\(\int \frac {1+x+4 x^2}{1-x^3} \, dx\) [52]

Optimal result

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

[Out]

-2*ln(1-x)-ln(x^2+x+1)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1889, 31, 642} \[ \int \frac {1+x+4 x^2}{1-x^3} \, dx=-\log \left (x^2+x+1\right )-2 \log (1-x) \]

[In]

Int[(1 + x + 4*x^2)/(1 - x^3),x]

[Out]

-2*Log[1 - x] - Log[1 + x + x^2]

Rule 31

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

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 1889

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2], q = (-a/b)^(1/3)}, Dist[q*((A + B*q + C*q^2)/(3*a)), Int[1/(q - x), x], x] + Dist[q/(3*a), Int[(q*(2*A -
B*q - C*q^2) + (A + B*q - 2*C*q^2)*x)/(q^2 + q*x + x^2), x], x] /; NeQ[a*B^3 - b*A^3, 0] && NeQ[A + B*q + C*q^
2, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2] && LtQ[a/b, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \int \frac {-3-6 x}{1+x+x^2} \, dx+2 \int \frac {1}{1-x} \, dx \\ & = -2 \log (1-x)-\log \left (1+x+x^2\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1+x+4 x^2}{1-x^3} \, dx=-2 \log (1-x)-\log \left (1+x+x^2\right ) \]

[In]

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

[Out]

-2*Log[1 - x] - Log[1 + x + x^2]

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
default	\(-2 \ln \left (-1+x \right )-\ln \left (x^{2}+x +1\right )\)	\(17\)
norman	\(-2 \ln \left (-1+x \right )-\ln \left (x^{2}+x +1\right )\)	\(17\)
risch	\(-2 \ln \left (-1+x \right )-\ln \left (x^{2}+x +1\right )\)	\(17\)
parallelrisch	\(-2 \ln \left (-1+x \right )-\ln \left (x^{2}+x +1\right )\)	\(17\)
meijerg	\(-\frac {4 \ln \left (-x^{3}+1\right )}{3}-\frac {x^{2} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}-\frac {x \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}\)	\(135\)

[In]

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

[Out]

-2*ln(-1+x)-ln(x^2+x+1)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1+x+4 x^2}{1-x^3} \, dx=-\log \left (x^{2} + x + 1\right ) - 2 \, \log \left (x - 1\right ) \]

[In]

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

[Out]

-log(x^2 + x + 1) - 2*log(x - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1+x+4 x^2}{1-x^3} \, dx=- 2 \log {\left (x - 1 \right )} - \log {\left (x^{2} + x + 1 \right )} \]

[In]

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

[Out]

-2*log(x - 1) - log(x**2 + x + 1)

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1+x+4 x^2}{1-x^3} \, dx=-\log \left (x^{2} + x + 1\right ) - 2 \, \log \left (x - 1\right ) \]

[In]

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

[Out]

-log(x^2 + x + 1) - 2*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94 \[ \int \frac {1+x+4 x^2}{1-x^3} \, dx=-\log \left (x^{2} + x + 1\right ) - 2 \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

-log(x^2 + x + 1) - 2*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 10.06 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1+x+4 x^2}{1-x^3} \, dx=-\ln \left (x^2+x+1\right )-2\,\ln \left (x-1\right ) \]

[In]

int(-(x + 4*x^2 + 1)/(x^3 - 1),x)

[Out]

- log(x + x^2 + 1) - 2*log(x - 1)