3.304 \(\int \frac {(1-x) x^4}{1+x^3} \, dx\)

Optimal. Leaf size=54 \[ -\frac {x^3}{3}+\frac {x^2}{2}+\frac {1}{6} \log \left (x^2-x+1\right )+\frac {2}{3} \log (x+1)+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1887, 1874, 31, 634, 618, 204, 628} \[ -\frac {x^3}{3}+\frac {x^2}{2}+\frac {1}{6} \log \left (x^2-x+1\right )+\frac {2}{3} \log (x+1)+\frac {\tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}} \]

Antiderivative was successfully verified.

[In]

Int[((1 - x)*x^4)/(1 + x^3),x]

[Out]

x^2/2 - x^3/3 + ArcTan[(1 - 2*x)/Sqrt[3]]/Sqrt[3] + (2*Log[1 + x])/3 + 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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

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 628

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 634

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 1874

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]

Rule 1887

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*} \int \frac {(1-x) x^4}{1+x^3} \, dx &=\int \left (x-x^2+\frac {(-1+x) x}{1+x^3}\right ) \, dx\\ &=\frac {x^2}{2}-\frac {x^3}{3}+\int \frac {(-1+x) x}{1+x^3} \, dx\\ &=\frac {x^2}{2}-\frac {x^3}{3}+\frac {1}{3} \int \frac {-2+x}{1-x+x^2} \, dx+\frac {2}{3} \int \frac {1}{1+x} \, dx\\ &=\frac {x^2}{2}-\frac {x^3}{3}+\frac {2}{3} \log (1+x)+\frac {1}{6} \int \frac {-1+2 x}{1-x+x^2} \, dx-\frac {1}{2} \int \frac {1}{1-x+x^2} \, dx\\ &=\frac {x^2}{2}-\frac {x^3}{3}+\frac {2}{3} \log (1+x)+\frac {1}{6} \log \left (1-x+x^2\right )+\operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=\frac {x^2}{2}-\frac {x^3}{3}-\frac {\tan ^{-1}\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 )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 59, normalized size = 1.09 \[ \frac {1}{6} \left (-2 x^3+2 \log \left (x^3+1\right )+3 x^2-\log \left (x^2-x+1\right )+2 \log (x+1)-2 \sqrt {3} \tan ^{-1}\left (\frac {2 x-1}{\sqrt {3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((1 - x)*x^4)/(1 + x^3),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.73, size = 44, normalized size = 0.81 \[ -\frac {1}{3} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {2}{3} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 45, normalized size = 0.83 \[ -\frac {1}{3} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {2}{3} \, \log \left ({\left | x + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^3 + 1/2*x^2 - 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))

________________________________________________________________________________________

maple [A]  time = 0.05, size = 45, normalized size = 0.83 \[ -\frac {x^{3}}{3}+\frac {x^{2}}{2}-\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\frac {2 \ln \left (x +1\right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1-x)*x^4/(x^3+1),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.90, size = 44, normalized size = 0.81 \[ -\frac {1}{3} \, x^{3} + \frac {1}{2} \, x^{2} - \frac {1}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{6} \, \log \left (x^{2} - x + 1\right ) + \frac {2}{3} \, \log \left (x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.10, size = 56, normalized size = 1.04 \[ \frac {2\,\ln \left (x+1\right )}{3}+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{6}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\frac {x^2}{2}-\frac {x^3}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4*(x - 1))/(x^3 + 1),x)

[Out]

(2*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) + x^2/2 - x^3/3

________________________________________________________________________________________

sympy [A]  time = 0.18, size = 53, normalized size = 0.98 \[ - \frac {x^{3}}{3} + \frac {x^{2}}{2} + \frac {2 \log {\left (x + 1 \right )}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{6} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________