\(\int \frac {2 x^2+x^4}{1-x^3} \, dx\) [77]

Optimal result

Integrand size = 19, antiderivative size = 46 \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=-\frac {x^2}{2}-\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\log (1-x)-\frac {1}{2} \log \left (1+x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {1607, 1901, 1889, 31, 648, 632, 210, 642} \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=-\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {x^2}{2}-\frac {1}{2} \log \left (x^2+x+1\right )-\log (1-x) \]

[In]

Int[(2*x^2 + x^4)/(1 - x^3),x]

[Out]

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

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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1889

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] && LtQ[a/b, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.17 \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=\frac {1}{6} \left (-3 x^2-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \log (1-x)+\log \left (1+x+x^2\right )-4 \log \left (1-x^3\right )\right ) \]

[In]

Integrate[(2*x^2 + x^4)/(1 - x^3),x]

[Out]

(-3*x^2 - 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 = 1.49 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {x^{2}}{2}-\ln \left (-1+x \right )-\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x +\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{2}\)	\(36\)
default	\(-\frac {x^{2}}{2}-\ln \left (-1+x \right )-\frac {\ln \left (x^{2}+x +1\right )}{2}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(38\)
meijerg	\(\frac {\left (-1\right )^{\frac {1}{3}} \left (\frac {3 x^{2} \left (-1\right )^{\frac {2}{3}}}{2}+\frac {x^{2} \left (-1\right )^{\frac {2}{3}} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{\left (x^{3}\right )^{\frac {2}{3}}}\right )}{3}-\frac {2 \ln \left (-x^{3}+1\right )}{3}\)	\(90\)

[In]

int((x^4+2*x^2)/(-x^3+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.59 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=-\frac {1}{2} \, x^{2} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \]

[In]

integrate((x^4+2*x^2)/(-x^3+1),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=- \frac {x^{2}}{2} - \log {\left (x - 1 \right )} - \frac {\log {\left (x^{2} + x + 1 \right )}}{2} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

integrate((x**4+2*x**2)/(-x**3+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80 \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=-\frac {1}{2} \, x^{2} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left (x - 1\right ) \]

[In]

integrate((x^4+2*x^2)/(-x^3+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=-\frac {1}{2} \, x^{2} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{2} \, \log \left (x^{2} + x + 1\right ) - \log \left ({\left | x - 1 \right |}\right ) \]

[In]

integrate((x^4+2*x^2)/(-x^3+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.11 \[ \int \frac {2 x^2+x^4}{1-x^3} \, dx=-\ln \left (x-1\right )+\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\frac {x^2}{2} \]

[In]

int(-(2*x^2 + x^4)/(x^3 - 1),x)

[Out]

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