\(\int \frac {-1+x}{(1+x) \sqrt [3]{-1+x^3}} \, dx\) [2685]

Optimal result

Integrand size = 18, antiderivative size = 243 \[ \int \frac {-1+x}{(1+x) \sqrt [3]{-1+x^3}} \, dx=\frac {\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )}{\sqrt {3}}-\frac {\sqrt {3} \arctan \left (\frac {-\frac {\sqrt [3]{2}}{\sqrt {3}}+\frac {\sqrt [3]{2} x}{\sqrt {3}}+\frac {\sqrt [3]{-1+x^3}}{\sqrt {3}}}{\sqrt [3]{-1+x^3}}\right )}{\sqrt [3]{2}}+\frac {\log \left (-\sqrt [3]{2}+\sqrt [3]{2} x-2 \sqrt [3]{-1+x^3}\right )}{\sqrt [3]{2}}-\frac {1}{3} \log \left (-x+\sqrt [3]{-1+x^3}\right )+\frac {1}{6} \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )-\frac {\log \left (2^{2/3}-2\ 2^{2/3} x+2^{2/3} x^2+\left (-2 \sqrt [3]{2}+2 \sqrt [3]{2} x\right ) \sqrt [3]{-1+x^3}+4 \left (-1+x^3\right )^{2/3}\right )}{2 \sqrt [3]{2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.57, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2177, 245, 2174} \[ \int \frac {-1+x}{(1+x) \sqrt [3]{-1+x^3}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {\sqrt [3]{2} (1-x)}{\sqrt [3]{x^3-1}}}{\sqrt {3}}\right )}{\sqrt [3]{2}}+\frac {\arctan \left (\frac {\frac {2 x}{\sqrt [3]{x^3-1}}+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{2} \log \left (\sqrt [3]{x^3-1}-x\right )+\frac {3 \log \left (2^{2/3} \sqrt [3]{x^3-1}-x+1\right )}{2 \sqrt [3]{2}}-\frac {\log \left ((1-x) (x+1)^2\right )}{2 \sqrt [3]{2}} \]

[In]

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

[Out]

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

Rule 245

Int[((a_) + (b_.)*(x_)^3)^(-1/3), x_Symbol] :> Simp[ArcTan[(1 + 2*Rt[b, 3]*(x/(a + b*x^3)^(1/3)))/Sqrt[3]]/(Sq
rt[3]*Rt[b, 3]), x] - Simp[Log[(a + b*x^3)^(1/3) - Rt[b, 3]*x]/(2*Rt[b, 3]), x] /; FreeQ[{a, b}, x]

Rule 2174

Int[1/(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Simp[Sqrt[3]*(ArcTan[(1 - 2^(1/3)*Rt[b,
3]*((c - d*x)/(d*(a + b*x^3)^(1/3))))/Sqrt[3]]/(2^(4/3)*Rt[b, 3]*c)), x] + (Simp[Log[(c + d*x)^2*(c - d*x)]/(2
^(7/3)*Rt[b, 3]*c), x] - Simp[(3*Log[Rt[b, 3]*(c - d*x) + 2^(2/3)*d*(a + b*x^3)^(1/3)])/(2^(7/3)*Rt[b, 3]*c),
x]) /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^3 + a*d^3, 0]

Rule 2177

Int[((e_.) + (f_.)*(x_))/(((c_.) + (d_.)*(x_))*((a_) + (b_.)*(x_)^3)^(1/3)), x_Symbol] :> Dist[f/d, Int[1/(a +
b*x^3)^(1/3), x], x] + Dist[(d*e - c*f)/d, Int[1/((c + d*x)*(a + b*x^3)^(1/3)), x], x] /; FreeQ[{a, b, c, d,
e, f}, x]

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

Mathematica [A] (verified)

Time = 5.32 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.93 \[ \int \frac {-1+x}{(1+x) \sqrt [3]{-1+x^3}} \, dx=\frac {1}{12} \left (-6 2^{2/3} \sqrt {3} \arctan \left (\frac {-\sqrt [3]{2}+\sqrt [3]{2} x+\sqrt [3]{-1+x^3}}{\sqrt {3} \sqrt [3]{-1+x^3}}\right )+4 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{-1+x^3}}\right )+6\ 2^{2/3} \log \left (-\sqrt [3]{2}+\sqrt [3]{2} x-2 \sqrt [3]{-1+x^3}\right )-4 \log \left (-x+\sqrt [3]{-1+x^3}\right )+2 \log \left (x^2+x \sqrt [3]{-1+x^3}+\left (-1+x^3\right )^{2/3}\right )-3\ 2^{2/3} \log \left (2^{2/3}-2\ 2^{2/3} x+2^{2/3} x^2+2 \sqrt [3]{2} (-1+x) \sqrt [3]{-1+x^3}+4 \left (-1+x^3\right )^{2/3}\right )\right ) \]

[In]

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

[Out]

(-6*2^(2/3)*Sqrt[3]*ArcTan[(-2^(1/3) + 2^(1/3)*x + (-1 + x^3)^(1/3))/(Sqrt[3]*(-1 + x^3)^(1/3))] + 4*Sqrt[3]*A
rcTan[(Sqrt[3]*x)/(x + 2*(-1 + x^3)^(1/3))] + 6*2^(2/3)*Log[-2^(1/3) + 2^(1/3)*x - 2*(-1 + x^3)^(1/3)] - 4*Log
[-x + (-1 + x^3)^(1/3)] + 2*Log[x^2 + x*(-1 + x^3)^(1/3) + (-1 + x^3)^(2/3)] - 3*2^(2/3)*Log[2^(2/3) - 2*2^(2/
3)*x + 2^(2/3)*x^2 + 2*2^(1/3)*(-1 + x)*(-1 + x^3)^(1/3) + 4*(-1 + x^3)^(2/3)])/12

Maple [C] (warning: unable to verify)

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

Time = 11.90 (sec) , antiderivative size = 1285, normalized size of antiderivative = 5.29

method	result	size
trager	\(\text {Expression too large to display}\)	\(1285\)

[In]

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

[Out]

-1/3*ln(-RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^4*x^3-6*(x^3-1)^(2/3)*RootOf(_Z^3-4)
^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x+2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(
_Z^3-4)^2*x^3-4*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)-24*x^2*(x^3-1)^(1/3)+8*x^3-8)
+1/2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*ln((2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*Roo
tOf(_Z^3-4)^3*x+5*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^2*x-4*(x^3-1)^(2/3)*RootOf(
_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)+4*(x^3-1)^(1/3)*RootOf(_Z^3-4)^2*x-9*RootOf(RootOf(_
Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)*(x^3-1)^(1/3)*x-4*(x^3-1)^(1/3)*RootOf(_Z^3-4)^2+9*RootOf(Root
Of(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)*(x^3-1)^(1/3)+14*RootOf(_Z^3-4)*x^2+35*RootOf(RootOf(_Z^3-
4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x^2+12*RootOf(_Z^3-4)*x+30*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x-52*(
x^3-1)^(2/3)+14*RootOf(_Z^3-4)+35*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2))/(1+x)^2)+1/2*RootOf(_Z^3-4)
*ln(-(5*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)^3*x+2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf
(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^2*x-4*(x^3-1)^(2/3)*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)
+_Z^2)+4*(x^3-1)^(1/3)*RootOf(_Z^3-4)^2*x+13*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4)*(x
^3-1)^(1/3)*x-4*(x^3-1)^(1/3)*RootOf(_Z^3-4)^2-13*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-
4)*(x^3-1)^(1/3)-35*RootOf(_Z^3-4)*x^2-14*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x^2-10*RootOf(_Z^3-4
)*x-4*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x+36*(x^3-1)^(2/3)-35*RootOf(_Z^3-4)-14*RootOf(RootOf(_Z
^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2))/(1+x)^2)+1/3*ln(RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3
-4)^4*x^3+12*(x^3-1)^(2/3)*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x-12*RootOf(_Z^3-4
)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*(x^3-1)^(1/3)*x^2+4*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3
-4)+_Z^2)*RootOf(_Z^3-4)^2*x^3-4*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)+48*x*(x^3-1)
^(2/3)-32*x^3+16)+1/12*ln(RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)^2*RootOf(_Z^3-4)^4*x^3+12*(x^3-1)^(2
/3)*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*x-12*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-
4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*(x^3-1)^(1/3)*x^2+4*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)*RootOf(_Z^3-4
)^2*x^3-4*RootOf(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)+48*x*(x^3-1)^(2/3)-32*x^3+16)*RootO
f(_Z^3-4)^2*RootOf(RootOf(_Z^3-4)^2+_Z*RootOf(_Z^3-4)+_Z^2)

Fricas [A] (verification not implemented)

none

Time = 1.62 (sec) , antiderivative size = 370, normalized size of antiderivative = 1.52 \[ \int \frac {-1+x}{(1+x) \sqrt [3]{-1+x^3}} \, dx=\frac {1}{6} \cdot 4^{\frac {1}{3}} \sqrt {3} \arctan \left (\frac {4 \cdot 4^{\frac {2}{3}} \sqrt {3} {\left (x^{4} + 2 \, x^{3} + 2 \, x^{2} + 2 \, x + 1\right )} {\left (x^{3} - 1\right )}^{\frac {2}{3}} + 2 \cdot 4^{\frac {1}{3}} \sqrt {3} {\left (5 \, x^{5} - 5 \, x^{4} + 6 \, x^{3} - 6 \, x^{2} + 5 \, x - 5\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}} + \sqrt {3} {\left (13 \, x^{6} + 2 \, x^{5} + 19 \, x^{4} - 4 \, x^{3} + 19 \, x^{2} + 2 \, x + 13\right )}}{3 \, {\left (3 \, x^{6} - 18 \, x^{5} - 3 \, x^{4} - 28 \, x^{3} - 3 \, x^{2} - 18 \, x + 3\right )}}\right ) + \frac {1}{3} \, \sqrt {3} \arctan \left (-\frac {4 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} - 2 \, \sqrt {3} {\left (x^{3} - 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (x^{3} - 1\right )}}{9 \, x^{3} - 1}\right ) - \frac {1}{12} \cdot 4^{\frac {1}{3}} \log \left (\frac {8 \cdot 4^{\frac {1}{3}} {\left (x^{3} - 1\right )}^{\frac {2}{3}} {\left (x^{2} + 1\right )} + 4^{\frac {2}{3}} {\left (5 \, x^{4} + 6 \, x^{2} + 5\right )} + 4 \, {\left (3 \, x^{3} - x^{2} + x - 3\right )} {\left (x^{3} - 1\right )}^{\frac {1}{3}}}{x^{4} + 4 \, x^{3} + 6 \, x^{2} + 4 \, x + 1}\right ) + \frac {1}{6} \cdot 4^{\frac {1}{3}} \log \left (\frac {4^{\frac {2}{3}} {\left (x^{3} - 1\right )}^{\frac {1}{3}} {\left (x - 1\right )} + 4^{\frac {1}{3}} {\left (x^{2} + 2 \, x + 1\right )} - 4 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}}}{x^{2} + 2 \, x + 1}\right ) - \frac {1}{6} \, \log \left (-3 \, {\left (x^{3} - 1\right )}^{\frac {1}{3}} x^{2} + 3 \, {\left (x^{3} - 1\right )}^{\frac {2}{3}} x + 1\right ) \]

[In]

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

[Out]

1/6*4^(1/3)*sqrt(3)*arctan(1/3*(4*4^(2/3)*sqrt(3)*(x^4 + 2*x^3 + 2*x^2 + 2*x + 1)*(x^3 - 1)^(2/3) + 2*4^(1/3)*
sqrt(3)*(5*x^5 - 5*x^4 + 6*x^3 - 6*x^2 + 5*x - 5)*(x^3 - 1)^(1/3) + sqrt(3)*(13*x^6 + 2*x^5 + 19*x^4 - 4*x^3 +
19*x^2 + 2*x + 13))/(3*x^6 - 18*x^5 - 3*x^4 - 28*x^3 - 3*x^2 - 18*x + 3)) + 1/3*sqrt(3)*arctan(-(4*sqrt(3)*(x
^3 - 1)^(1/3)*x^2 - 2*sqrt(3)*(x^3 - 1)^(2/3)*x + sqrt(3)*(x^3 - 1))/(9*x^3 - 1)) - 1/12*4^(1/3)*log((8*4^(1/3
)*(x^3 - 1)^(2/3)*(x^2 + 1) + 4^(2/3)*(5*x^4 + 6*x^2 + 5) + 4*(3*x^3 - x^2 + x - 3)*(x^3 - 1)^(1/3))/(x^4 + 4*
x^3 + 6*x^2 + 4*x + 1)) + 1/6*4^(1/3)*log((4^(2/3)*(x^3 - 1)^(1/3)*(x - 1) + 4^(1/3)*(x^2 + 2*x + 1) - 4*(x^3
- 1)^(2/3))/(x^2 + 2*x + 1)) - 1/6*log(-3*(x^3 - 1)^(1/3)*x^2 + 3*(x^3 - 1)^(2/3)*x + 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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