\(\int \frac {(-1+x^3)^{2/3} (2+x^3)}{x^6 (-4-2 x^3+x^6)} \, dx\) [1169]

Optimal result

Integrand size = 30, antiderivative size = 86 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=\frac {\left (1-x^3\right ) \left (-1+x^3\right )^{2/3}}{10 x^5}-\frac {1}{12} \text {RootSum}\left [5-10 \text {$\#$1}^3+4 \text {$\#$1}^6\&,\frac {-\log (x) \text {$\#$1}^2+\log \left (\sqrt [3]{-1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-5+4 \text {$\#$1}^3}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(319\) vs. \(2(86)=172\).

Time = 0.45 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.71, number of steps used = 10, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6860, 270, 1442, 399, 245, 384} \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=-\frac {\sqrt [3]{\frac {1}{5} \left (3 \sqrt {5}-5\right )} \arctan \left (\frac {\frac {\sqrt [3]{2 \left (5-\sqrt {5}\right )} x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {\sqrt [3]{\frac {1}{5} \left (5+3 \sqrt {5}\right )} \arctan \left (\frac {\frac {\sqrt [3]{2 \left (5+\sqrt {5}\right )} x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{48} \sqrt [3]{\frac {1}{5} \left (5+3 \sqrt {5}\right )} \log \left (2 x^3-2 \left (1-\sqrt {5}\right )\right )-\frac {1}{48} \sqrt [3]{\frac {1}{5} \left (3 \sqrt {5}-5\right )} \log \left (2 x^3-2 \left (1+\sqrt {5}\right )\right )+\frac {1}{16} \sqrt [3]{\frac {1}{5} \left (3 \sqrt {5}-5\right )} \log \left (\frac {\sqrt [3]{5-\sqrt {5}} x}{2^{2/3}}-\sqrt [3]{x^3-1}\right )-\frac {1}{16} \sqrt [3]{\frac {1}{5} \left (5+3 \sqrt {5}\right )} \log \left (\frac {\sqrt [3]{5+\sqrt {5}} x}{2^{2/3}}-\sqrt [3]{x^3-1}\right )-\frac {\left (x^3-1\right )^{5/3}}{10 x^5} \]

[In]

Int[((-1 + x^3)^(2/3)*(2 + x^3))/(x^6*(-4 - 2*x^3 + x^6)),x]

[Out]

-1/10*(-1 + x^3)^(5/3)/x^5 - (((-5 + 3*Sqrt[5])/5)^(1/3)*ArcTan[(1 + ((2*(5 - Sqrt[5]))^(1/3)*x)/(-1 + x^3)^(1
/3))/Sqrt[3]])/(8*Sqrt[3]) + (((5 + 3*Sqrt[5])/5)^(1/3)*ArcTan[(1 + ((2*(5 + Sqrt[5]))^(1/3)*x)/(-1 + x^3)^(1/
3))/Sqrt[3]])/(8*Sqrt[3]) + (((5 + 3*Sqrt[5])/5)^(1/3)*Log[-2*(1 - Sqrt[5]) + 2*x^3])/48 - (((-5 + 3*Sqrt[5])/
5)^(1/3)*Log[-2*(1 + Sqrt[5]) + 2*x^3])/48 + (((-5 + 3*Sqrt[5])/5)^(1/3)*Log[((5 - Sqrt[5])^(1/3)*x)/2^(2/3) -
(-1 + x^3)^(1/3)])/16 - (((5 + 3*Sqrt[5])/5)^(1/3)*Log[((5 + Sqrt[5])^(1/3)*x)/2^(2/3) - (-1 + x^3)^(1/3)])/1
6

Rule 245

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

Rule 270

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

Rule 384

Int[1/(((a_) + (b_.)*(x_)^3)^(1/3)*((c_) + (d_.)*(x_)^3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/c, 3]}, Simp[
ArcTan[(1 + (2*q*x)/(a + b*x^3)^(1/3))/Sqrt[3]]/(Sqrt[3]*c*q), x] + (-Simp[Log[q*x - (a + b*x^3)^(1/3)]/(2*c*q
), x] + Simp[Log[c + d*x^3]/(6*c*q), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 399

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

Rule 1442

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[2*(c/r), Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[2*(c/r), Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\left (-1+x^3\right )^{2/3}}{2 x^6}+\frac {\left (-1+x^3\right )^{2/3}}{2 \left (-4-2 x^3+x^6\right )}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {\left (-1+x^3\right )^{2/3}}{x^6} \, dx\right )+\frac {1}{2} \int \frac {\left (-1+x^3\right )^{2/3}}{-4-2 x^3+x^6} \, dx \\ & = -\frac {\left (-1+x^3\right )^{5/3}}{10 x^5}+\frac {\int \frac {\left (-1+x^3\right )^{2/3}}{-2-2 \sqrt {5}+2 x^3} \, dx}{2 \sqrt {5}}-\frac {\int \frac {\left (-1+x^3\right )^{2/3}}{-2+2 \sqrt {5}+2 x^3} \, dx}{2 \sqrt {5}} \\ & = -\frac {\left (-1+x^3\right )^{5/3}}{10 x^5}+\frac {1}{2} \int \frac {1}{\sqrt [3]{-1+x^3} \left (-2-2 \sqrt {5}+2 x^3\right )} \, dx+\frac {1}{2} \int \frac {1}{\sqrt [3]{-1+x^3} \left (-2+2 \sqrt {5}+2 x^3\right )} \, dx \\ & = -\frac {\left (-1+x^3\right )^{5/3}}{10 x^5}-\frac {\sqrt [3]{\frac {1}{5} \left (-5+3 \sqrt {5}\right )} \arctan \left (\frac {1+\frac {\sqrt [3]{2 \left (5-\sqrt {5}\right )} x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {\sqrt [3]{\frac {1}{5} \left (5+3 \sqrt {5}\right )} \arctan \left (\frac {1+\frac {\sqrt [3]{2 \left (5+\sqrt {5}\right )} x}{\sqrt [3]{-1+x^3}}}{\sqrt {3}}\right )}{8 \sqrt {3}}+\frac {1}{48} \sqrt [3]{\frac {1}{5} \left (5+3 \sqrt {5}\right )} \log \left (-2 \left (1-\sqrt {5}\right )+2 x^3\right )-\frac {1}{48} \sqrt [3]{\frac {1}{5} \left (-5+3 \sqrt {5}\right )} \log \left (-2 \left (1+\sqrt {5}\right )+2 x^3\right )+\frac {1}{16} \sqrt [3]{\frac {1}{5} \left (-5+3 \sqrt {5}\right )} \log \left (\frac {\sqrt [3]{5-\sqrt {5}} x}{2^{2/3}}-\sqrt [3]{-1+x^3}\right )-\frac {1}{16} \sqrt [3]{\frac {1}{5} \left (5+3 \sqrt {5}\right )} \log \left (\frac {\sqrt [3]{5+\sqrt {5}} x}{2^{2/3}}-\sqrt [3]{-1+x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=-\frac {6 \left (-1+x^3\right )^{5/3}+5 x^5 \text {RootSum}\left [5-10 \text {$\#$1}^3+4 \text {$\#$1}^6\&,\frac {-\log (x) \text {$\#$1}^2+\log \left (\sqrt [3]{-1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-5+4 \text {$\#$1}^3}\&\right ]}{60 x^5} \]

[In]

Integrate[((-1 + x^3)^(2/3)*(2 + x^3))/(x^6*(-4 - 2*x^3 + x^6)),x]

[Out]

-1/60*(6*(-1 + x^3)^(5/3) + 5*x^5*RootSum[5 - 10*#1^3 + 4*#1^6 & , (-(Log[x]*#1^2) + Log[(-1 + x^3)^(1/3) - x*
#1]*#1^2)/(-5 + 4*#1^3) & ])/x^5

Maple [N/A] (verified)

Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92

method	result	size
pseudoelliptic	\(\frac {-5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{6}-10 \textit {\_Z}^{3}+5\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (\frac {-\textit {\_R} x +\left (x^{3}-1\right )^{\frac {1}{3}}}{x}\right )}{4 \textit {\_R}^{3}-5}\right ) x^{5}-6 x^{3} \left (x^{3}-1\right )^{\frac {2}{3}}+6 \left (x^{3}-1\right )^{\frac {2}{3}}}{60 x^{5}}\)	\(79\)
risch	\(\text {Expression too large to display}\)	\(6931\)
trager	\(\text {Expression too large to display}\)	\(10910\)

[In]

int((x^3-1)^(2/3)*(x^3+2)/x^6/(x^6-2*x^3-4),x,method=_RETURNVERBOSE)

[Out]

1/60*(-5*sum(_R^2*ln((-_R*x+(x^3-1)^(1/3))/x)/(4*_R^3-5),_R=RootOf(4*_Z^6-10*_Z^3+5))*x^5-6*x^3*(x^3-1)^(2/3)+
6*(x^3-1)^(2/3))/x^5

Fricas [F(-2)]

Exception generated. \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((x^3-1)^(2/3)*(x^3+2)/x^6/(x^6-2*x^3-4),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=\text {Timed out} \]

[In]

integrate((x**3-1)**(2/3)*(x**3+2)/x**6/(x**6-2*x**3-4),x)

[Out]

Timed out

Maxima [N/A]

Not integrable

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.35 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} + 2\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{{\left (x^{6} - 2 \, x^{3} - 4\right )} x^{6}} \,d x } \]

[In]

integrate((x^3-1)^(2/3)*(x^3+2)/x^6/(x^6-2*x^3-4),x, algorithm="maxima")

[Out]

integrate((x^3 + 2)*(x^3 - 1)^(2/3)/((x^6 - 2*x^3 - 4)*x^6), x)

Giac [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.35 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=\int { \frac {{\left (x^{3} + 2\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{{\left (x^{6} - 2 \, x^{3} - 4\right )} x^{6}} \,d x } \]

[In]

integrate((x^3-1)^(2/3)*(x^3+2)/x^6/(x^6-2*x^3-4),x, algorithm="giac")

[Out]

integrate((x^3 + 2)*(x^3 - 1)^(2/3)/((x^6 - 2*x^3 - 4)*x^6), x)

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.38 \[ \int \frac {\left (-1+x^3\right )^{2/3} \left (2+x^3\right )}{x^6 \left (-4-2 x^3+x^6\right )} \, dx=\int -\frac {{\left (x^3-1\right )}^{2/3}\,\left (x^3+2\right )}{x^6\,\left (-x^6+2\,x^3+4\right )} \,d x \]

[In]

int(-((x^3 - 1)^(2/3)*(x^3 + 2))/(x^6*(2*x^3 - x^6 + 4)),x)

[Out]

int(-((x^3 - 1)^(2/3)*(x^3 + 2))/(x^6*(2*x^3 - x^6 + 4)), x)