\(\int \frac {(1+x^3)^{2/3} (1+x^3+2 x^6)}{x^6 (-1+2 x^6)} \, dx\) [1540]

Optimal result

Integrand size = 32, antiderivative size = 106 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (1+x^3+2 x^6\right )}{x^6 \left (-1+2 x^6\right )} \, dx=\frac {\left (1+x^3\right )^{2/3} \left (2+7 x^3\right )}{10 x^5}-\frac {1}{6} \text {RootSum}\left [-1-2 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {\log (x)-\log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+5 \log (x) \text {$\#$1}^3-5 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-\text {$\#$1}+\text {$\#$1}^4}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(465\) vs. \(2(106)=212\).

Time = 0.34 (sec) , antiderivative size = 465, normalized size of antiderivative = 4.39, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6857, 270, 283, 245, 399, 384} \[ \int \frac {\left (1+x^3\right )^{2/3} \left (1+x^3+2 x^6\right )}{x^6 \left (-1+2 x^6\right )} \, dx=\frac {\left (1+2 \sqrt {2}\right ) \arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\left (1-2 \sqrt {2}\right ) \arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\sqrt [3]{116 \sqrt {2}-163} \arctan \left (\frac {1-\frac {2 \sqrt [3]{\sqrt {2}-1} x}{\sqrt [3]{x^3+1}}}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\sqrt [3]{163+116 \sqrt {2}} \arctan \left (\frac {\frac {2 \sqrt [3]{1+\sqrt {2}} x}{\sqrt [3]{x^3+1}}+1}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{12} \sqrt [3]{163+116 \sqrt {2}} \log \left (1-\sqrt {2} x^3\right )+\frac {1}{12} \sqrt [3]{116 \sqrt {2}-163} \log \left (\sqrt {2} x^3+1\right )-\frac {1}{4} \sqrt [3]{116 \sqrt {2}-163} \log \left (-\sqrt [3]{x^3+1}-\sqrt [3]{\sqrt {2}-1} x\right )+\frac {1}{4} \sqrt [3]{163+116 \sqrt {2}} \log \left (\sqrt [3]{1+\sqrt {2}} x-\sqrt [3]{x^3+1}\right )-\frac {1}{4} \left (1+2 \sqrt {2}\right ) \log \left (\sqrt [3]{x^3+1}-x\right )-\frac {1}{4} \left (1-2 \sqrt {2}\right ) \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {1}{2} \log \left (\sqrt [3]{x^3+1}-x\right )+\frac {\left (x^3+1\right )^{5/3}}{5 x^5}+\frac {\left (x^3+1\right )^{2/3}}{2 x^2} \]

[In]

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

[Out]

(1 + x^3)^(2/3)/(2*x^2) + (1 + x^3)^(5/3)/(5*x^5) - ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]]/Sqrt[3] + ((1
- 2*Sqrt[2])*ArcTan[(1 + (2*x)/(1 + x^3)^(1/3))/Sqrt[3]])/(2*Sqrt[3]) + ((1 + 2*Sqrt[2])*ArcTan[(1 + (2*x)/(1
+ x^3)^(1/3))/Sqrt[3]])/(2*Sqrt[3]) + ((-163 + 116*Sqrt[2])^(1/3)*ArcTan[(1 - (2*(-1 + Sqrt[2])^(1/3)*x)/(1 +
x^3)^(1/3))/Sqrt[3]])/(2*Sqrt[3]) - ((163 + 116*Sqrt[2])^(1/3)*ArcTan[(1 + (2*(1 + Sqrt[2])^(1/3)*x)/(1 + x^3)
^(1/3))/Sqrt[3]])/(2*Sqrt[3]) - ((163 + 116*Sqrt[2])^(1/3)*Log[1 - Sqrt[2]*x^3])/12 + ((-163 + 116*Sqrt[2])^(1
/3)*Log[1 + Sqrt[2]*x^3])/12 - ((-163 + 116*Sqrt[2])^(1/3)*Log[-((-1 + Sqrt[2])^(1/3)*x) - (1 + x^3)^(1/3)])/4
+ ((163 + 116*Sqrt[2])^(1/3)*Log[(1 + Sqrt[2])^(1/3)*x - (1 + x^3)^(1/3)])/4 + Log[-x + (1 + x^3)^(1/3)]/2 -
((1 - 2*Sqrt[2])*Log[-x + (1 + x^3)^(1/3)])/4 - ((1 + 2*Sqrt[2])*Log[-x + (1 + x^3)^(1/3)])/4

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 283

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

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 6857

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

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {\left (1+x^3\right )^{2/3}}{x^6}-\frac {\left (1+x^3\right )^{2/3}}{x^3}+\frac {2 \left (1+x^3\right )^{2/3} \left (2+x^3\right )}{-1+2 x^6}\right ) \, dx \\ & = 2 \int \frac {\left (1+x^3\right )^{2/3} \left (2+x^3\right )}{-1+2 x^6} \, dx-\int \frac {\left (1+x^3\right )^{2/3}}{x^6} \, dx-\int \frac {\left (1+x^3\right )^{2/3}}{x^3} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\frac {\left (1+x^3\right )^{5/3}}{5 x^5}+2 \int \left (-\frac {\left (1+2 \sqrt {2}\right ) \left (1+x^3\right )^{2/3}}{2 \sqrt {2} \left (1-\sqrt {2} x^3\right )}+\frac {\left (1-2 \sqrt {2}\right ) \left (1+x^3\right )^{2/3}}{2 \sqrt {2} \left (1+\sqrt {2} x^3\right )}\right ) \, dx-\int \frac {1}{\sqrt [3]{1+x^3}} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\frac {\left (1+x^3\right )^{5/3}}{5 x^5}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{2} \left (-4+\sqrt {2}\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1+\sqrt {2} x^3} \, dx-\frac {1}{2} \left (4+\sqrt {2}\right ) \int \frac {\left (1+x^3\right )^{2/3}}{1-\sqrt {2} x^3} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\frac {\left (1+x^3\right )^{5/3}}{5 x^5}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \log \left (-x+\sqrt [3]{1+x^3}\right )+\frac {1}{2} \left (1-2 \sqrt {2}\right ) \int \frac {1}{\sqrt [3]{1+x^3}} \, dx+\frac {1}{2} \left (1+2 \sqrt {2}\right ) \int \frac {1}{\sqrt [3]{1+x^3}} \, dx+\frac {1}{2} \left (-5+3 \sqrt {2}\right ) \int \frac {1}{\sqrt [3]{1+x^3} \left (1+\sqrt {2} x^3\right )} \, dx-\frac {1}{2} \left (5+3 \sqrt {2}\right ) \int \frac {1}{\sqrt [3]{1+x^3} \left (1-\sqrt {2} x^3\right )} \, dx \\ & = \frac {\left (1+x^3\right )^{2/3}}{2 x^2}+\frac {\left (1+x^3\right )^{5/3}}{5 x^5}-\frac {\arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\left (1-2 \sqrt {2}\right ) \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\left (1+2 \sqrt {2}\right ) \arctan \left (\frac {1+\frac {2 x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\sqrt [3]{-163+116 \sqrt {2}} \arctan \left (\frac {1-\frac {2 \sqrt [3]{-1+\sqrt {2}} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {\sqrt [3]{163+116 \sqrt {2}} \arctan \left (\frac {1+\frac {2 \sqrt [3]{1+\sqrt {2}} x}{\sqrt [3]{1+x^3}}}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{12} \sqrt [3]{163+116 \sqrt {2}} \log \left (1-\sqrt {2} x^3\right )+\frac {1}{12} \sqrt [3]{-163+116 \sqrt {2}} \log \left (1+\sqrt {2} x^3\right )-\frac {1}{4} \sqrt [3]{-163+116 \sqrt {2}} \log \left (-\sqrt [3]{-1+\sqrt {2}} x-\sqrt [3]{1+x^3}\right )+\frac {1}{4} \sqrt [3]{163+116 \sqrt {2}} \log \left (\sqrt [3]{1+\sqrt {2}} x-\sqrt [3]{1+x^3}\right )+\frac {1}{2} \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{4} \left (1-2 \sqrt {2}\right ) \log \left (-x+\sqrt [3]{1+x^3}\right )-\frac {1}{4} \left (1+2 \sqrt {2}\right ) \log \left (-x+\sqrt [3]{1+x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (1+x^3+2 x^6\right )}{x^6 \left (-1+2 x^6\right )} \, dx=\frac {\left (1+x^3\right )^{2/3} \left (2+7 x^3\right )}{10 x^5}-\frac {1}{6} \text {RootSum}\left [-1-2 \text {$\#$1}^3+\text {$\#$1}^6\&,\frac {\log (x)-\log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right )+5 \log (x) \text {$\#$1}^3-5 \log \left (\sqrt [3]{1+x^3}-x \text {$\#$1}\right ) \text {$\#$1}^3}{-\text {$\#$1}+\text {$\#$1}^4}\&\right ] \]

[In]

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

[Out]

((1 + x^3)^(2/3)*(2 + 7*x^3))/(10*x^5) - RootSum[-1 - 2*#1^3 + #1^6 & , (Log[x] - Log[(1 + x^3)^(1/3) - x*#1]
+ 5*Log[x]*#1^3 - 5*Log[(1 + x^3)^(1/3) - x*#1]*#1^3)/(-#1 + #1^4) & ]/6

Maple [N/A] (verified)

Time = 100.73 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(\frac {5 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-2 \textit {\_Z}^{3}-1\right )}{\sum }\frac {\left (5 \textit {\_R}^{3}+1\right ) \ln \left (\frac {-\textit {\_R} x +\left (x^{3}+1\right )^{\frac {1}{3}}}{x}\right )}{\textit {\_R}^{4}-\textit {\_R}}\right ) x^{5}+21 x^{3} \left (x^{3}+1\right )^{\frac {2}{3}}+6 \left (x^{3}+1\right )^{\frac {2}{3}}}{30 x^{5}}\)	\(81\)
risch	\(\text {Expression too large to display}\)	\(8584\)
trager	\(\text {Expression too large to display}\)	\(15487\)

[In]

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

[Out]

1/30*(5*sum((5*_R^3+1)*ln((-_R*x+(x^3+1)^(1/3))/x)/(_R^4-_R),_R=RootOf(_Z^6-2*_Z^3-1))*x^5+21*x^3*(x^3+1)^(2/3
)+6*(x^3+1)^(2/3))/x^5

Fricas [F(-2)]

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

[In]

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

[Out]

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

Sympy [N/A]

Not integrable

Time = 23.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.32 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (1+x^3+2 x^6\right )}{x^6 \left (-1+2 x^6\right )} \, dx=\int \frac {\left (\left (x + 1\right ) \left (x^{2} - x + 1\right )\right )^{\frac {2}{3}} \cdot \left (2 x^{6} + x^{3} + 1\right )}{x^{6} \cdot \left (2 x^{6} - 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.33 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.30 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (1+x^3+2 x^6\right )}{x^6 \left (-1+2 x^6\right )} \, dx=\int { \frac {{\left (2 \, x^{6} + x^{3} + 1\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (2 \, x^{6} - 1\right )} x^{6}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.32 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.30 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (1+x^3+2 x^6\right )}{x^6 \left (-1+2 x^6\right )} \, dx=\int { \frac {{\left (2 \, x^{6} + x^{3} + 1\right )} {\left (x^{3} + 1\right )}^{\frac {2}{3}}}{{\left (2 \, x^{6} - 1\right )} x^{6}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.30 \[ \int \frac {\left (1+x^3\right )^{2/3} \left (1+x^3+2 x^6\right )}{x^6 \left (-1+2 x^6\right )} \, dx=\int \frac {{\left (x^3+1\right )}^{2/3}\,\left (2\,x^6+x^3+1\right )}{x^6\,\left (2\,x^6-1\right )} \,d x \]

[In]

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

[Out]

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