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

Optimal. Leaf size=69 \[ \frac{1}{6} \log \left (x^2-x+1\right )-\frac{1}{6} \log \left (x^2+x+1\right )+x+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{2}{3} \tanh ^{-1}(x) \]

[Out]

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

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Rubi [A]  time = 0.123517, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.474, Rules used = {1593, 1584, 388, 210, 634, 618, 204, 628, 206} \[ \frac{1}{6} \log \left (x^2-x+1\right )-\frac{1}{6} \log \left (x^2+x+1\right )+x+\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{\sqrt{3}}-\frac{2}{3} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(x^(-3) + x^3)/(-x^(-3) + x^3),x]

[Out]

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

Rule 1593

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 1584

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

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 210

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt[-(a/b
), n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r +
s*Cos[(2*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; (2*r^2*Int[1/(r^2 - s^2*x^2), x])/(a*n) +
Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[a/b]

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]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.0056237, size = 78, normalized size = 1.13 \[ \frac{1}{6} \left (\log \left (x^2-x+1\right )-\log \left (x^2+x+1\right )+6 x+2 \log (1-x)-2 \log (x+1)-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )-2 \sqrt{3} \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 67, normalized size = 1. \begin{align*} x+{\frac{\ln \left ( x-1 \right ) }{3}}-{\frac{\ln \left ({x}^{2}+x+1 \right ) }{6}}-{\frac{\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }+{\frac{\ln \left ({x}^{2}-x+1 \right ) }{6}}-{\frac{\sqrt{3}}{3}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{\ln \left ( 1+x \right ) }{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.45959, size = 89, normalized size = 1.29 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + x - \frac{1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac{1}{3} \, \log \left (x + 1\right ) + \frac{1}{3} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.985839, size = 232, normalized size = 3.36 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + x - \frac{1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac{1}{3} \, \log \left (x + 1\right ) + \frac{1}{3} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.226974, size = 85, normalized size = 1.23 \begin{align*} x + \frac{\log{\left (x - 1 \right )}}{3} - \frac{\log{\left (x + 1 \right )}}{3} + \frac{\log{\left (x^{2} - x + 1 \right )}}{6} - \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} - \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} + \frac{\sqrt{3}}{3} \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.10676, size = 92, normalized size = 1.33 \begin{align*} -\frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - \frac{1}{3} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + x - \frac{1}{6} \, \log \left (x^{2} + x + 1\right ) + \frac{1}{6} \, \log \left (x^{2} - x + 1\right ) - \frac{1}{3} \, \log \left ({\left | x + 1 \right |}\right ) + \frac{1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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