3.345 \(\int \frac{x^3}{(1-x^3) (1+x^3)^2} \, dx\)

Optimal. Leaf size=97 \[ -\frac{x}{6 \left (x^3+1\right )}+\frac{1}{72} \log \left (x^2-x+1\right )+\frac{1}{24} \log \left (x^2+x+1\right )-\frac{1}{12} \log (1-x)-\frac{1}{36} \log (x+1)+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

[Out]

-x/(6*(1 + x^3)) + ArcTan[(1 - 2*x)/Sqrt[3]]/(12*Sqrt[3]) + ArcTan[(1 + 2*x)/Sqrt[3]]/(4*Sqrt[3]) - Log[1 - x]
/12 - Log[1 + x]/36 + Log[1 - x + x^2]/72 + Log[1 + x + x^2]/24

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Rubi [A]  time = 0.0707737, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {471, 522, 200, 31, 634, 618, 204, 628} \[ -\frac{x}{6 \left (x^3+1\right )}+\frac{1}{72} \log \left (x^2-x+1\right )+\frac{1}{24} \log \left (x^2+x+1\right )-\frac{1}{12} \log (1-x)-\frac{1}{36} \log (x+1)+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{4 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((1 - x^3)*(1 + x^3)^2),x]

[Out]

-x/(6*(1 + x^3)) + ArcTan[(1 - 2*x)/Sqrt[3]]/(12*Sqrt[3]) + ArcTan[(1 + 2*x)/Sqrt[3]]/(4*Sqrt[3]) - Log[1 - x]
/12 - Log[1 + x]/36 + Log[1 - x + x^2]/72 + Log[1 + x + x^2]/24

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[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]

Rule 31

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

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{x^3}{\left (1-x^3\right ) \left (1+x^3\right )^2} \, dx &=-\frac{x}{6 \left (1+x^3\right )}+\frac{1}{6} \int \frac{1+2 x^3}{\left (1-x^3\right ) \left (1+x^3\right )} \, dx\\ &=-\frac{x}{6 \left (1+x^3\right )}-\frac{1}{12} \int \frac{1}{1+x^3} \, dx+\frac{1}{4} \int \frac{1}{1-x^3} \, dx\\ &=-\frac{x}{6 \left (1+x^3\right )}-\frac{1}{36} \int \frac{1}{1+x} \, dx-\frac{1}{36} \int \frac{2-x}{1-x+x^2} \, dx+\frac{1}{12} \int \frac{1}{1-x} \, dx+\frac{1}{12} \int \frac{2+x}{1+x+x^2} \, dx\\ &=-\frac{x}{6 \left (1+x^3\right )}-\frac{1}{12} \log (1-x)-\frac{1}{36} \log (1+x)+\frac{1}{72} \int \frac{-1+2 x}{1-x+x^2} \, dx-\frac{1}{24} \int \frac{1}{1-x+x^2} \, dx+\frac{1}{24} \int \frac{1+2 x}{1+x+x^2} \, dx+\frac{1}{8} \int \frac{1}{1+x+x^2} \, dx\\ &=-\frac{x}{6 \left (1+x^3\right )}-\frac{1}{12} \log (1-x)-\frac{1}{36} \log (1+x)+\frac{1}{72} \log \left (1-x+x^2\right )+\frac{1}{24} \log \left (1+x+x^2\right )+\frac{1}{12} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )-\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=-\frac{x}{6 \left (1+x^3\right )}+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{\tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{4 \sqrt{3}}-\frac{1}{12} \log (1-x)-\frac{1}{36} \log (1+x)+\frac{1}{72} \log \left (1-x+x^2\right )+\frac{1}{24} \log \left (1+x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0465136, size = 85, normalized size = 0.88 \[ \frac{1}{72} \left (-\frac{12 x}{x^3+1}+\log \left (x^2-x+1\right )+3 \log \left (x^2+x+1\right )-6 \log (1-x)-2 \log (x+1)-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )+6 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((1 - x^3)*(1 + x^3)^2),x]

[Out]

((-12*x)/(1 + x^3) - 2*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + 6*Sqrt[3]*ArcTan[(1 + 2*x)/Sqrt[3]] - 6*Log[1 - x]
- 2*Log[1 + x] + Log[1 - x + x^2] + 3*Log[1 + x + x^2])/72

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Maple [A]  time = 0.012, size = 90, normalized size = 0.9 \begin{align*} -{\frac{\ln \left ( x-1 \right ) }{12}}+{\frac{-2\,x-2}{36\,{x}^{2}-36\,x+36}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{72}}-{\frac{\sqrt{3}}{36}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{1}{18+18\,x}}-{\frac{\ln \left ( 1+x \right ) }{36}}+{\frac{\ln \left ({x}^{2}+x+1 \right ) }{24}}+{\frac{\sqrt{3}}{12}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(-x^3+1)/(x^3+1)^2,x)

[Out]

-1/12*ln(x-1)+1/36*(-2*x-2)/(x^2-x+1)+1/72*ln(x^2-x+1)-1/36*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))+1/18/(1+x)-1/3
6*ln(1+x)+1/24*ln(x^2+x+1)+1/12*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)

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Maxima [A]  time = 1.87009, size = 101, normalized size = 1.04 \begin{align*} \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{36} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{x}{6 \,{\left (x^{3} + 1\right )}} + \frac{1}{24} \, \log \left (x^{2} + x + 1\right ) + \frac{1}{72} \, \log \left (x^{2} - x + 1\right ) - \frac{1}{36} \, \log \left (x + 1\right ) - \frac{1}{12} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/36*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*x/(x^3 + 1) + 1/
24*log(x^2 + x + 1) + 1/72*log(x^2 - x + 1) - 1/36*log(x + 1) - 1/12*log(x - 1)

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Fricas [A]  time = 1.51337, size = 320, normalized size = 3.3 \begin{align*} \frac{6 \, \sqrt{3}{\left (x^{3} + 1\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt{3}{\left (x^{3} + 1\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 3 \,{\left (x^{3} + 1\right )} \log \left (x^{2} + x + 1\right ) +{\left (x^{3} + 1\right )} \log \left (x^{2} - x + 1\right ) - 2 \,{\left (x^{3} + 1\right )} \log \left (x + 1\right ) - 6 \,{\left (x^{3} + 1\right )} \log \left (x - 1\right ) - 12 \, x}{72 \,{\left (x^{3} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/72*(6*sqrt(3)*(x^3 + 1)*arctan(1/3*sqrt(3)*(2*x + 1)) - 2*sqrt(3)*(x^3 + 1)*arctan(1/3*sqrt(3)*(2*x - 1)) +
3*(x^3 + 1)*log(x^2 + x + 1) + (x^3 + 1)*log(x^2 - x + 1) - 2*(x^3 + 1)*log(x + 1) - 6*(x^3 + 1)*log(x - 1) -
12*x)/(x^3 + 1)

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Sympy [A]  time = 0.329192, size = 92, normalized size = 0.95 \begin{align*} - \frac{x}{6 x^{3} + 6} - \frac{\log{\left (x - 1 \right )}}{12} - \frac{\log{\left (x + 1 \right )}}{36} + \frac{\log{\left (x^{2} - x + 1 \right )}}{72} + \frac{\log{\left (x^{2} + x + 1 \right )}}{24} - \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{36} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} + \frac{\sqrt{3}}{3} \right )}}{12} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(6*x**3 + 6) - log(x - 1)/12 - log(x + 1)/36 + log(x**2 - x + 1)/72 + log(x**2 + x + 1)/24 - sqrt(3)*atan(2
*sqrt(3)*x/3 - sqrt(3)/3)/36 + sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/12

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Giac [A]  time = 1.1593, size = 104, normalized size = 1.07 \begin{align*} \frac{1}{12} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{36} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - \frac{x}{6 \,{\left (x^{3} + 1\right )}} + \frac{1}{24} \, \log \left (x^{2} + x + 1\right ) + \frac{1}{72} \, \log \left (x^{2} - x + 1\right ) - \frac{1}{36} \, \log \left ({\left | x + 1 \right |}\right ) - \frac{1}{12} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) - 1/36*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) - 1/6*x/(x^3 + 1) + 1/
24*log(x^2 + x + 1) + 1/72*log(x^2 - x + 1) - 1/36*log(abs(x + 1)) - 1/12*log(abs(x - 1))