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]

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

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Rubi [A]  time = 0.07, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, 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 31

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

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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.70, size = 106, normalized size = 1.09 \[ \frac {6 \, \sqrt {3} {\left (x^{3} + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt {3} {\left (x^{3} + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 3 \, {\left (x^{3} + 1\right )} \log \left (x^{2} + x + 1\right ) + {\left (x^{3} + 1\right )} \log \left (x^{2} - x + 1\right ) - 2 \, {\left (x^{3} + 1\right )} \log \left (x + 1\right ) - 6 \, {\left (x^{3} + 1\right )} \log \left (x - 1\right ) - 12 \, x}{72 \, {\left (x^{3} + 1\right )}} \]

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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giac [A]  time = 0.31, size = 77, normalized size = 0.79 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{36} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x}{6 \, {\left (x^{3} + 1\right )}} + \frac {1}{24} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{72} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{36} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{12} \, \log \left ({\left | x - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.01, size = 90, normalized size = 0.93 \[ -\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{36}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x -1\right )}{12}-\frac {\ln \left (x +1\right )}{36}+\frac {\ln \left (x^{2}-x +1\right )}{72}+\frac {\ln \left (x^{2}+x +1\right )}{24}+\frac {-2 x -2}{36 x^{2}-36 x +36}+\frac {1}{18 x +18} \]

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

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maxima [A]  time = 1.93, size = 75, normalized size = 0.77 \[ \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - \frac {1}{36} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {x}{6 \, {\left (x^{3} + 1\right )}} + \frac {1}{24} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{72} \, \log \left (x^{2} - x + 1\right ) - \frac {1}{36} \, \log \left (x + 1\right ) - \frac {1}{12} \, \log \left (x - 1\right ) \]

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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mupad [B]  time = 0.18, size = 103, normalized size = 1.06 \[ -\frac {\ln \left (x-1\right )}{12}-\frac {\ln \left (x+1\right )}{36}-\frac {x}{6\,\left (x^3+1\right )}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{24}+\frac {\sqrt {3}\,1{}\mathrm {i}}{24}\right )+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{72}+\frac {\sqrt {3}\,1{}\mathrm {i}}{72}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{72}+\frac {\sqrt {3}\,1{}\mathrm {i}}{72}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^3/((x^3 - 1)*(x^3 + 1)^2),x)

[Out]

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

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sympy [A]  time = 0.36, size = 92, normalized size = 0.95 \[ - \frac {x}{6 x^{3} + 6} - \frac {\log {\left (x - 1 \right )}}{12} - \frac {\log {\left (x + 1 \right )}}{36} + \frac {\log {\left (x^{2} - x + 1 \right )}}{72} + \frac {\log {\left (x^{2} + x + 1 \right )}}{24} - \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{36} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{12} \]

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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