\(\int \frac {1+2 x-x^2+8 x^3+x^4}{(x+x^2) (1+x^3)} \, dx\) [275]

Optimal result

Integrand size = 33, antiderivative size = 44 \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=-\frac {3}{1+x}-\frac {2 \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-2 \log (1+x)+\log \left (1-x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1607, 6857, 648, 632, 210, 642} \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=-\frac {2 \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log \left (x^2-x+1\right )-\frac {3}{x+1}+\log (x)-2 \log (x+1) \]

[In]

Int[(1 + 2*x - x^2 + 8*x^3 + x^4)/((x + x^2)*(1 + x^3)),x]

[Out]

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

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 1607

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 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1+2 x-x^2+8 x^3+x^4}{x (1+x) \left (1+x^3\right )} \, dx \\ & = \int \left (\frac {1}{x}+\frac {3}{(1+x)^2}-\frac {2}{1+x}+\frac {2 x}{1-x+x^2}\right ) \, dx \\ & = -\frac {3}{1+x}+\log (x)-2 \log (1+x)+2 \int \frac {x}{1-x+x^2} \, dx \\ & = -\frac {3}{1+x}+\log (x)-2 \log (1+x)+\int \frac {1}{1-x+x^2} \, dx+\int \frac {-1+2 x}{1-x+x^2} \, dx \\ & = -\frac {3}{1+x}+\log (x)-2 \log (1+x)+\log \left (1-x+x^2\right )-2 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x\right ) \\ & = -\frac {3}{1+x}-\frac {2 \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-2 \log (1+x)+\log \left (1-x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=-\frac {3}{1+x}+\frac {2 \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\log (x)-2 \log (1+x)+\log \left (1-x+x^2\right ) \]

[In]

Integrate[(1 + 2*x - x^2 + 8*x^3 + x^4)/((x + x^2)*(1 + x^3)),x]

[Out]

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

Maple [A] (verified)

Time = 0.80 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.95

method	result	size
default	\(\ln \left (x \right )-\frac {3}{x +1}-2 \ln \left (x +1\right )+\ln \left (x^{2}-x +1\right )+\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\)	\(42\)
risch	\(-\frac {3}{x +1}+\ln \left (4 x^{2}-4 x +4\right )+\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}-2 \ln \left (x +1\right )+\ln \left (x \right )\)	\(44\)

[In]

int((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.32 \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=\frac {2 \, \sqrt {3} {\left (x + 1\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 3 \, {\left (x + 1\right )} \log \left (x^{2} - x + 1\right ) - 6 \, {\left (x + 1\right )} \log \left (x + 1\right ) + 3 \, {\left (x + 1\right )} \log \left (x\right ) - 9}{3 \, {\left (x + 1\right )}} \]

[In]

integrate((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x, algorithm="fricas")

[Out]

1/3*(2*sqrt(3)*(x + 1)*arctan(1/3*sqrt(3)*(2*x - 1)) + 3*(x + 1)*log(x^2 - x + 1) - 6*(x + 1)*log(x + 1) + 3*(
x + 1)*log(x) - 9)/(x + 1)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=\log {\left (x \right )} - 2 \log {\left (x + 1 \right )} + \log {\left (x^{2} - x + 1 \right )} + \frac {2 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} - \frac {3}{x + 1} \]

[In]

integrate((x**4+8*x**3-x**2+2*x+1)/(x**2+x)/(x**3+1),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {3}{x + 1} + \log \left (x^{2} - x + 1\right ) - 2 \, \log \left (x + 1\right ) + \log \left (x\right ) \]

[In]

integrate((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=\frac {2}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {3}{x + 1} + \log \left (x^{2} - x + 1\right ) - 2 \, \log \left ({\left | x + 1 \right |}\right ) + \log \left ({\left | x \right |}\right ) \]

[In]

integrate((x^4+8*x^3-x^2+2*x+1)/(x^2+x)/(x^3+1),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.25 \[ \int \frac {1+2 x-x^2+8 x^3+x^4}{\left (x+x^2\right ) \left (1+x^3\right )} \, dx=\ln \left (x\right )-2\,\ln \left (x+1\right )-\frac {3}{x+1}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-1+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (1+\frac {\sqrt {3}\,1{}\mathrm {i}}{3}\right ) \]

[In]

int((2*x - x^2 + 8*x^3 + x^4 + 1)/((x^3 + 1)*(x + x^2)),x)

[Out]

log(x) - 2*log(x + 1) - 3/(x + 1) - log(x - (3^(1/2)*1i)/2 - 1/2)*((3^(1/2)*1i)/3 - 1) + log(x + (3^(1/2)*1i)/
2 - 1/2)*((3^(1/2)*1i)/3 + 1)