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

Optimal. Leaf size=64 \[ \frac{x \left (x-x^2\right )}{3 \left (x^3+1\right )}-\frac{5}{18} \log \left (x^2-x+1\right )+\log (x)-\frac{4}{9} \log (x+1)-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

(x*(x - x^2))/(3*(1 + x^3)) - ArcTan[(1 - 2*x)/Sqrt[3]]/(3*Sqrt[3]) + Log[x] - (4*Log[1 + x])/9 - (5*Log[1 - x
+ x^2])/18

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Rubi [A]  time = 0.0730219, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {1829, 1834, 634, 618, 204, 628} \[ \frac{x \left (x-x^2\right )}{3 \left (x^3+1\right )}-\frac{5}{18} \log \left (x^2-x+1\right )+\log (x)-\frac{4}{9} \log (x+1)-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^2)/(x*(1 + x^3)^2),x]

[Out]

(x*(x - x^2))/(3*(1 + x^3)) - ArcTan[(1 - 2*x)/Sqrt[3]]/(3*Sqrt[3]) + Log[x] - (4*Log[1 + x])/9 - (5*Log[1 - x
+ x^2])/18

Rule 1829

Int[(Pq_)*(x_)^(m_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, Module[{Q = Polynomi
alQuotient[a*b^(Floor[(q - 1)/n] + 1)*x^m*Pq, a + b*x^n, x], R = PolynomialRemainder[a*b^(Floor[(q - 1)/n] + 1
)*x^m*Pq, a + b*x^n, x], i}, Dist[1/(a*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), Int[x^m*(a + b*x^n)^(p + 1)*Expand
ToSum[(n*(p + 1)*Q)/x^m + Sum[((n*(p + 1) + i + 1)*Coeff[R, x, i]*x^(i - m))/a, {i, 0, n - 1}], x], x], x] - S
imp[(x*R*(a + b*x^n)^(p + 1))/(a^2*n*(p + 1)*b^(Floor[(q - 1)/n] + 1)), x]]] /; FreeQ[{a, b}, x] && PolyQ[Pq,
x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1834

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[((c*x)^m*Pq)/(a + b*
x^n), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 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 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{1+x^2}{x \left (1+x^3\right )^2} \, dx &=\frac{x \left (x-x^2\right )}{3 \left (1+x^3\right )}-\frac{1}{3} \int \frac{-3-x^2}{x \left (1+x^3\right )} \, dx\\ &=\frac{x \left (x-x^2\right )}{3 \left (1+x^3\right )}-\frac{1}{3} \int \left (-\frac{3}{x}+\frac{4}{3 (1+x)}+\frac{-4+5 x}{3 \left (1-x+x^2\right )}\right ) \, dx\\ &=\frac{x \left (x-x^2\right )}{3 \left (1+x^3\right )}+\log (x)-\frac{4}{9} \log (1+x)-\frac{1}{9} \int \frac{-4+5 x}{1-x+x^2} \, dx\\ &=\frac{x \left (x-x^2\right )}{3 \left (1+x^3\right )}+\log (x)-\frac{4}{9} \log (1+x)+\frac{1}{6} \int \frac{1}{1-x+x^2} \, dx-\frac{5}{18} \int \frac{-1+2 x}{1-x+x^2} \, dx\\ &=\frac{x \left (x-x^2\right )}{3 \left (1+x^3\right )}+\log (x)-\frac{4}{9} \log (1+x)-\frac{5}{18} \log \left (1-x+x^2\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )\\ &=\frac{x \left (x-x^2\right )}{3 \left (1+x^3\right )}-\frac{\tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{3 \sqrt{3}}+\log (x)-\frac{4}{9} \log (1+x)-\frac{5}{18} \log \left (1-x+x^2\right )\\ \end{align*}

Mathematica [A]  time = 0.0330718, size = 65, normalized size = 1.02 \[ \frac{1}{18} \left (\frac{6 \left (x^2+1\right )}{x^3+1}+\log \left (x^2-x+1\right )-6 \log \left (x^3+1\right )+18 \log (x)-2 \log (x+1)+2 \sqrt{3} \tan ^{-1}\left (\frac{2 x-1}{\sqrt{3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*(1 + x^2))/(1 + x^3) + 2*Sqrt[3]*ArcTan[(-1 + 2*x)/Sqrt[3]] + 18*Log[x] - 2*Log[1 + x] + Log[1 - x + x^2]
- 6*Log[1 + x^3])/18

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Maple [A]  time = 0.011, size = 61, normalized size = 1. \begin{align*} \ln \left ( x \right ) +{\frac{2}{9+9\,x}}-{\frac{4\,\ln \left ( 1+x \right ) }{9}}-{\frac{-1-x}{9\,{x}^{2}-9\,x+9}}-{\frac{5\,\ln \left ({x}^{2}-x+1 \right ) }{18}}+{\frac{\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

ln(x)+2/9/(1+x)-4/9*ln(1+x)-1/9*(-1-x)/(x^2-x+1)-5/18*ln(x^2-x+1)+1/9*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))

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Maxima [A]  time = 1.41201, size = 68, normalized size = 1.06 \begin{align*} \frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{x^{2} + 1}{3 \,{\left (x^{3} + 1\right )}} - \frac{5}{18} \, \log \left (x^{2} - x + 1\right ) - \frac{4}{9} \, \log \left (x + 1\right ) + \log \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/3*(x^2 + 1)/(x^3 + 1) - 5/18*log(x^2 - x + 1) - 4/9*log(x + 1) +
log(x)

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Fricas [A]  time = 2.02832, size = 213, normalized size = 3.33 \begin{align*} \frac{2 \, \sqrt{3}{\left (x^{3} + 1\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + 6 \, x^{2} - 5 \,{\left (x^{3} + 1\right )} \log \left (x^{2} - x + 1\right ) - 8 \,{\left (x^{3} + 1\right )} \log \left (x + 1\right ) + 18 \,{\left (x^{3} + 1\right )} \log \left (x\right ) + 6}{18 \,{\left (x^{3} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(2*sqrt(3)*(x^3 + 1)*arctan(1/3*sqrt(3)*(2*x - 1)) + 6*x^2 - 5*(x^3 + 1)*log(x^2 - x + 1) - 8*(x^3 + 1)*l
og(x + 1) + 18*(x^3 + 1)*log(x) + 6)/(x^3 + 1)

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Sympy [A]  time = 0.197632, size = 60, normalized size = 0.94 \begin{align*} \frac{x^{2} + 1}{3 x^{3} + 3} + \log{\left (x \right )} - \frac{4 \log{\left (x + 1 \right )}}{9} - \frac{5 \log{\left (x^{2} - x + 1 \right )}}{18} + \frac{\sqrt{3} \operatorname{atan}{\left (\frac{2 \sqrt{3} x}{3} - \frac{\sqrt{3}}{3} \right )}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x**2 + 1)/(3*x**3 + 3) + log(x) - 4*log(x + 1)/9 - 5*log(x**2 - x + 1)/18 + sqrt(3)*atan(2*sqrt(3)*x/3 - sqrt
(3)/3)/9

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Giac [A]  time = 1.05661, size = 81, normalized size = 1.27 \begin{align*} \frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{x^{2} + 1}{3 \,{\left (x^{2} - x + 1\right )}{\left (x + 1\right )}} - \frac{5}{18} \, \log \left (x^{2} - x + 1\right ) - \frac{4}{9} \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*x - 1)) + 1/3*(x^2 + 1)/((x^2 - x + 1)*(x + 1)) - 5/18*log(x^2 - x + 1) - 4/
9*log(abs(x + 1)) + log(abs(x))