\(\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx\) [1527]

Optimal result

Integrand size = 18, antiderivative size = 106 \[ \int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=\frac {\left ((1+x)^2\right )^{2/3} \left (-\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x}}{3^{5/6}}\right )}{\sqrt [6]{3}}+\frac {\log \left (-3+3^{2/3} \sqrt [3]{1+x}\right )}{3^{2/3}}-\frac {\log \left (3+3^{2/3} \sqrt [3]{1+x}+\sqrt [3]{3} (1+x)^{2/3}\right )}{2\ 3^{2/3}}\right )}{(1+x)^{4/3}} \] Output:

((1+x)^2)^(2/3)*(-1/3*arctan(1/3*3^(1/2)+2/3*(1+x)^(1/3)*3^(1/6))*3^(5/6)+ 
1/3*ln(-3+3^(2/3)*(1+x)^(1/3))*3^(1/3)-1/6*ln(3+3^(2/3)*(1+x)^(1/3)+3^(1/3 
)*(1+x)^(2/3))*3^(1/3))/(1+x)^(4/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=-\frac {(1+x)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x}}{3^{5/6}}\right )-2 \log \left (-3+3^{2/3} \sqrt [3]{1+x}\right )+\log \left (3+3^{2/3} \sqrt [3]{1+x}+\sqrt [3]{3} (1+x)^{2/3}\right )\right )}{2\ 3^{2/3} \sqrt [3]{(1+x)^2}} \] Input:

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

Output:

-1/2*((1 + x)^(2/3)*(2*Sqrt[3]*ArcTan[1/Sqrt[3] + (2*(1 + x)^(1/3))/3^(5/6 
)] - 2*Log[-3 + 3^(2/3)*(1 + x)^(1/3)] + Log[3 + 3^(2/3)*(1 + x)^(1/3) + 3 
^(1/3)*(1 + x)^(2/3)]))/(3^(2/3)*((1 + x)^2)^(1/3))
 

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.86, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1102, 25, 69, 16, 1082, 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.

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

\(\Big \downarrow \) 1102

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

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 25

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

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1102

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.87 (sec) , antiderivative size = 1406, normalized size of antiderivative = 13.26

method	result	size
trager	\(\text {Expression too large to display}\)	\(1406\)

Input:

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

Output:

-1/3*ln(-(15*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z 
^3-3)^3*x^2+63*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)^2*RootO 
f(_Z^3-3)^2*x^2+15*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*Roo 
tOf(_Z^3-3)^3*x-63*(x^2+2*x+1)^(2/3)*RootOf(_Z^3-3)^2*RootOf(RootOf(_Z^3-3 
)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)+63*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^ 
3-3)+9*_Z^2)^2*RootOf(_Z^3-3)^2*x-45*(x^2+2*x+1)^(1/3)*RootOf(_Z^3-3)^2*x- 
72*(x^2+2*x+1)^(1/3)*RootOf(_Z^3-3)*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z 
^3-3)+9*_Z^2)*x-45*(x^2+2*x+1)^(1/3)*RootOf(_Z^3-3)^2-72*(x^2+2*x+1)^(1/3) 
*RootOf(_Z^3-3)*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)+20*Roo 
tOf(_Z^3-3)*x^2+84*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*x^2 
+85*RootOf(_Z^3-3)*x+72*(x^2+2*x+1)^(2/3)+357*RootOf(RootOf(_Z^3-3)^2+3*_Z 
*RootOf(_Z^3-3)+9*_Z^2)*x+65*RootOf(_Z^3-3)+273*RootOf(RootOf(_Z^3-3)^2+3* 
_Z*RootOf(_Z^3-3)+9*_Z^2))/(1+x)/(-2+x))*RootOf(_Z^3-3)-ln(-(15*RootOf(Roo 
tOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)^3*x^2+63*RootOf(R 
ootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)^2*RootOf(_Z^3-3)^2*x^2+15*Root 
Of(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)^3*x-63*(x^2 
+2*x+1)^(2/3)*RootOf(_Z^3-3)^2*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3) 
+9*_Z^2)+63*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)^2*RootOf(_ 
Z^3-3)^2*x-45*(x^2+2*x+1)^(1/3)*RootOf(_Z^3-3)^2*x-72*(x^2+2*x+1)^(1/3)*Ro 
otOf(_Z^3-3)*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*x-45*(...
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.27 \[ \int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=9^{\frac {1}{6}} \sqrt {\frac {1}{3}} \arctan \left (\frac {9^{\frac {1}{6}} \sqrt {\frac {1}{3}} {\left (9^{\frac {1}{3}} {\left (x + 1\right )} + 6 \, {\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (x + 1\right )}}\right ) - \frac {1}{18} \cdot 9^{\frac {2}{3}} \log \left (\frac {9^{\frac {2}{3}} {\left (x^{2} + 2 \, x + 1\right )} + 3 \cdot 9^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + 9 \, {\left (x^{2} + 2 \, x + 1\right )}^{\frac {2}{3}}}{x^{2} + 2 \, x + 1}\right ) + \frac {1}{9} \cdot 9^{\frac {2}{3}} \log \left (-\frac {9^{\frac {1}{3}} {\left (x + 1\right )} - 3 \, {\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}}}{x + 1}\right ) \] Input:

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

Output:

9^(1/6)*sqrt(1/3)*arctan(1/3*9^(1/6)*sqrt(1/3)*(9^(1/3)*(x + 1) + 6*(x^2 + 
 2*x + 1)^(1/3))/(x + 1)) - 1/18*9^(2/3)*log((9^(2/3)*(x^2 + 2*x + 1) + 3* 
9^(1/3)*(x^2 + 2*x + 1)^(1/3)*(x + 1) + 9*(x^2 + 2*x + 1)^(2/3))/(x^2 + 2* 
x + 1)) + 1/9*9^(2/3)*log(-(9^(1/3)*(x + 1) - 3*(x^2 + 2*x + 1)^(1/3))/(x 
+ 1))
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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