3.2.20 \(\int \frac {1+x+4 x^2}{-1+x^3} \, dx\) [120]

Optimal. Leaf size=16 \[ 2 \log (1-x)+\log \left (1+x+x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1889, 31, 642} \begin {gather*} \log \left (x^2+x+1\right )+2 \log (1-x) \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1+x+4 x^2}{-1+x^3} \, dx &=-\left (\frac {1}{3} \int \frac {-3-6 x}{1+x+x^2} \, dx\right )-2 \int \frac {1}{1-x} \, dx\\ &=2 \log (1-x)+\log \left (1+x+x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} 2 \log (1-x)+\log \left (1+x+x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Log[1 - x] + Log[1 + x + x^2]

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Maple [A]
time = 0.07, size = 15, normalized size = 0.94

method	result	size
default	\(2 \ln \left (-1+x \right )+\ln \left (x^{2}+x +1\right )\)	\(15\)
norman	\(2 \ln \left (-1+x \right )+\ln \left (x^{2}+x +1\right )\)	\(15\)
risch	\(2 \ln \left (-1+x \right )+\ln \left (x^{2}+x +1\right )\)	\(15\)
meijerg	\(\frac {x \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {1}{3}}}+\frac {4 \ln \left (-x^{3}+1\right )}{3}+\frac {x^{2} \left (\ln \left (1-\left (x^{3}\right )^{\frac {1}{3}}\right )-\frac {\ln \left (1+\left (x^{3}\right )^{\frac {1}{3}}+\left (x^{3}\right )^{\frac {2}{3}}\right )}{2}+\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x^{3}\right )^{\frac {1}{3}}}{2+\left (x^{3}\right )^{\frac {1}{3}}}\right )\right )}{3 \left (x^{3}\right )^{\frac {2}{3}}}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 1.31, size = 14, normalized size = 0.88 \begin {gather*} \log \left (x^{2} + x + 1\right ) + 2 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2 + x + 1) + 2*log(x - 1)

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Fricas [A]
time = 0.90, size = 14, normalized size = 0.88 \begin {gather*} \log \left (x^{2} + x + 1\right ) + 2 \, \log \left (x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2 + x + 1) + 2*log(x - 1)

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Sympy [A]
time = 0.03, size = 14, normalized size = 0.88 \begin {gather*} 2 \log {\left (x - 1 \right )} + \log {\left (x^{2} + x + 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*log(x - 1) + log(x**2 + x + 1)

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Giac [A]
time = 0.46, size = 15, normalized size = 0.94 \begin {gather*} \log \left (x^{2} + x + 1\right ) + 2 \, \log \left ({\left | x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2 + x + 1) + 2*log(abs(x - 1))

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Mupad [B]
time = 0.04, size = 14, normalized size = 0.88 \begin {gather*} \ln \left (x^2+x+1\right )+2\,\ln \left (x-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 4*x^2 + 1)/(x^3 - 1),x)

[Out]

log(x + x^2 + 1) + 2*log(x - 1)

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