\(\int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx\) [1199]

Optimal result

Integrand size = 18, antiderivative size = 88 \[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=-\frac {1}{4} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {\sqrt [3]{-4-4 x+x^2}}{\sqrt {3}}\right )-\frac {1}{4} \log \left (2+\sqrt [3]{-4-4 x+x^2}\right )+\frac {1}{8} \log \left (4-2 \sqrt [3]{-4-4 x+x^2}+\left (-4-4 x+x^2\right )^{2/3}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\frac {1}{8} \left (2 \sqrt {3} \arctan \left (\frac {-1+\sqrt [3]{-4-4 x+x^2}}{\sqrt {3}}\right )-2 \log \left (2+\sqrt [3]{-4-4 x+x^2}\right )+\log \left (4-2 \sqrt [3]{-4-4 x+x^2}+\left (-4-4 x+x^2\right )^{2/3}\right )\right ) \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.58, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1118, 243, 68, 16, 1083, 217}

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.

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

\(\Big \downarrow \) 1118

\(\displaystyle \int \frac {1}{\sqrt [3]{(x-2)^2-8} (x-2)}d(x-2)\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \int \frac {1}{\sqrt [3]{x-10} (x-2)^2}d(x-2)^2\)

\(\Big \downarrow \) 68

\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{(x-2)^4-2 \sqrt [3]{x-10}+4}d\sqrt [3]{x-10}-\frac {3}{4} \int \frac {1}{x}d\sqrt [3]{x-10}+\frac {1}{4} \log \left ((x-2)^2\right )\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} \left (\frac {3}{2} \int \frac {1}{(x-2)^4-2 \sqrt [3]{x-10}+4}d\sqrt [3]{x-10}+\frac {1}{4} \log \left ((x-2)^2\right )-\frac {3 \log (x)}{4}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (-3 \int \frac {1}{-(x-2)^4-12}d\left (2 \sqrt [3]{x-10}-2\right )+\frac {1}{4} \log \left ((x-2)^2\right )-\frac {3 \log (x)}{4}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x-10}-2}{2 \sqrt {3}}\right )+\frac {1}{4} \log \left ((x-2)^2\right )-\frac {3 \log (x)}{4}\right )\)

Input:

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

Output:

((Sqrt[3]*ArcTan[(-2 + 2*(-10 + x)^(1/3))/(2*Sqrt[3])])/2 + Log[(-2 + x)^2 
]/4 - (3*Log[x])/4)/2
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 68

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

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1118

Int[((d_) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_S 
ymbol] :> Simp[1/e   Subst[Int[x^m*(a - b^2/(4*c) + (c*x^2)/e^2)^p, x], x, 
d + e*x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[2*c*d - b*e, 0]
 

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.76

method	result	size
pseudoelliptic	\(-\frac {\ln \left (2+\left (x^{2}-4 x -4\right )^{\frac {1}{3}}\right )}{4}+\frac {\ln \left (4-2 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+\left (x^{2}-4 x -4\right )^{\frac {2}{3}}\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (\left (x^{2}-4 x -4\right )^{\frac {1}{3}}-1\right ) \sqrt {3}}{3}\right )}{4}\)	\(67\)
trager	\(\frac {\operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \ln \left (-\frac {-40 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x^{2}+96 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {2}{3}}+160 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x -6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x^{2}-18 \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-192 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+24 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x +x^{2}+36 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+280 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-4 x -28}{\left (-2+x \right )^{2}}\right )}{2}+\frac {\ln \left (\frac {20 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x^{2}+48 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-80 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x -23 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x^{2}-15 \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-96 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+92 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x +6 x^{2}+30 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+140 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-24 x -56}{\left (-2+x \right )^{2}}\right )}{4}-\frac {\ln \left (\frac {20 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x^{2}+48 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-80 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )^{2} x -23 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x^{2}-15 \left (x^{2}-4 x -4\right )^{\frac {2}{3}}-96 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+92 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) x +6 x^{2}+30 \left (x^{2}-4 x -4\right )^{\frac {1}{3}}+140 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-24 x -56}{\left (-2+x \right )^{2}}\right ) \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )}{2}\)	\(521\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(-2+x) \sqrt [3]{-4-4 x+x^2}} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {1}{8} \, \log \left ({\left (x^{2} - 4 \, x - 4\right )}^{\frac {2}{3}} - 2 \, {\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} + 4\right ) - \frac {1}{4} \, \log \left ({\left (x^{2} - 4 \, x - 4\right )}^{\frac {1}{3}} + 2\right ) \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(1/((x - 2)*(x**2 - 4*x - 4)**(1/3)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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