\(\int \frac {1+x^2}{x (1+x^3)^2} \, dx\) [181]

Optimal result

Integrand size = 16, antiderivative size = 64 \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=\frac {x \left (x-x^2\right )}{3 \left (1+x^3\right )}-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}+\log (x)-\frac {4}{9} \log (1+x)-\frac {5}{18} \log \left (1-x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1843, 1848, 648, 632, 210, 642} \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {5}{18} \log \left (x^2-x+1\right )+\frac {x \left (x-x^2\right )}{3 \left (x^3+1\right )}+\log (x)-\frac {4}{9} \log (x+1) \]

[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 210

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

Rule 632

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 642

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 648

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 1843

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)/a)*Coeff[R, x, i]*x^(i - m), {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[P
q, x] && IGtQ[n, 0] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1848

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]

Rubi steps \begin{align*} \text {integral}& = \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} \text {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 {\arctan \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] (verified)

Time = 0.03 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=\frac {1}{18} \left (\frac {6 \left (1+x^2\right )}{1+x^3}+2 \sqrt {3} \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )+18 \log (x)-2 \log (1+x)+\log \left (1-x+x^2\right )-6 \log \left (1+x^3\right )\right ) \]

[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

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {\frac {x^{2}}{3}+\frac {1}{3}}{x^{3}+1}+\ln \left (x \right )-\frac {4 \ln \left (1+x \right )}{9}-\frac {5 \ln \left (x^{2}-x +1\right )}{18}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x -\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{9}\)	\(50\)
default	\(-\frac {-1-x}{9 \left (x^{2}-x +1\right )}-\frac {5 \ln \left (x^{2}-x +1\right )}{18}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{9}+\ln \left (x \right )+\frac {2}{9 \left (1+x \right )}-\frac {4 \ln \left (1+x \right )}{9}\)	\(61\)
meijerg	\(\frac {x^{2}}{3 x^{3}+3}-\frac {x^{2} \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{9 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{18 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{9 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {1}{3}+\ln \left (x \right )-\frac {2 x^{3}}{3 \left (2 x^{3}+2\right )}-\frac {\ln \left (x^{3}+1\right )}{3}\)	\(118\)

[In]

int((x^2+1)/x/(x^3+1)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.14 \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=\frac {2 \, \sqrt {3} {\left (x^{3} + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 6 \, x^{2} - 5 \, {\left (x^{3} + 1\right )} \log \left (x^{2} - x + 1\right ) - 8 \, {\left (x^{3} + 1\right )} \log \left (x + 1\right ) + 18 \, {\left (x^{3} + 1\right )} \log \left (x\right ) + 6}{18 \, {\left (x^{3} + 1\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=\frac {x^{2} + 1}{3 x^{3} + 3} + \log {\left (x \right )} - \frac {4 \log {\left (x + 1 \right )}}{9} - \frac {5 \log {\left (x^{2} - x + 1 \right )}}{18} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{9} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.78 \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {x^{2} + 1}{3 \, {\left (x^{3} + 1\right )}} - \frac {5}{18} \, \log \left (x^{2} - x + 1\right ) - \frac {4}{9} \, \log \left (x + 1\right ) + \log \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.94 \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=\frac {1}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {x^{2} + 1}{3 \, {\left (x^{2} - x + 1\right )} {\left (x + 1\right )}} - \frac {5}{18} \, \log \left (x^{2} - x + 1\right ) - \frac {4}{9} \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.98 \[ \int \frac {1+x^2}{x \left (1+x^3\right )^2} \, dx=\ln \left (x\right )-\frac {4\,\ln \left (x+1\right )}{9}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {5}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {5}{18}+\frac {\sqrt {3}\,1{}\mathrm {i}}{18}\right )+\frac {\frac {x^2}{3}+\frac {1}{3}}{x^3+1} \]

[In]

int((x^2 + 1)/(x*(x^3 + 1)^2),x)

[Out]

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