Integrand size = 11, antiderivative size = 41 \[ \int \frac {x^3}{1+x^3} \, dx=x+\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \] Output:
x-1/3*ln(1+x)+1/6*ln(x^2-x+1)+1/3*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)
Time = 0.01 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{1+x^3} \, dx=x-\frac {\arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right ) \] Input:
Integrate[x^3/(1 + x^3),x]
Output:
x - ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] - Log[1 + x]/3 + Log[1 - x + x^2]/6
Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.727, Rules used = {843, 750, 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.
\(\displaystyle \int \frac {x^3}{x^3+1} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle x-\int \frac {1}{x^3+1}dx\) |
\(\Big \downarrow \) 750 |
\(\displaystyle -\frac {1}{3} \int \frac {2-x}{x^2-x+1}dx-\frac {1}{3} \int \frac {1}{x+1}dx+x\) |
\(\Big \downarrow \) 16 |
\(\displaystyle -\frac {1}{3} \int \frac {2-x}{x^2-x+1}dx+x-\frac {1}{3} \log (x+1)\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx-\frac {3}{2} \int \frac {1}{x^2-x+1}dx\right )+x-\frac {1}{3} \log (x+1)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (-\frac {3}{2} \int \frac {1}{x^2-x+1}dx-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+x-\frac {1}{3} \log (x+1)\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{3} \left (3 \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx\right )+x-\frac {1}{3} \log (x+1)\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )\right )+x-\frac {1}{3} \log (x+1)\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \log \left (x^2-x+1\right )-\sqrt {3} \arctan \left (\frac {2 x-1}{\sqrt {3}}\right )\right )+x-\frac {1}{3} \log (x+1)\) |
Input:
Int[x^3/(1 + x^3),x]
Output:
x - 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_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2) Int[1/ (Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2) Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; FreeQ[{a, b}, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
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]
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83
method | result | size |
risch | \(x -\frac {\ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}\) | \(34\) |
default | \(x -\frac {\ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\) | \(36\) |
meijerg | \(x -\frac {x \left (\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}-\frac {\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3}\) | \(74\) |
Input:
int(x^3/(x^3+1),x,method=_RETURNVERBOSE)
Output:
x-1/3*ln(1+x)+1/6*ln(x^2-x+1)-1/3*3^(1/2)*arctan(2/3*(x-1/2)*3^(1/2))
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) \] Input:
integrate(x^3/(x^3+1),x, algorithm="fricas")
Output:
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + x + 1/6*log(x^2 - x + 1) - 1/ 3*log(x + 1)
Time = 0.05 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{1+x^3} \, dx=x - \frac {\log {\left (x + 1 \right )}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \] Input:
integrate(x**3/(x**3+1),x)
Output:
x - 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 = 35, normalized size of antiderivative = 0.85 \[ \int \frac {x^3}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left (x + 1\right ) \] Input:
integrate(x^3/(x^3+1),x, algorithm="maxima")
Output:
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + x + 1/6*log(x^2 - x + 1) - 1/ 3*log(x + 1)
Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {x^3}{1+x^3} \, dx=-\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + x + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{3} \, \log \left ({\left | x + 1 \right |}\right ) \] Input:
integrate(x^3/(x^3+1),x, algorithm="giac")
Output:
-1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + x + 1/6*log(x^2 - x + 1) - 1/ 3*log(abs(x + 1))
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15 \[ \int \frac {x^3}{1+x^3} \, dx=x-\frac {\ln \left (x+1\right )}{3}+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \] Input:
int(x^3/(x^3 + 1),x)
Output:
x - log(x + 1)/3 + log(x - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/6 + 1/6) - log(x + (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/6 - 1/6)
Time = 0.15 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x^3}{1+x^3} \, dx=-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{3}+\frac {\mathrm {log}\left (x^{2}-x +1\right )}{6}-\frac {\mathrm {log}\left (x +1\right )}{3}+x \] Input:
int(x^3/(x^3+1),x)
Output:
( - 2*sqrt(3)*atan((2*x - 1)/sqrt(3)) + log(x**2 - x + 1) - 2*log(x + 1) + 6*x)/6