\(\int \frac {(1-x) x^3}{1+x^3} \, dx\) [305]

Optimal result

Integrand size = 16, antiderivative size = 30 \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=x-\frac {x^2}{2}-\frac {2}{3} \log (1+x)+\frac {1}{3} \log \left (1-x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1901, 1874, 31, 642} \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=-\frac {x^2}{2}+\frac {1}{3} \log \left (x^2-x+1\right )+x-\frac {2}{3} \log (x+1) \]

[In]

Int[((1 - x)*x^3)/(1 + x^3),x]

[Out]

x - x^2/2 - (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]

Rule 1901

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (1-x-\frac {1-x}{1+x^3}\right ) \, dx \\ & = x-\frac {x^2}{2}-\int \frac {1-x}{1+x^3} \, dx \\ & = x-\frac {x^2}{2}-\frac {1}{3} \int \frac {1-2 x}{1-x+x^2} \, dx-\frac {2}{3} \int \frac {1}{1+x} \, dx \\ & = x-\frac {x^2}{2}-\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 = 30, normalized size of antiderivative = 1.00 \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=x-\frac {x^2}{2}-\frac {2}{3} \log (1+x)+\frac {1}{3} \log \left (1-x+x^2\right ) \]

[In]

Integrate[((1 - x)*x^3)/(1 + x^3),x]

[Out]

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

Maple [A] (verified)

Time = 1.50 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83

method	result	size
default	\(x -\frac {x^{2}}{2}-\frac {2 \ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(25\)
norman	\(x -\frac {x^{2}}{2}-\frac {2 \ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(25\)
risch	\(x -\frac {x^{2}}{2}-\frac {2 \ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(25\)
parallelrisch	\(x -\frac {x^{2}}{2}-\frac {2 \ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{3}\)	\(25\)
meijerg	\(x -\frac {x \left (\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}-\frac {\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3}-\frac {x^{2}}{2}+\frac {x^{2} \left (-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{\left (x^{3}\right )^{\frac {2}{3}}}+\frac {\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{\left (x^{3}\right )^{\frac {2}{3}}}\right )}{3}\)	\(153\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=-\frac {1}{2} \, x^{2} + x + \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=- \frac {x^{2}}{2} + x - \frac {2 \log {\left (x + 1 \right )}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{3} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=-\frac {1}{2} \, x^{2} + x + \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.83 \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=-\frac {1}{2} \, x^{2} + x + \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.80 \[ \int \frac {(1-x) x^3}{1+x^3} \, dx=x-\frac {2\,\ln \left (x+1\right )}{3}+\frac {\ln \left (x^2-x+1\right )}{3}-\frac {x^2}{2} \]

[In]

int(-(x^3*(x - 1))/(x^3 + 1),x)

[Out]

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