Integrand size = 14, antiderivative size = 41 \[ \int \frac {(1-x) x}{1+x^3} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log (1+x)-\frac {1}{6} \log \left (1-x+x^2\right ) \] Output:
-1/3*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)-2/3*ln(1+x)-1/6*ln(x^2-x+1)
Time = 0.01 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.22 \[ \int \frac {(1-x) x}{1+x^3} \, dx=\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right )-\frac {1}{3} \log \left (1+x^3\right ) \] Input:
Integrate[((1 - x)*x)/(1 + x^3),x]
Output:
ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] - Log[1 + x]/3 + Log[1 - x + x^2]/6 - L og[1 + x^3]/3
Time = 0.39 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2413, 16, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(1-x) x}{x^3+1} \, dx\) |
\(\Big \downarrow \) 2413 |
\(\displaystyle \frac {1}{3} \int \frac {2-x}{x^2-x+1}dx-\frac {2}{3} \int \frac {1}{x+1}dx\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{3} \int \frac {2-x}{x^2-x+1}dx-\frac {2}{3} \log (x+1)\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx\right )-\frac {2}{3} \log (x+1)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )-\frac {2}{3} \log (x+1)\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-3 \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )-\frac {2}{3} \log (x+1)\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx+\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )\right )-\frac {2}{3} \log (x+1)\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )-\frac {1}{2} \log \left (x^2-x+1\right )\right )-\frac {2}{3} \log (x+1)\) |
Input:
Int[((1 - x)*x)/(1 + x^3),x]
Output:
(-2*Log[1 + x])/3 + (Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] - Log[1 - x + x^2] /2)/3
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[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]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x, 2], q = (a/b)^(1/3)}, Simp[q*((A - B*q + C*q^2)/(3*a)) Int[1/(q + x), x], x] + Simp[q/(3*a) Int[(q*(2*A + B*q - C*q^2) - (A - B*q - 2*C*q^2)*x)/(q^2 - q*x + x^2), x], x] /; NeQ[a*B^3 - b*A^3, 0] && NeQ[A - B*q + C*q^2, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2 ] && GtQ[a/b, 0]
Time = 0.10 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85
method | result | size |
default | \(-\frac {\ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {2 \ln \left (x +1\right )}{3}\) | \(35\) |
risch | \(-\frac {\ln \left (4 x^{2}-4 x +4\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-\frac {2 \ln \left (x +1\right )}{3}\) | \(37\) |
meijerg | \(-\frac {x^{2} \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{3}+1\right )}{3}\) | \(88\) |
Input:
int((1-x)*x/(x^3+1),x,method=_RETURNVERBOSE)
Output:
-1/6*ln(x^2-x+1)+1/3*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))-2/3*ln(x+1)
Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {(1-x) x}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) \] Input:
integrate((1-x)*x/(x^3+1),x, algorithm="fricas")
Output:
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*log(x^2 - x + 1) - 2/3*log (x + 1)
Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {(1-x) x}{1+x^3} \, dx=- \frac {2 \log {\left (x + 1 \right )}}{3} - \frac {\log {\left (x^{2} - x + 1 \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \] Input:
integrate((1-x)*x/(x**3+1),x)
Output:
-2*log(x + 1)/3 - log(x**2 - x + 1)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt( 3)/3)/3
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {(1-x) x}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) \] Input:
integrate((1-x)*x/(x^3+1),x, algorithm="maxima")
Output:
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*log(x^2 - x + 1) - 2/3*log (x + 1)
Time = 0.12 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {(1-x) x}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left ({\left | x + 1 \right |}\right ) \] Input:
integrate((1-x)*x/(x^3+1),x, algorithm="giac")
Output:
1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*log(x^2 - x + 1) - 2/3*log (abs(x + 1))
Time = 0.10 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.54 \[ \int \frac {(1-x) x}{1+x^3} \, dx=-\frac {\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{6}-\frac {\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{6}-\frac {2\,\ln \left (x+1\right )}{3}-\frac {\sqrt {3}\,\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{6} \] Input:
int(-(x*(x - 1))/(x^3 + 1),x)
Output:
(3^(1/2)*log(x + (3^(1/2)*1i)/2 - 1/2)*1i)/6 - log(x + (3^(1/2)*1i)/2 - 1/ 2)/6 - (2*log(x + 1))/3 - (3^(1/2)*log(x - (3^(1/2)*1i)/2 - 1/2)*1i)/6 - l og(x - (3^(1/2)*1i)/2 - 1/2)/6
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {(1-x) x}{1+x^3} \, dx=\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{3}-\frac {\mathrm {log}\left (x^{2}-x +1\right )}{6}-\frac {2 \,\mathrm {log}\left (x +1\right )}{3} \] Input:
int((1-x)*x/(x^3+1),x)
Output:
(2*sqrt(3)*atan((2*x - 1)/sqrt(3)) - log(x**2 - x + 1) - 4*log(x + 1))/6