\(\int \frac {1-x+4 x^3}{1+x^3} \, dx\) [78]

Optimal result

Integrand size = 18, antiderivative size = 44 \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=4 x+\frac {4 \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log (1+x)+\frac {1}{3} \log \left (1-x+x^2\right ) \]

[Out]

4*x-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 = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {1901, 1874, 31, 648, 632, 210, 642} \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=\frac {4 \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{3} \log \left (x^2-x+1\right )+4 x-\frac {2}{3} \log (x+1) \]

[In]

Int[(1 - x + 4*x^3)/(1 + x^3),x]

[Out]

4*x + (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 1874

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[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] && PosQ[
a/b]

Rule 1901

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \int \left (4-\frac {3+x}{1+x^3}\right ) \, dx \\ & = 4 x-\int \frac {3+x}{1+x^3} \, dx \\ & = 4 x-\frac {1}{3} \int \frac {7-2 x}{1-x+x^2} \, dx-\frac {2}{3} \int \frac {1}{1+x} \, dx \\ & = 4 x-\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 \\ & = 4 x-\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 ) \\ & = 4 x+\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 = 44, normalized size of antiderivative = 1.00 \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=4 x-\frac {4 \arctan \left (\frac {-1+2 x}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {2}{3} \log (1+x)+\frac {1}{3} \log \left (1-x+x^2\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.45 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86

method	result	size
default	\(4 x -\frac {2 \ln \left (1+x \right )}{3}+\frac {\ln \left (x^{2}-x +1\right )}{3}-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (-1+2 x \right ) \sqrt {3}}{3}\right )}{3}\)	\(38\)
risch	\(4 x -\frac {2 \ln \left (1+x \right )}{3}+\frac {\ln \left (16 x^{2}-16 x +16\right )}{3}-\frac {4 \sqrt {3}\, \arctan \left (\frac {\left (-2+4 x \right ) \sqrt {3}}{6}\right )}{3}\)	\(40\)
meijerg	\(4 x -\frac {4 x \left (\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}-\frac {\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3}+\frac {x^{2} \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {x^{2} \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {x^{2} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}+\frac {x \ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}-\frac {x \ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{6 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2-\left (x^{3}\right )^{\frac {1}{3}}}\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\)	\(226\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=-\frac {4}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 4 \, x + \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.09 \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=4 x - \frac {2 \log {\left (x + 1 \right )}}{3} + \frac {\log {\left (x^{2} - x + 1 \right )}}{3} - \frac {4 \sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{3} \]

[In]

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

[Out]

4*x - 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 = 0.30 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.84 \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=-\frac {4}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 4 \, x + \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left (x + 1\right ) \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.86 \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=-\frac {4}{3} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + 4 \, x + \frac {1}{3} \, \log \left (x^{2} - x + 1\right ) - \frac {2}{3} \, \log \left ({\left | x + 1 \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.38 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.11 \[ \int \frac {1-x+4 x^3}{1+x^3} \, dx=4\,x-\frac {2\,\ln \left (x+1\right )}{3}+\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{3}+\frac {\sqrt {3}\,2{}\mathrm {i}}{3}\right )-\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{3}+\frac {\sqrt {3}\,2{}\mathrm {i}}{3}\right ) \]

[In]

int((4*x^3 - x + 1)/(x^3 + 1),x)

[Out]

4*x - (2*log(x + 1))/3 + 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)