^3)^(1/3)/(x^3 + 1),x)

[Out]

int((1 - x^3)^(1/3)/(x^3 + 1), x)

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