\(\int \frac {1-x+x^2}{(-1+x^2) \sqrt [3]{x^2+x^4}} \, dx\) [2737]

Optimal result

Integrand size = 27, antiderivative size = 254 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=-\frac {3 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{-x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{x^2+x^4}}\right )}{4 \sqrt [3]{2}}+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}-\frac {3 \log \left (2 x+2^{2/3} \sqrt [3]{x^2+x^4}\right )}{4 \sqrt [3]{2}}+\frac {3 \log \left (-2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}-\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{x^2+x^4}+\sqrt [3]{2} \left (x^2+x^4\right )^{2/3}\right )}{8 \sqrt [3]{2}} \] Output:

-3/8*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)+1/8*ln(-2*x+2^ 
(2/3)*(x^4+x^2)^(1/3))*2^(2/3)-3/8*ln(2*x+2^(2/3)*(x^4+x^2)^(1/3))*2^(2/3) 
+3/16*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) 
-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)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.48 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.04 \[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\frac {x^{2/3} \sqrt [3]{1+x^2} \left (6 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}-2^{2/3} \sqrt [3]{1+x^2}}\right )-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{\sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}}\right )+2 \log \left (-2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )-6 \log \left (2 \sqrt [3]{x}+2^{2/3} \sqrt [3]{1+x^2}\right )+3 \log \left (-2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}-\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )-\log \left (2 x^{2/3}+2^{2/3} \sqrt [3]{x} \sqrt [3]{1+x^2}+\sqrt [3]{2} \left (1+x^2\right )^{2/3}\right )\right )}{8 \sqrt [3]{2} \sqrt [3]{x^2+x^4}} \] Input:

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

Output:

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

Rubi [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 1.40 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.63, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {2467, 25, 2035, 7276, 2009}

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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 x^{2/3} \sqrt [3]{x^2+1} \left (2 \sqrt [3]{x} \operatorname {AppellF1}\left (\frac {1}{6},1,\frac {1}{3},\frac {7}{6},x^2,-x^2\right )-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2}}+\frac {\arctan \left (\frac {1-\frac {2 \sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}}{\sqrt {3}}\right )}{4 \sqrt [3]{2} \sqrt {3}}-\frac {\sqrt {3} \arctan \left (\frac {\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{8 \sqrt [3]{2}}+\frac {\arctan \left (\frac {\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1}{\sqrt {3}}\right )}{8 \sqrt [3]{2} \sqrt {3}}-\sqrt [3]{x} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-x^2\right )-\frac {\log \left (\left (1-x^{2/3}\right )^2 \left (x^{2/3}+1\right )\right )}{24 \sqrt [3]{2}}-\frac {\log \left (\frac {2^{2/3} \left (x^{2/3}+1\right )^2}{\left (x^2+1\right )^{2/3}}-\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{12 \sqrt [3]{2}}+\frac {\log \left (\frac {\sqrt [3]{2} \left (x^{2/3}+1\right )}{\sqrt [3]{x^2+1}}+1\right )}{6 \sqrt [3]{2}}+\frac {\log \left (x^{2/3}-2^{2/3} \sqrt [3]{x^2+1}+1\right )}{8 \sqrt [3]{2}}\right )}{\sqrt [3]{x^4+x^2}}\)

Input:

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

Output:

(-3*x^(2/3)*(1 + x^2)^(1/3)*(2*x^(1/3)*AppellF1[1/6, 1, 1/3, 7/6, x^2, -x^ 
2] + ArcTan[(1 - (2*2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3]]/(4*2^ 
(1/3)*Sqrt[3]) - (Sqrt[3]*ArcTan[(1 - (2*2^(1/3)*(1 + x^(2/3)))/(1 + x^2)^ 
(1/3))/Sqrt[3]])/(4*2^(1/3)) + ArcTan[(1 + (2^(1/3)*(1 + x^(2/3)))/(1 + x^ 
2)^(1/3))/Sqrt[3]]/(8*2^(1/3)*Sqrt[3]) - (Sqrt[3]*ArcTan[(1 + (2^(1/3)*(1 
+ x^(2/3)))/(1 + x^2)^(1/3))/Sqrt[3]])/(8*2^(1/3)) - x^(1/3)*Hypergeometri 
c2F1[1/6, 1/3, 7/6, -x^2] - Log[(1 - x^(2/3))^2*(1 + x^(2/3))]/(24*2^(1/3) 
) - Log[1 + (2^(2/3)*(1 + x^(2/3))^2)/(1 + x^2)^(2/3) - (2^(1/3)*(1 + x^(2 
/3)))/(1 + x^2)^(1/3)]/(12*2^(1/3)) + Log[1 + (2^(1/3)*(1 + x^(2/3)))/(1 + 
 x^2)^(1/3)]/(6*2^(1/3)) + Log[1 + x^(2/3) - 2^(2/3)*(1 + x^2)^(1/3)]/(8*2 
^(1/3))))/(x^2 + x^4)^(1/3)
 

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]
 

Maple [A] (verified)

Time = 42.58 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.80

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {2}{3}} \left (6 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}+x \right )}{3 x}\right )-2 \ln \left (\frac {-2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )+\ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )+6 \ln \left (\frac {2^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}}}{x}\right )-3 \ln \left (\frac {2^{\frac {2}{3}} x^{2}-2^{\frac {1}{3}} \left (x^{2} \left (x^{2}+1\right )\right )^{\frac {1}{3}} x +\left (x^{2} \left (x^{2}+1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )\right )}{16}\)	\(202\)
trager	\(\text {Expression too large to display}\)	\(5848\)

Input:

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

Output:

-1/16*2^(2/3)*(6*3^(1/2)*arctan(1/3*3^(1/2)*(-2^(2/3)*(x^2*(x^2+1))^(1/3)+ 
x)/x)-2*3^(1/2)*arctan(1/3*3^(1/2)*(2^(2/3)*(x^2*(x^2+1))^(1/3)+x)/x)-2*ln 
((-2^(1/3)*x+(x^2*(x^2+1))^(1/3))/x)+ln((2^(2/3)*x^2+2^(1/3)*(x^2*(x^2+1)) 
^(1/3)*x+(x^2*(x^2+1))^(2/3))/x^2)+6*ln((2^(1/3)*x+(x^2*(x^2+1))^(1/3))/x) 
-3*ln((2^(2/3)*x^2-2^(1/3)*(x^2*(x^2+1))^(1/3)*x+(x^2*(x^2+1))^(2/3))/x^2) 
)
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1-x+x^2}{\left (-1+x^2\right ) \sqrt [3]{x^2+x^4}} \, dx=\int \frac {x^{2}}{x^{\frac {8}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}-x^{\frac {2}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}d x -\left (\int \frac {x}{x^{\frac {8}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}-x^{\frac {2}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}d x \right )+\int \frac {1}{x^{\frac {8}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}-x^{\frac {2}{3}} \left (x^{2}+1\right )^{\frac {1}{3}}}d x \] Input:

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

Output:

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