\(\int \frac {1-x}{1+x^3} \, dx\) [16]

Optimal result

Integrand size = 13, antiderivative size = 22 \[ \int \frac {1-x}{1+x^3} \, dx=\frac {2}{3} \log (1+x)-\frac {1}{3} \log \left (1-x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1874, 31, 642} \[ \int \frac {1-x}{1+x^3} \, dx=\frac {2}{3} \log (x+1)-\frac {1}{3} \log \left (x^2-x+1\right ) \]

[In]

Int[(1 - x)/(1 + x^3),x]

[Out]

(2*Log[1 + x])/3 - Log[1 - x + x^2]/3

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1874

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, Dist[(-r)*((B*r - A*s)/(3*a*s)), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) +
s*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[
a/b]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1-x}{1+x^3} \, dx=\frac {2}{3} \log (1+x)-\frac {1}{3} \log \left (1-x+x^2\right ) \]

[In]

Integrate[(1 - x)/(1 + x^3),x]

[Out]

(2*Log[1 + x])/3 - Log[1 - x + x^2]/3

Maple [A] (verified)

Time = 1.46 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
default	\(\frac {2 \ln \left (1+x \right )}{3}-\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(19\)
norman	\(\frac {2 \ln \left (1+x \right )}{3}-\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(19\)
risch	\(\frac {2 \ln \left (1+x \right )}{3}-\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(19\)
parallelrisch	\(\frac {2 \ln \left (1+x \right )}{3}-\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(19\)
meijerg	\(\frac {x \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}-\frac {x \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x^{2} \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {x^{2} \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}\)	\(152\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1-x}{1+x^3} \, dx=-\frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {2}{3} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {1-x}{1+x^3} \, dx=\frac {2 \log {\left (x + 1 \right )}}{3} - \frac {\log {\left (x^{2} - x + 1 \right )}}{3} \]

[In]

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

[Out]

2*log(x + 1)/3 - log(x**2 - x + 1)/3

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1-x}{1+x^3} \, dx=-\frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {2}{3} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1-x}{1+x^3} \, dx=-\frac {1}{3} \, \log \left (x^{2} - x + 1\right ) + \frac {2}{3} \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1-x}{1+x^3} \, dx=\frac {2\,\ln \left (x+1\right )}{3}-\frac {\ln \left (x^2-x+1\right )}{3} \]

[In]

int(-(x - 1)/(x^3 + 1),x)

[Out]

(2*log(x + 1))/3 - log(x^2 - x + 1)/3