\(\int \frac {3-x}{1-x^3} \, dx\) [17]

Optimal result

Integrand size = 15, antiderivative size = 41 \[ \int \frac {3-x}{1-x^3} \, dx=\frac {4 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log (1-x)+\frac {1}{3} \log \left (1+x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1875, 31, 648, 632, 210, 642} \[ \int \frac {3-x}{1-x^3} \, dx=\frac {4 \arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (x^2+x+1\right )-\frac {2}{3} \log (1-x) \]

[In]

Int[(3 - x)/(1 - x^3),x]

[Out]

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

Rule 31

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

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 1875

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 3]], s = Denominator[
Rt[-a/b, 3]]}, Dist[r*((B*r + A*s)/(3*a*s)), Int[1/(r - s*x), x], x] - Dist[r/(3*a*s), Int[(r*(B*r - 2*A*s) -
s*(B*r + A*s)*x)/(r^2 + r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && NegQ[a
/b]

Rubi steps \begin{align*} \text {integral}& = -\left (\frac {1}{3} \int \frac {-7-2 x}{1+x+x^2} \, dx\right )+\frac {2}{3} \int \frac {1}{1-x} \, dx \\ & = -\frac {2}{3} \log (1-x)+\frac {1}{3} \int \frac {1+2 x}{1+x+x^2} \, dx+2 \int \frac {1}{1+x+x^2} \, dx \\ & = -\frac {2}{3} \log (1-x)+\frac {1}{3} \log \left (1+x+x^2\right )-4 \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x\right ) \\ & = \frac {4 \tan ^{-1}\left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log (1-x)+\frac {1}{3} \log \left (1+x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {3-x}{1-x^3} \, dx=\frac {4 \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log (1-x)+\frac {1}{3} \log \left (1+x+x^2\right ) \]

[In]

Integrate[(3 - x)/(1 - x^3),x]

[Out]

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

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {2 \ln \left (-1+x \right )}{3}+\frac {\ln \left (x^{2}+x +1\right )}{3}+\frac {4 \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}\)	\(33\)
risch	\(-\frac {2 \ln \left (-1+x \right )}{3}+\frac {\ln \left (16 x^{2}+16 x +16\right )}{3}+\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (2+4 x \right ) \sqrt {3}}{6}\right )}{3}\)	\(37\)
meijerg	\(-\frac {x \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{\left (x^{3}\right )^{\frac {1}{3}}}+\frac {x^{2} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}\)	\(125\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {3-x}{1-x^3} \, dx=\frac {4}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{3} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.07 \[ \int \frac {3-x}{1-x^3} \, dx=- \frac {2 \log {\left (x - 1 \right )}}{3} + \frac {\log {\left (x^{2} + x + 1 \right )}}{3} + \frac {4 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.78 \[ \int \frac {3-x}{1-x^3} \, dx=\frac {4}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{3} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80 \[ \int \frac {3-x}{1-x^3} \, dx=\frac {4}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{3} \, \log \left (x^{2} + x + 1\right ) - \frac {2}{3} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {3-x}{1-x^3} \, dx=-\frac {2\,\ln \left (x-1\right )}{3}-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,2{}\mathrm {i}}{3}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,2{}\mathrm {i}}{3}\right ) \]

[In]

int((x - 3)/(x^3 - 1),x)

[Out]

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