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

Optimal. Leaf size=105 \[ \frac {\log \left (x^3+1\right )}{6 \sqrt [3]{2}}-\frac {\log \left (-\sqrt [3]{1-x^3}-\sqrt [3]{2} x\right )}{2 \sqrt [3]{2}}+\frac {\tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{2} x}{\sqrt [3]{1-x^3}}}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\left (1-x^3\right )^{2/3}}{2 x^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 139, normalized size of antiderivative = 1.32, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {494, 453, 200, 31, 634, 617, 204, 628} \begin {gather*} -\frac {\left (1-x^3\right )^{2/3}}{2 x^2}+\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1 - x^3)^(2/3)/(2*x^2) + 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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 494

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{k = Denominato
r[p]}, Dist[(k*a^(p + (m + 1)/n))/n, Subst[Int[(x^((k*(m + 1))/n - 1)*(c - (b*c - a*d)*x^k)^q)/(1 - b*x^k)^(p
+ q + (m + 1)/n + 1), x], x, x^(n/k)/(a + b*x^n)^(1/k)], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && Ration
alQ[m, p] && IntegersQ[p + (m + 1)/n, q] && LtQ[-1, p, 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]

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

________________________________________________________________________________________

Mathematica [C]  time = 0.60, size = 82, normalized size = 0.78 \begin {gather*} -\frac {\left (-6 x^6+4 x^3+2\right ) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {2 x^3}{x^3-1}\right )-3 x^3 \left (x^3+1\right ) \, _2F_1\left (\frac {4}{3},2;\frac {7}{3};\frac {2 x^3}{x^3-1}\right )}{4 x^2 \left (1-x^3\right )^{4/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/4*((2 + 4*x^3 - 6*x^6)*Hypergeometric2F1[1/3, 1, 4/3, (2*x^3)/(-1 + x^3)] - 3*x^3*(1 + x^3)*Hypergeometric2
F1[4/3, 2, 7/3, (2*x^3)/(-1 + x^3)])/(x^2*(1 - x^3)^(4/3))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.29, size = 145, normalized size = 1.38 \begin {gather*} -\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3}+2 x\right )}{3 \sqrt [3]{2}}-\frac {\tan ^{-1}\left (\frac {\sqrt {3} x}{2^{2/3} \sqrt [3]{1-x^3}-x}\right )}{\sqrt [3]{2} \sqrt {3}}-\frac {\left (1-x^3\right )^{2/3}}{2 x^2}+\frac {\log \left (2^{2/3} \sqrt [3]{1-x^3} x-\sqrt [3]{2} \left (1-x^3\right )^{2/3}-2 x^2\right )}{6 \sqrt [3]{2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 2.58, size = 307, normalized size = 2.92 \begin {gather*} -\frac {2 \, \sqrt {6} 2^{\frac {1}{6}} \left (-1\right )^{\frac {1}{3}} x^{2} \arctan \left (\frac {2^{\frac {1}{6}} {\left (6 \, \sqrt {6} 2^{\frac {2}{3}} \left (-1\right )^{\frac {2}{3}} {\left (5 \, x^{7} + 4 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 12 \, \sqrt {6} \left (-1\right )^{\frac {1}{3}} {\left (19 \, x^{8} - 16 \, x^{5} + x^{2}\right )} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} - \sqrt {6} 2^{\frac {1}{3}} {\left (71 \, x^{9} - 111 \, x^{6} + 33 \, x^{3} - 1\right )}\right )}}{6 \, {\left (109 \, x^{9} - 105 \, x^{6} + 3 \, x^{3} + 1\right )}}\right ) - 2 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x^{2} \log \left (\frac {6 \cdot 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{3}} {\left (-x^{3} + 1\right )}^{\frac {1}{3}} x^{2} - 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (x^{3} + 1\right )} + 6 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}} x}{x^{3} + 1}\right ) + 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} x^{2} \log \left (-\frac {3 \cdot 2^{\frac {2}{3}} \left (-1\right )^{\frac {1}{3}} {\left (5 \, x^{4} - x\right )} {\left (-x^{3} + 1\right )}^{\frac {2}{3}} - 2^{\frac {1}{3}} \left (-1\right )^{\frac {2}{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 ) + 18 \, {\left (-x^{3} + 1\right )}^{\frac {2}{3}}}{36 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 4.49, size = 929, normalized size = 8.85

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x^3-1)/x^2/(-x^3+1)^(1/3)-1/6*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+3*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(_Z^3+4)*RootOf(Root
Of(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^2-(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x^2+36*RootOf(RootOf(_Z^3+4)^2+6
*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^3-3*RootOf(_Z^3+4)*x^3-10*(-x^3+1)^(2/3)*x-12*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootO
f(_Z^3+4)+36*_Z^2)+RootOf(_Z^3+4))/(x+1)/(x^2-x+1))*RootOf(_Z^3+4)-ln(-(-36*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootO
f(_Z^3+4)+36*_Z^2)^2*RootOf(_Z^3+4)^2*x^3+3*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(_Z^3+4)*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^2-(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*
x^2+36*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^3-3*RootOf(_Z^3+4)*x^3-10*(-x^3+1)^(2/3)*x-12*Ro
otOf(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)+1/6*RootOf(_Z^3+4)*ln((-54*RootOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)^2*R
ootOf(_Z^3+4)^2*x^3-3*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+6*(-x^3+1)^(1/3)*RootOf(_Z^3+4)*Ro
otOf(RootOf(_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^2+5*(-x^3+1)^(1/3)*RootOf(_Z^3+4)^2*x^2-18*RootOf(RootOf(
_Z^3+4)^2+6*_Z*RootOf(_Z^3+4)+36*_Z^2)*x^3-RootOf(_Z^3+4)*x^3+2*(-x^3+1)^(2/3)*x+18*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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________