\(\int \frac {1}{1-x+x^2-x^3} \, dx\) [155]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int \frac {1}{1-x+x^2-x^3} \, dx=\frac {\arctan (x)}{2}-\frac {1}{2} \log (1-x)+\frac {1}{4} \log \left (1+x^2\right ) \]

[Out]

1/2*arctan(x)-1/2*ln(1-x)+1/4*ln(x^2+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2083, 649, 209, 266} \[ \int \frac {1}{1-x+x^2-x^3} \, dx=\frac {\arctan (x)}{2}+\frac {1}{4} \log \left (x^2+1\right )-\frac {1}{2} \log (1-x) \]

[In]

Int[(1 - x + x^2 - x^3)^(-1),x]

[Out]

ArcTan[x]/2 - Log[1 - x]/2 + Log[1 + x^2]/4

Rule 209

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

Rule 266

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

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 2083

Int[(P_)^(p_), x_Symbol] :> With[{u = Factor[P]}, Int[ExpandIntegrand[u^p, x], x] /;  !SumQ[NonfreeFactors[u,
x]]] /; PolyQ[P, x] && ILtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {1}{1-x+x^2-x^3} \, dx=\frac {\arctan (x)}{2}-\frac {1}{2} \log (1-x)+\frac {1}{4} \log \left (1+x^2\right ) \]

[In]

Integrate[(1 - x + x^2 - x^3)^(-1),x]

[Out]

ArcTan[x]/2 - Log[1 - x]/2 + Log[1 + x^2]/4

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74

method	result	size
default	\(-\frac {\ln \left (-1+x \right )}{2}+\frac {\ln \left (x^{2}+1\right )}{4}+\frac {\arctan \left (x \right )}{2}\)	\(20\)
risch	\(-\frac {\ln \left (-1+x \right )}{2}+\frac {\ln \left (x^{2}+1\right )}{4}+\frac {\arctan \left (x \right )}{2}\)	\(20\)
parallelrisch	\(-\frac {\ln \left (-1+x \right )}{2}+\frac {\ln \left (x -i\right )}{4}-\frac {i \ln \left (x -i\right )}{4}+\frac {\ln \left (i+x \right )}{4}+\frac {i \ln \left (i+x \right )}{4}\)	\(38\)

[In]

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

[Out]

-1/2*ln(-1+x)+1/4*ln(x^2+1)+1/2*arctan(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {1}{1-x+x^2-x^3} \, dx=\frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

1/2*arctan(x) + 1/4*log(x^2 + 1) - 1/2*log(x - 1)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {1}{1-x+x^2-x^3} \, dx=- \frac {\log {\left (x - 1 \right )}}{2} + \frac {\log {\left (x^{2} + 1 \right )}}{4} + \frac {\operatorname {atan}{\left (x \right )}}{2} \]

[In]

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

[Out]

-log(x - 1)/2 + log(x**2 + 1)/4 + atan(x)/2

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.70 \[ \int \frac {1}{1-x+x^2-x^3} \, dx=\frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

1/2*arctan(x) + 1/4*log(x^2 + 1) - 1/2*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {1}{1-x+x^2-x^3} \, dx=\frac {1}{2} \, \arctan \left (x\right ) + \frac {1}{4} \, \log \left (x^{2} + 1\right ) - \frac {1}{2} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

1/2*arctan(x) + 1/4*log(x^2 + 1) - 1/2*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {1}{1-x+x^2-x^3} \, dx=-\frac {\ln \left (x-1\right )}{2}+\ln \left (x-\mathrm {i}\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right )+\ln \left (x+1{}\mathrm {i}\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right ) \]

[In]

int(-1/(x - x^2 + x^3 - 1),x)

[Out]

log(x - 1i)*(1/4 - 1i/4) - log(x - 1)/2 + log(x + 1i)*(1/4 + 1i/4)