\(\int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx\) [226]

Optimal result

Integrand size = 13, antiderivative size = 67 \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\sqrt {3} \arctan \left (\frac {1+\frac {2 (-1+x)}{\sqrt [3]{(-1+x)^2 (1+x)}}}{\sqrt {3}}\right )-\frac {1}{2} \log (1+x)-\frac {3}{2} \log \left (1-\frac {-1+x}{\sqrt [3]{(-1+x)^2 (1+x)}}\right ) \]

[Out]

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

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(188\) vs. \(2(67)=134\).

Time = 0.09 (sec) , antiderivative size = 188, normalized size of antiderivative = 2.81, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2092, 2089, 62} \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=-\frac {(3-3 x)^{2/3} \sqrt [3]{x+1} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x+1}}{\sqrt [6]{3} \sqrt [3]{3-3 x}}\right )}{\sqrt [6]{3} \sqrt [3]{x^3-x^2-x+1}}-\frac {(3-3 x)^{2/3} \sqrt [3]{x+1} \log \left (-\frac {8}{3} (x-1)\right )}{2\ 3^{2/3} \sqrt [3]{x^3-x^2-x+1}}-\frac {\sqrt [3]{3} (3-3 x)^{2/3} \sqrt [3]{x+1} \log \left (\frac {\sqrt [3]{3} \sqrt [3]{x+1}}{\sqrt [3]{3-3 x}}+1\right )}{2 \sqrt [3]{x^3-x^2-x+1}} \]

[In]

Int[((-1 + x)^2*(1 + x))^(-1/3),x]

[Out]

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

Rule 62

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[-d/b, 3]}, Simp[Sqrt[
3]*(q/d)*ArcTan[1/Sqrt[3] - 2*q*((a + b*x)^(1/3)/(Sqrt[3]*(c + d*x)^(1/3)))], x] + (Simp[3*(q/(2*d))*Log[q*((a
+ b*x)^(1/3)/(c + d*x)^(1/3)) + 1], x] + Simp[(q/(2*d))*Log[c + d*x], x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0] && NegQ[d/b]

Rule 2089

Int[((a_.) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/((3*a - b*x)^p*(3*a + 2*b*
x)^(2*p)), Int[(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b, d, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0]
&&  !IntegerQ[p]

Rule 2092

Int[(P3_)^(p_), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = Coeff[P3, x, 2], d = Coeff[P3
, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c*d + 27*a*d^2)/(27*d^2) - (c^2 - 3*b*d)*(x/(3*d)) + d*x^3, x]^p, x], x,
x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[p, x] && PolyQ[P3, x, 3]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\sqrt [3]{\frac {16}{27}-\frac {4 x}{3}+x^3}} \, dx,x,-\frac {1}{3}+x\right ) \\ & = \frac {\left (4\ 2^{2/3} (1-x)^{2/3} \sqrt [3]{1+x}\right ) \text {Subst}\left (\int \frac {1}{\left (\frac {16}{9}-\frac {8 x}{3}\right )^{2/3} \sqrt [3]{\frac {16}{9}+\frac {4 x}{3}}} \, dx,x,-\frac {1}{3}+x\right )}{3 \sqrt [3]{1-x-x^2+x^3}} \\ & = -\frac {\sqrt {3} (1-x)^{2/3} \sqrt [3]{1+x} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{1+x}}{\sqrt {3} \sqrt [3]{1-x}}\right )}{\sqrt [3]{1-x-x^2+x^3}}-\frac {(1-x)^{2/3} \sqrt [3]{1+x} \log (1-x)}{2 \sqrt [3]{1-x-x^2+x^3}}-\frac {3 (1-x)^{2/3} \sqrt [3]{1+x} \log \left (\frac {\sqrt [3]{1-x}+\sqrt [3]{1+x}}{\sqrt [3]{1-x}}\right )}{2 \sqrt [3]{1-x-x^2+x^3}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.84 \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=\frac {(-1+x)^{2/3} \sqrt [3]{1+x} \left (-2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{1+x}}{2 \sqrt [3]{-1+x}+\sqrt [3]{1+x}}\right )-2 \log \left (\sqrt [3]{-1+x}-\sqrt [3]{1+x}\right )+\log \left ((-1+x)^{2/3}+\sqrt [3]{-1+x} \sqrt [3]{1+x}+(1+x)^{2/3}\right )\right )}{2 \sqrt [3]{(-1+x)^2 (1+x)}} \]

[In]

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

[Out]

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

Maple [C] (verified)

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

Time = 0.53 (sec) , antiderivative size = 370, normalized size of antiderivative = 5.52

method	result	size
trager	\(-\ln \left (\frac {4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -4 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +x^{2}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-1}{-1+x}\right )+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{2}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x -5 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{2}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {2}{3}}+3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}} x +6 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x +2 x^{2}-3 \left (x^{3}-x^{2}-x +1\right )^{\frac {1}{3}}-\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )-4 x +2}{-1+x}\right )\)	\(370\)

[In]

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

[Out]

-ln((4*RootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(2/3)-3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(1/3)
*x-4*RootOf(_Z^2-_Z+1)^2*x-4*RootOf(_Z^2-_Z+1)*x^2+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(1/3)+3*(x^3-x^2-x+1)^(1/
3)*x+2*RootOf(_Z^2-_Z+1)*x+x^2-3*(x^3-x^2-x+1)^(1/3)+2*RootOf(_Z^2-_Z+1)-1)/(-1+x))+RootOf(_Z^2-_Z+1)*ln(-(2*R
ootOf(_Z^2-_Z+1)^2*x^2+3*RootOf(_Z^2-_Z+1)*(x^3-x^2-x+1)^(2/3)-2*RootOf(_Z^2-_Z+1)^2*x-5*RootOf(_Z^2-_Z+1)*x^2
-3*(x^3-x^2-x+1)^(2/3)+3*(x^3-x^2-x+1)^(1/3)*x+6*RootOf(_Z^2-_Z+1)*x+2*x^2-3*(x^3-x^2-x+1)^(1/3)-RootOf(_Z^2-_
Z+1)-4*x+2)/(-1+x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (56) = 112\).

Time = 0.23 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.91 \[ \int \frac {1}{\sqrt [3]{(-1+x)^2 (1+x)}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (x - 1\right )} + 2 \, \sqrt {3} {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}}}{3 \, {\left (x - 1\right )}}\right ) + \frac {1}{2} \, \log \left (\frac {x^{2} + {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} - 2 \, x + {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {2}{3}} + 1}{x^{2} - 2 \, x + 1}\right ) - \log \left (-\frac {x - {\left (x^{3} - x^{2} - x + 1\right )}^{\frac {1}{3}} - 1}{x - 1}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/((x - 1)^2*(x + 1))^(1/3),x)

[Out]

int(1/((x - 1)^2*(x + 1))^(1/3), x)