\(\int \frac {x (1+2 x)}{1+x^3} \, dx\) [311]

Optimal result

Integrand size = 14, antiderivative size = 41 \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1+x)+\frac {5}{6} \log \left (1-x+x^2\right ) \]

[Out]

1/3*ln(1+x)+5/6*ln(x^2-x+1)-1/3*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {1888, 31, 648, 632, 210, 642} \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {5}{6} \log \left (x^2-x+1\right )+\frac {1}{3} \log (x+1) \]

[In]

Int[(x*(1 + 2*x))/(1 + x^3),x]

[Out]

-(ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3]) + Log[1 + x]/3 + (5*Log[1 - x + x^2])/6

Rule 31

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

Rule 210

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])

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

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 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1888

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)}, Dist[q*((A - B*q + C*q^2)/(3*a)), Int[1/(q + x), x], x] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {1}{1+x} \, dx+\frac {1}{3} \int \frac {-1+5 x}{1-x+x^2} \, dx \\ & = \frac {1}{3} \log (1+x)+\frac {1}{2} \int \frac {1}{1-x+x^2} \, dx+\frac {5}{6} \int \frac {-1+2 x}{1-x+x^2} \, dx \\ & = \frac {1}{3} \log (1+x)+\frac {5}{6} \log \left (1-x+x^2\right )-\text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right ) \\ & = \frac {\tan ^{-1}\left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log (1+x)+\frac {5}{6} \log \left (1-x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.15 \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=\frac {1}{6} \left (2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )-2 \log (1+x)+\log \left (1-x+x^2\right )+4 \log \left (1+x^3\right )\right ) \]

[In]

Integrate[(x*(1 + 2*x))/(1 + x^3),x]

[Out]

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

Maple [A] (verified)

Time = 1.68 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85

method	result	size
default	\(\frac {\ln \left (1+x \right )}{3}+\frac {5 \ln \left (x^{2}-x +1\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{3}\)	\(35\)
risch	\(\frac {5 \ln \left (4 x^{2}-4 x +4\right )}{6}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{3}+\frac {\ln \left (1+x \right )}{3}\)	\(37\)
meijerg	\(-\frac {x^{2} \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {2 \ln \left (x^{3}+1\right )}{3}\)	\(88\)

[In]

int(x*(1+2*x)/(x^3+1),x,method=_RETURNVERBOSE)

[Out]

1/3*ln(1+x)+5/6*ln(x^2-x+1)+1/3*3^(1/2)*arctan(1/3*(-1+2*x)*3^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {5}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{3} \, \log \left (x + 1\right ) \]

[In]

integrate(x*(1+2*x)/(x^3+1),x, algorithm="fricas")

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 5/6*log(x^2 - x + 1) + 1/3*log(x + 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.02 \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=\frac {\log {\left (x + 1 \right )}}{3} + \frac {5 \log {\left (x^{2} - x + 1 \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

integrate(x*(1+2*x)/(x**3+1),x)

[Out]

log(x + 1)/3 + 5*log(x**2 - x + 1)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt(3)/3)/3

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {5}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{3} \, \log \left (x + 1\right ) \]

[In]

integrate(x*(1+2*x)/(x^3+1),x, algorithm="maxima")

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 5/6*log(x^2 - x + 1) + 1/3*log(x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.85 \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=\frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {5}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {1}{3} \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

integrate(x*(1+2*x)/(x^3+1),x, algorithm="giac")

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 5/6*log(x^2 - x + 1) + 1/3*log(abs(x + 1))

Mupad [B] (verification not implemented)

Time = 9.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.54 \[ \int \frac {x (1+2 x)}{1+x^3} \, dx=\frac {5\,\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{6}+\frac {5\,\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{6}+\frac {\ln \left (x+1\right )}{3}-\frac {\sqrt {3}\,\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{6} \]

[In]

int((x*(2*x + 1))/(x^3 + 1),x)

[Out]

(5*log(x - (3^(1/2)*1i)/2 - 1/2))/6 + (5*log(x + (3^(1/2)*1i)/2 - 1/2))/6 + log(x + 1)/3 - (3^(1/2)*log(x - (3
^(1/2)*1i)/2 - 1/2)*1i)/6 + (3^(1/2)*log(x + (3^(1/2)*1i)/2 - 1/2)*1i)/6