\(\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx\) [1527]
Optimal result
Integrand size = 18, antiderivative size = 106 \[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=\frac {\left ((1+x)^2\right )^{2/3} \left (-\frac {\arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x}}{3^{5/6}}\right )}{\sqrt [6]{3}}+\frac {\log \left (-3+3^{2/3} \sqrt [3]{1+x}\right )}{3^{2/3}}-\frac {\log \left (3+3^{2/3} \sqrt [3]{1+x}+\sqrt [3]{3} (1+x)^{2/3}\right )}{2\ 3^{2/3}}\right )}{(1+x)^{4/3}}
\]
[Out]
((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)
Rubi [A] (verified)
Time = 0.03 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.20, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {660, 59, 631, 210, 31}
\[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=-\frac {(x+1)^{2/3} \arctan \left (\frac {2 \sqrt [3]{x+1}+\sqrt [3]{3}}{3^{5/6}}\right )}{\sqrt [6]{3} \sqrt [3]{x^2+2 x+1}}-\frac {(x+1)^{2/3} \log (2-x)}{2\ 3^{2/3} \sqrt [3]{x^2+2 x+1}}+\frac {\sqrt [3]{3} (x+1)^{2/3} \log \left (\sqrt [3]{3}-\sqrt [3]{x+1}\right )}{2 \sqrt [3]{x^2+2 x+1}}
\]
[In]
Int[1/((-2 + x)*(1 + 2*x + x^2)^(1/3)),x]
[Out]
-(((1 + x)^(2/3)*ArcTan[(3^(1/3) + 2*(1 + x)^(1/3))/3^(5/6)])/(3^(1/6)*(1 + 2*x + x^2)^(1/3))) - ((1 + x)^(2/3
)*Log[2 - x])/(2*3^(2/3)*(1 + 2*x + x^2)^(1/3)) + (3^(1/3)*(1 + x)^(2/3)*Log[3^(1/3) - (1 + x)^(1/3)])/(2*(1 +
2*x + x^2)^(1/3))
Rule 31
Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]
Rule 59
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, Simp[-L
og[RemoveContent[a + b*x, x]]/(2*b*q^2), x] + (-Dist[3/(2*b*q), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x
)^(1/3)], x] - Dist[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 210
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 631
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 660
Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[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] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]
Rubi steps \begin{align*}
\text {integral}& = \frac {(1+x)^{2/3} \int \frac {1}{(-2+x) (1+x)^{2/3}} \, dx}{\sqrt [3]{1+2 x+x^2}} \\ & = -\frac {(1+x)^{2/3} \log (2-x)}{2\ 3^{2/3} \sqrt [3]{1+2 x+x^2}}-\frac {\left (\sqrt [3]{3} (1+x)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [3]{3}-x} \, dx,x,\sqrt [3]{1+x}\right )}{2 \sqrt [3]{1+2 x+x^2}}-\frac {\left (3^{2/3} (1+x)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{3^{2/3}+\sqrt [3]{3} x+x^2} \, dx,x,\sqrt [3]{1+x}\right )}{2 \sqrt [3]{1+2 x+x^2}} \\ & = -\frac {(1+x)^{2/3} \log (2-x)}{2\ 3^{2/3} \sqrt [3]{1+2 x+x^2}}+\frac {\sqrt [3]{3} (1+x)^{2/3} \log \left (\sqrt [3]{3}-\sqrt [3]{1+x}\right )}{2 \sqrt [3]{1+2 x+x^2}}+\frac {\left (\sqrt [3]{3} (1+x)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+\frac {2 \sqrt [3]{1+x}}{\sqrt [3]{3}}\right )}{\sqrt [3]{1+2 x+x^2}} \\ & = -\frac {(1+x)^{2/3} \arctan \left (\frac {1}{3} \left (\sqrt {3}+2 \sqrt [6]{3} \sqrt [3]{1+x}\right )\right )}{\sqrt [6]{3} \sqrt [3]{1+2 x+x^2}}-\frac {(1+x)^{2/3} \log (2-x)}{2\ 3^{2/3} \sqrt [3]{1+2 x+x^2}}+\frac {\sqrt [3]{3} (1+x)^{2/3} \log \left (\sqrt [3]{3}-\sqrt [3]{1+x}\right )}{2 \sqrt [3]{1+2 x+x^2}} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.10 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.95
\[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=-\frac {(1+x)^{2/3} \left (2 \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x}}{3^{5/6}}\right )-2 \log \left (-3+3^{2/3} \sqrt [3]{1+x}\right )+\log \left (3+3^{2/3} \sqrt [3]{1+x}+\sqrt [3]{3} (1+x)^{2/3}\right )\right )}{2\ 3^{2/3} \sqrt [3]{(1+x)^2}}
\]
[In]
Integrate[1/((-2 + x)*(1 + 2*x + x^2)^(1/3)),x]
[Out]
-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))
Maple [C] (warning: unable to verify)
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.63 (sec) , antiderivative size = 942, normalized size of antiderivative = 8.89
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| method | result | size |
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| trager |
\(\text {Expression too large to display}\) |
\(942\) |
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[In]
int(1/(-2+x)/(x^2+2*x+1)^(1/3),x,method=_RETURNVERBOSE)
[Out]
RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*ln(-(6*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*RootOf(_Z^3-3)^2*x^2+6*RootOf(R
ootOf(_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(RootOf(_Z^3-3)^2+3*
_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)^2+63*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)^2*RootOf(_Z^
3-3)^2*x-21*(x^2+2*x+1)^(1/3)*RootOf(_Z^3-3)^2*x+72*(x^2+2*x+1)^(1/3)*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3
-3)+9*_Z^2)*RootOf(_Z^3-3)*x-21*(x^2+2*x+1)^(1/3)*RootOf(_Z^3-3)^2+72*(x^2+2*x+1)^(1/3)*RootOf(RootOf(_Z^3-3)^
2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)-2*RootOf(_Z^3-3)*x^2-21*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-
3)+9*_Z^2)*x^2-28*RootOf(_Z^3-3)*x+135*(x^2+2*x+1)^(2/3)-294*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^
2)*x-26*RootOf(_Z^3-3)-273*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2))/(-2+x)/(1+x))+1/3*RootOf(_Z^3-
3)*ln((21*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)^3*x^2+18*RootOf(RootOf(_Z^3-3)^2+
3*_Z*RootOf(_Z^3-3)+9*_Z^2)^2*RootOf(_Z^3-3)^2*x^2+21*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*Root
Of(_Z^3-3)^3*x+63*(x^2+2*x+1)^(2/3)*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)^2+18*Ro
otOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)^2*RootOf(_Z^3-3)^2*x-21*(x^2+2*x+1)^(1/3)*RootOf(_Z^3-3)^2*x
-135*(x^2+2*x+1)^(1/3)*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)*x-21*(x^2+2*x+1)^(1/
3)*RootOf(_Z^3-3)^2-135*(x^2+2*x+1)^(1/3)*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*RootOf(_Z^3-3)+2
8*RootOf(_Z^3-3)*x^2+24*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*x^2+119*RootOf(_Z^3-3)*x-72*(x^2+2
*x+1)^(2/3)+102*RootOf(RootOf(_Z^3-3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2)*x+91*RootOf(_Z^3-3)+78*RootOf(RootOf(_Z^3-
3)^2+3*_Z*RootOf(_Z^3-3)+9*_Z^2))/(-2+x)/(1+x))
Fricas [A] (verification not implemented)
none
Time = 0.28 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.31
\[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=\frac {1}{3} \cdot 9^{\frac {1}{6}} \sqrt {3} \arctan \left (\frac {9^{\frac {1}{6}} {\left (9^{\frac {1}{3}} \sqrt {3} {\left (x + 1\right )} + 6 \, \sqrt {3} {\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}}\right )}}{9 \, {\left (x + 1\right )}}\right ) - \frac {1}{18} \cdot 9^{\frac {2}{3}} \log \left (\frac {9^{\frac {2}{3}} {\left (x^{2} + 2 \, x + 1\right )} + 3 \cdot 9^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} {\left (x + 1\right )} + 9 \, {\left (x^{2} + 2 \, x + 1\right )}^{\frac {2}{3}}}{x^{2} + 2 \, x + 1}\right ) + \frac {1}{9} \cdot 9^{\frac {2}{3}} \log \left (-\frac {9^{\frac {1}{3}} {\left (x + 1\right )} - 3 \, {\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}}}{x + 1}\right )
\]
[In]
integrate(1/(-2+x)/(x^2+2*x+1)^(1/3),x, algorithm="fricas")
[Out]
1/3*9^(1/6)*sqrt(3)*arctan(1/9*9^(1/6)*(9^(1/3)*sqrt(3)*(x + 1) + 6*sqrt(3)*(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]
\[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=\int \frac {1}{\left (x - 2\right ) \sqrt [3]{\left (x + 1\right )^{2}}}\, dx
\]
[In]
integrate(1/(-2+x)/(x**2+2*x+1)**(1/3),x)
[Out]
Integral(1/((x - 2)*((x + 1)**2)**(1/3)), x)
Maxima [F]
\[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} {\left (x - 2\right )}} \,d x }
\]
[In]
integrate(1/(-2+x)/(x^2+2*x+1)^(1/3),x, algorithm="maxima")
[Out]
integrate(1/((x^2 + 2*x + 1)^(1/3)*(x - 2)), x)
Giac [F]
\[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=\int { \frac {1}{{\left (x^{2} + 2 \, x + 1\right )}^{\frac {1}{3}} {\left (x - 2\right )}} \,d x }
\]
[In]
integrate(1/(-2+x)/(x^2+2*x+1)^(1/3),x, algorithm="giac")
[Out]
integrate(1/((x^2 + 2*x + 1)^(1/3)*(x - 2)), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {1}{(-2+x) \sqrt [3]{1+2 x+x^2}} \, dx=\int \frac {1}{\left (x-2\right )\,{\left (x^2+2\,x+1\right )}^{1/3}} \,d x
\]
[In]
int(1/((x - 2)*(2*x + x^2 + 1)^(1/3)),x)
[Out]
int(1/((x - 2)*(2*x + x^2 + 1)^(1/3)), x)