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

Optimal. Leaf size=88 \[ -\frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \]

[Out]

-1/12*ln(x^3+1)*2^(2/3)+1/4*ln(-2^(1/3)*x-(-x^3+1)^(1/3))*2^(2/3)-1/6*arctan(1/3*(1-2*2^(1/3)*x/(-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 = 122, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {377, 200, 31, 634, 617, 204, 628} \[ -\frac {\log \left (\frac {2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}-\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )}{3 \sqrt [3]{2}}-\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - x^3)^(1/3)*(1 + x^3)),x]

[Out]

-(ArcTan[(1 - (2*2^(1/3)*x)/(1 - x^3)^(1/3))/Sqrt[3]]/(2^(1/3)*Sqrt[3])) - Log[1 + (2^(2/3)*x^2)/(1 - x^3)^(2/
3) - (2^(1/3)*x)/(1 - x^3)^(1/3)]/(6*2^(1/3)) + Log[1 + (2^(1/3)*x)/(1 - x^3)^(1/3)]/(3*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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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 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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[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]

Rule 628

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 634

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

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Mathematica [A]  time = 0.07, size = 112, normalized size = 1.27 \[ \frac {2 \log \left (\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+1\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}-1}{\sqrt {3}}\right )-\log \left (-\frac {\sqrt [3]{2} x}{\sqrt [3]{1-x^3}}+\frac {2^{2/3} x^2}{\left (1-x^3\right )^{2/3}}+1\right )}{6 \sqrt [3]{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 4.67, size = 253, normalized size = 2.88 \[ -\frac {1}{18} \, \sqrt {6} 2^{\frac {1}{6}} \arctan \left (\frac {2^{\frac {1}{6}} {\left (6 \, \sqrt {6} 2^{\frac {2}{3}} {\left (5 \, x^{7} + 4 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - \sqrt {6} 2^{\frac {1}{3}} {\left (71 \, x^{9} - 111 \, x^{6} + 33 \, x^{3} - 1\right )} + 12 \, \sqrt {6} {\left (19 \, x^{8} - 16 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}\right )}}{6 \, {\left (109 \, x^{9} - 105 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) + \frac {1}{18} \cdot 2^{\frac {2}{3}} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 2^{\frac {2}{3}} {\left (x^{3} + 1\right )} + 6 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} + 1}\right ) - \frac {1}{36} \cdot 2^{\frac {2}{3}} \log \left (\frac {3 \cdot 2^{\frac {2}{3}} {\left (5 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (19 \, x^{6} - 16 \, x^{3} + 1\right )} - 12 \, {\left (2 \, x^{5} - x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}}}{x^{6} + 2 \, x^{3} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/18*sqrt(6)*2^(1/6)*arctan(1/6*2^(1/6)*(6*sqrt(6)*2^(2/3)*(5*x^7 + 4*x^4 - x)*(-x^3 + 1)^(2/3) - sqrt(6)*2^(
1/3)*(71*x^9 - 111*x^6 + 33*x^3 - 1) + 12*sqrt(6)*(19*x^8 - 16*x^5 + x^2)*(-x^3 + 1)^(1/3))/(109*x^9 - 105*x^6
+ 3*x^3 + 1)) + 1/18*2^(2/3)*log((6*2^(1/3)*(-x^3 + 1)^(1/3)*x^2 + 2^(2/3)*(x^3 + 1) + 6*(-x^3 + 1)^(2/3)*x)/
(x^3 + 1)) - 1/36*2^(2/3)*log((3*2^(2/3)*(5*x^4 - x)*(-x^3 + 1)^(2/3) + 2^(1/3)*(19*x^6 - 16*x^3 + 1) - 12*(2*
x^5 - x^2)*(-x^3 + 1)^(1/3))/(x^6 + 2*x^3 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.42, size = 614, normalized size = 6.98 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*RootOf(_Z^3-4)*ln((-27*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-3*RootO
f(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^3*x^3+6*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-4)^2+
6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)^2*x-3*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+3
6*_Z^2)*RootOf(_Z^3-4)*x^2+2*(-x^3+1)^(1/3)*RootOf(_Z^3-4)^2*x^2-27*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4
)+36*_Z^2)*x^3-3*RootOf(_Z^3-4)*x^3+5*(-x^3+1)^(2/3)*x+9*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)+
RootOf(_Z^3-4))/(x+1)/(x^2-x+1))+RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*ln(-(-36*RootOf(RootOf(_
Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)^2*RootOf(_Z^3-4)^2*x^3-9*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*
_Z^2)*RootOf(_Z^3-4)^3*x^3+12*(-x^3+1)^(2/3)*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-
4)^2*x+30*(-x^3+1)^(1/3)*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*RootOf(_Z^3-4)*x^2+4*(-x^3+1)^(1
/3)*RootOf(_Z^3-4)^2*x^2+12*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)*x^3+3*RootOf(_Z^3-4)*x^3-2*(-
x^3+1)^(2/3)*x-12*RootOf(RootOf(_Z^3-4)^2+6*_Z*RootOf(_Z^3-4)+36*_Z^2)-3*RootOf(_Z^3-4))/(x+1)/(x^2-x+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\left (1-x^3\right )}^{1/3}\,\left (x^3+1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1 - x^3)^(1/3)*(x^3 + 1)),x)

[Out]

int(1/((1 - x^3)^(1/3)*(x^3 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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