3.31.26 \(\int \frac {1+x^3+x^6}{\sqrt [3]{x^2+x^4} (-1+x^6)} \, dx\) [3026]

3.31.26.1 Optimal result

Integrand size = 27, antiderivative size = 429 \[ \int \frac {1+x^3+x^6}{\sqrt [3]{x^2+x^4} \left (-1+x^6\right )} \, dx=-\frac {1}{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{-x+2 \sqrt [3]{x^2+x^4}}\right )-\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{x^2+x^4}}\right )}{2 \sqrt {3}}-\frac {\arctan \left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2} \sqrt {3}}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}+\frac {1}{6} \log \left (-x+\sqrt [3]{x^2+x^4}\right )-\frac {1}{2} \log \left (x+\sqrt [3]{x^2+x^4}\right )+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}-\frac {\log \left (2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{12 \sqrt [3]{2}}+\frac {1}{4} \log \left (x^2-x \sqrt [3]{x^2+x^4}+\left (x^2+x^4\right )^{2/3}\right )-\frac {1}{12} \log \left (x^2+x \sqrt [3]{x^2+x^4}+\left (x^2+x^4\right )^{2/3}\right )+\frac {\log \left (-2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}-\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{24 \sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}} \]

output

-1/2*3^(1/2)*arctan(3^(1/2)*x/(-x+2*(x^4+x^2)^(1/3)))-1/6*arctan(3^(1/2)*x 
/(x+2*(x^4+x^2)^(1/3)))*3^(1/2)-1/24*3^(1/2)*arctan(3^(1/2)*x/(-x+2^(2/3)* 
(x^4+x^2)^(1/3)))*2^(2/3)-1/8*3^(1/2)*arctan(3^(1/2)*x/(x+2^(2/3)*(x^4+x^2 
)^(1/3)))*2^(2/3)+1/6*ln(-x+(x^4+x^2)^(1/3))-1/2*ln(x+(x^4+x^2)^(1/3))+1/8 
*ln(-2*x+2^(2/3)*(x^4+x^2)^(1/3))*2^(2/3)-1/24*ln(2*x+2^(2/3)*(x^4+x^2)^(1 
/3))*2^(2/3)+1/4*ln(x^2-x*(x^4+x^2)^(1/3)+(x^4+x^2)^(2/3))-1/12*ln(x^2+x*( 
x^4+x^2)^(1/3)+(x^4+x^2)^(2/3))+1/48*ln(-2*x^2+2^(2/3)*x*(x^4+x^2)^(1/3)-2 
^(1/3)*(x^4+x^2)^(2/3))*2^(2/3)-1/16*ln(2*x^2+2^(2/3)*x*(x^4+x^2)^(1/3)+2^ 
(1/3)*(x^4+x^2)^(2/3))*2^(2/3)
 

3.31.26.2 Mathematica [A] (verified)

Time = 3.54 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.10 \[ \int \frac {1+x^3+x^6}{\sqrt [3]{x^2+x^4} \left (-1+x^6\right )} \, dx=\frac {x^{2/3} \sqrt [3]{1+x^2} \left (24 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}-2 \sqrt [3]{1+x^2}}\right )-8 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2 \sqrt [3]{1+x^2}}\right )+2\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}-2^{2/3} \sqrt [3]{1+x^2}}\right )-6\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )+8 \log \left (-\sqrt [3]{x}+\sqrt [3]{1+x^2}\right )-24 \log \left (\sqrt [3]{x}+\sqrt [3]{1+x^2}\right )+6\ 2^{2/3} \log \left (-2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )-2\ 2^{2/3} \log \left (2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )+12 \log \left (x^{2/3}-\sqrt [3]{x} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )-4 \log \left (x^{2/3}+\sqrt [3]{x} \sqrt [3]{1+x^2}+\left (1+x^2\right )^{2/3}\right )+2^{2/3} \log \left (-2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}-\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )-3\ 2^{2/3} \log \left (2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}+\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )\right )}{48 \sqrt [3]{x^2+x^4}} \]

input

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

output

(x^(2/3)*(1 + x^2)^(1/3)*(24*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) - 2 
*(1 + x^2)^(1/3))] - 8*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) + 2*(1 + 
x^2)^(1/3))] + 2*2^(2/3)*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) - 2^(2/ 
3)*(1 + x^2)^(1/3))] - 6*2^(2/3)*Sqrt[3]*ArcTan[(Sqrt[3]*x^(1/3))/(x^(1/3) 
 + 2^(2/3)*(1 + x^2)^(1/3))] + 8*Log[-x^(1/3) + (1 + x^2)^(1/3)] - 24*Log[ 
x^(1/3) + (1 + x^2)^(1/3)] + 6*2^(2/3)*Log[-2*x^(1/3) + 2^(2/3)*(1 + x^2)^ 
(1/3)] - 2*2^(2/3)*Log[2*x^(1/3) + 2^(2/3)*(1 + x^2)^(1/3)] + 12*Log[x^(2/ 
3) - x^(1/3)*(1 + x^2)^(1/3) + (1 + x^2)^(2/3)] - 4*Log[x^(2/3) + x^(1/3)* 
(1 + x^2)^(1/3) + (1 + x^2)^(2/3)] + 2^(2/3)*Log[-2*x^(2/3) + 2^(2/3)*x^(1 
/3)*(1 + x^2)^(1/3) - 2^(1/3)*(1 + x^2)^(2/3)] - 3*2^(2/3)*Log[2*x^(2/3) + 
 2^(2/3)*x^(1/3)*(1 + x^2)^(1/3) + 2^(1/3)*(1 + x^2)^(2/3)]))/(48*(x^2 + x 
^4)^(1/3))
 

3.31.26.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^6+x^3+1}{\sqrt [3]{x^4+x^2} \left (x^6-1\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {x^{2/3} \sqrt [3]{x^2+1} \int -\frac {x^6+x^3+1}{x^{2/3} \sqrt [3]{x^2+1} \left (1-x^6\right )}dx}{\sqrt [3]{x^4+x^2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {x^{2/3} \sqrt [3]{x^2+1} \int \frac {x^6+x^3+1}{x^{2/3} \sqrt [3]{x^2+1} \left (1-x^6\right )}dx}{\sqrt [3]{x^4+x^2}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \int \frac {x^6+x^3+1}{\sqrt [3]{x^2+1} \left (1-x^6\right )}d\sqrt [3]{x}}{\sqrt [3]{x^4+x^2}}\)

\(\Big \downarrow \) 7276

\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \int \left (\frac {x^3+2}{\sqrt [3]{x^2+1} \left (1-x^6\right )}-\frac {1}{\sqrt [3]{x^2+1}}\right )d\sqrt [3]{x}}{\sqrt [3]{x^4+x^2}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \left (-\frac {1}{18} \int \frac {1}{\left (-\sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left (1-\sqrt [3]{x}\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left (\sqrt [9]{-1} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left (\sqrt [9]{-1} \sqrt [3]{x}+1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left (-(-1)^{2/9} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left (1-(-1)^{2/9} \sqrt [3]{x}\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left (\sqrt [3]{-1} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left (\sqrt [3]{-1} \sqrt [3]{x}+1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left (-(-1)^{4/9} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left (1-(-1)^{4/9} \sqrt [3]{x}\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left ((-1)^{5/9} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left ((-1)^{5/9} \sqrt [3]{x}+1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left (-(-1)^{2/3} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left (1-(-1)^{2/3} \sqrt [3]{x}\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left ((-1)^{7/9} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left ((-1)^{7/9} \sqrt [3]{x}+1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\frac {1}{18} \int \frac {1}{\left (-(-1)^{8/9} \sqrt [3]{x}-1\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}+\frac {1}{6} \int \frac {1}{\left (1-(-1)^{8/9} \sqrt [3]{x}\right ) \sqrt [3]{x^2+1}}d\sqrt [3]{x}-\sqrt [3]{x} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )\right )}{\sqrt [3]{x^4+x^2}}\)

input

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

output

$Aborted
 

3.31.26.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

rule 7276

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

3.31.26.4 Maple [F(-1)]

Timed out.

\[\int \frac {x^{6}+x^{3}+1}{\left (x^{4}+x^{2}\right )^{\frac {1}{3}} \left (x^{6}-1\right )}d x\]

input

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

output

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

3.31.26.5 Fricas [F(-2)]

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

input

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

output

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (residue poly has multiple non-linear fac 
tors)
 

3.31.26.6 Sympy [F]

\[ \int \frac {1+x^3+x^6}{\sqrt [3]{x^2+x^4} \left (-1+x^6\right )} \, dx=\int \frac {x^{6} + x^{3} + 1}{\sqrt [3]{x^{2} \left (x^{2} + 1\right )} \left (x - 1\right ) \left (x + 1\right ) \left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )}\, dx \]

input

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

output

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

3.31.26.7 Maxima [F]

\[ \int \frac {1+x^3+x^6}{\sqrt [3]{x^2+x^4} \left (-1+x^6\right )} \, dx=\int { \frac {x^{6} + x^{3} + 1}{{\left (x^{6} - 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}} \,d x } \]

input

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

output

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

3.31.26.8 Giac [F]

\[ \int \frac {1+x^3+x^6}{\sqrt [3]{x^2+x^4} \left (-1+x^6\right )} \, dx=\int { \frac {x^{6} + x^{3} + 1}{{\left (x^{6} - 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{3}}} \,d x } \]

input

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

output

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

3.31.26.9 Mupad [F(-1)]

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

input

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

output

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