\(\int \frac {1}{(-1+x^2) \sqrt [3]{-x^2+x^3}} \, dx\) [2165]

Optimal result

Integrand size = 21, antiderivative size = 159 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {3 \left (-x^2+x^3\right )^{2/3}}{2 (-1+x) x}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2^{2/3} \sqrt [3]{-x^2+x^3}}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2 x+2^{2/3} \sqrt [3]{-x^2+x^3}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2 x^2+2^{2/3} x \sqrt [3]{-x^2+x^3}+\sqrt [3]{2} \left (-x^2+x^3\right )^{2/3}\right )}{4 \sqrt [3]{2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.06 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^3}} \, dx=-\frac {x^{2/3} \left (12 \sqrt [3]{x}+2\ 2^{2/3} \sqrt {3} \sqrt [3]{-1+x} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x}}{2^{2/3} \sqrt [3]{-1+x}+\sqrt [3]{x}}\right )-2\ 2^{2/3} \sqrt [3]{-1+x} \log \left (2^{2/3} \sqrt [3]{-1+x}-2 \sqrt [3]{x}\right )+2^{2/3} \sqrt [3]{-1+x} \log \left (\sqrt [3]{2} (-1+x)^{2/3}+2^{2/3} \sqrt [3]{-1+x} \sqrt [3]{x}+2 x^{2/3}\right )\right )}{8 \sqrt [3]{(-1+x) x^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2467, 25, 516, 107, 25, 102}

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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 516

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

\(\Big \downarrow \) 107

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 102

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

Input:

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

Output:

-(((-1 + x)^(1/3)*x^(2/3)*((3*x^(1/3))/(2*(-1 + x)^(1/3)) + (-((Sqrt[3]*Ar 
cTan[1/Sqrt[3] + (2^(2/3)*(-1 + x)^(1/3))/(Sqrt[3]*x^(1/3))])/2^(1/3)) - ( 
3*Log[(-1 + x)^(1/3)/2^(1/3) - x^(1/3)])/(2*2^(1/3)) + Log[1 + x]/(2*2^(1/ 
3)))/2))/(-x^2 + x^3)^(1/3))
 

Defintions of rubi rules used

rule 25

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

rule 102

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

rule 107

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[b*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 
)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Simp[(a*d*f*(m + 1) + b*c*f*(n + 
 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 
 1)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x 
] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || SumSimplerQ[m, 1])
 

rule 516

Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. 
), x_Symbol] :> Int[(e*x)^m*(c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; Free 
Q[{a, b, c, d, e, m, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || 
(GtQ[a, 0] && GtQ[c, 0] &&  !IntegerQ[n]))
 

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]
 

Maple [A] (verified)

Time = 4.70 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {-\frac {2^{\frac {2}{3}} \ln \left (\frac {2^{\frac {2}{3}} x^{2}+2^{\frac {1}{3}} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {2}{3}}}{x^{2}}\right ) \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}{2}+\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+x \right )}{3 x}\right ) \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}+2^{\frac {2}{3}} \ln \left (\frac {-2^{\frac {1}{3}} x +\left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}{x}\right ) \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}-6 x}{4 \left (\left (-1+x \right ) x^{2}\right )^{\frac {1}{3}}}\)	\(140\)
trager	\(\text {Expression too large to display}\)	\(774\)
risch	\(\text {Expression too large to display}\)	\(1072\)

Input:

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

Output:

1/4/((-1+x)*x^2)^(1/3)*(-1/2*2^(2/3)*ln((2^(2/3)*x^2+2^(1/3)*((-1+x)*x^2)^ 
(1/3)*x+((-1+x)*x^2)^(2/3))/x^2)*((-1+x)*x^2)^(1/3)+3^(1/2)*2^(2/3)*arctan 
(1/3*3^(1/2)*(2^(2/3)*((-1+x)*x^2)^(1/3)+x)/x)*((-1+x)*x^2)^(1/3)+2^(2/3)* 
ln((-2^(1/3)*x+((-1+x)*x^2)^(1/3))/x)*((-1+x)*x^2)^(1/3)-6*x)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^3}} \, dx=\frac {4 \cdot 2^{\frac {1}{6}} \sqrt {\frac {3}{2}} {\left (x^{2} - x\right )} \arctan \left (\frac {2^{\frac {1}{6}} \sqrt {\frac {3}{2}} {\left (2^{\frac {1}{3}} x + 2 \, {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}\right )}}{3 \, x}\right ) + 2 \cdot 2^{\frac {2}{3}} {\left (x^{2} - x\right )} \log \left (-\frac {2^{\frac {1}{3}} x - {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}}}{x}\right ) - 2^{\frac {2}{3}} {\left (x^{2} - x\right )} \log \left (\frac {2^{\frac {2}{3}} x^{2} + 2^{\frac {1}{3}} {\left (x^{3} - x^{2}\right )}^{\frac {1}{3}} x + {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{x^{2}}\right ) - 12 \, {\left (x^{3} - x^{2}\right )}^{\frac {2}{3}}}{8 \, {\left (x^{2} - x\right )}} \] Input:

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

Output:

1/8*(4*2^(1/6)*sqrt(3/2)*(x^2 - x)*arctan(1/3*2^(1/6)*sqrt(3/2)*(2^(1/3)*x 
 + 2*(x^3 - x^2)^(1/3))/x) + 2*2^(2/3)*(x^2 - x)*log(-(2^(1/3)*x - (x^3 - 
x^2)^(1/3))/x) - 2^(2/3)*(x^2 - x)*log((2^(2/3)*x^2 + 2^(1/3)*(x^3 - x^2)^ 
(1/3)*x + (x^3 - x^2)^(2/3))/x^2) - 12*(x^3 - x^2)^(2/3))/(x^2 - x)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.62 \[ \int \frac {1}{\left (-1+x^2\right ) \sqrt [3]{-x^2+x^3}} \, dx=\frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} + 2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left ({\left | -2^{\frac {1}{3}} + {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}} \right |}\right ) - \frac {3}{2 \, {\left (-\frac {1}{x} + 1\right )}^{\frac {1}{3}}} \] Input:

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

Output:

1/4*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3) + 2*(-1/x + 1)^(1/ 
3))) - 1/8*2^(2/3)*log(2^(2/3) + 2^(1/3)*(-1/x + 1)^(1/3) + (-1/x + 1)^(2/ 
3)) + 1/4*2^(2/3)*log(abs(-2^(1/3) + (-1/x + 1)^(1/3))) - 3/2/(-1/x + 1)^( 
1/3)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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