3.2.32 \(\int \frac {1+x}{-1+x^3} \, dx\) [132]

Optimal. Leaf size=22 \[ \frac {2}{3} \log (1-x)-\frac {1}{3} \log \left (1+x+x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1875, 31, 642} \begin {gather*} \frac {2}{3} \log (1-x)-\frac {1}{3} \log \left (x^2+x+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x)/(-1 + x^3),x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 22, normalized size = 1.00 \begin {gather*} \frac {2}{3} \log (1-x)-\frac {1}{3} \log \left (1+x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)/(-1 + x^3),x]

[Out]

(2*Log[1 - x])/3 - Log[1 + x + x^2]/3

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Maple [A]
time = 0.07, size = 17, normalized size = 0.77

method	result	size
default	\(\frac {2 \ln \left (-1+x \right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{3}\)	\(17\)
norman	\(\frac {2 \ln \left (-1+x \right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{3}\)	\(17\)
risch	\(\frac {2 \ln \left (-1+x \right )}{3}-\frac {\ln \left (x^{2}+x +1\right )}{3}\)	\(17\)
meijerg	\(\frac {x \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {x^{2} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 1.03, size = 16, normalized size = 0.73 \begin {gather*} -\frac {1}{3} \, \log \left (x^{2} + x + 1\right ) + \frac {2}{3} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.80, size = 16, normalized size = 0.73 \begin {gather*} -\frac {1}{3} \, \log \left (x^{2} + x + 1\right ) + \frac {2}{3} \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.03, size = 17, normalized size = 0.77 \begin {gather*} \frac {2 \log {\left (x - 1 \right )}}{3} - \frac {\log {\left (x^{2} + x + 1 \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x - 1)/3 - log(x**2 + x + 1)/3

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Giac [A]
time = 0.48, size = 17, normalized size = 0.77 \begin {gather*} -\frac {1}{3} \, \log \left (x^{2} + x + 1\right ) + \frac {2}{3} \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.16, size = 16, normalized size = 0.73 \begin {gather*} \frac {2\,\ln \left (x-1\right )}{3}-\frac {\ln \left (x^2+x+1\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 1)/(x^3 - 1),x)

[Out]

(2*log(x - 1))/3 - log(x + x^2 + 1)/3

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