3.73 \(\int x^2 \log (\frac{-1+x}{x}) \, dx\)

Optimal. Leaf size=36 \[ -\frac{x^2}{6}+\frac{1}{3} x^3 \log \left (\frac{x-1}{x}\right )-\frac{x}{3}-\frac{1}{3} \log (x-1) \]

[Out]

-x/3 - x^2/6 - Log[-1 + x]/3 + (x^3*Log[(-1 + x)/x])/3

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Rubi [A]  time = 0.0255435, antiderivative size = 38, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2461, 2455, 263, 43} \[ -\frac{x^2}{6}+\frac{1}{3} x^3 \log \left (1-\frac{1}{x}\right )-\frac{x}{3}-\frac{1}{3} \log (1-x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[(-1 + x)/x],x]

[Out]

-x/3 - x^2/6 + (x^3*Log[1 - x^(-1)])/3 - Log[1 - x]/3

Rule 2461

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.)*((f_.)*(x_))^(m_.), x_Symbol] :> Int[(f*x)^m*(a + b*Log[c*Expa
ndToSum[v, x]^p])^q, x] /; FreeQ[{a, b, c, f, m, p, q}, x] && BinomialQ[v, x] &&  !BinomialMatchQ[v, x]

Rule 2455

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[((f*x)^(m
+ 1)*(a + b*Log[c*(d + e*x^n)^p]))/(f*(m + 1)), x] - Dist[(b*e*n*p)/(f*(m + 1)), Int[(x^(n - 1)*(f*x)^(m + 1))
/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rule 263

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0056543, size = 38, normalized size = 1.06 \[ -\frac{x^2}{6}+\frac{1}{3} x^3 \log \left (\frac{x-1}{x}\right )-\frac{x}{3}-\frac{1}{3} \log (1-x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[(-1 + x)/x],x]

[Out]

-x/3 - x^2/6 - Log[1 - x]/3 + (x^3*Log[(-1 + x)/x])/3

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Maple [A]  time = 0.012, size = 53, normalized size = 1.5 \begin{align*} -{\frac{{x}^{2}}{6}}+{\frac{1}{3}\ln \left ( -{x}^{-1} \right ) }-{\frac{x}{3}}+{\frac{{x}^{3}}{3}\ln \left ( 1-{x}^{-1} \right ) \left ( 1-{x}^{-1} \right ) \left ( \left ( 1-{x}^{-1} \right ) ^{2}+3\,{x}^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*ln((-1+x)/x),x)

[Out]

-1/6*x^2+1/3*ln(-1/x)-1/3*x+1/3*ln(1-1/x)*(1-1/x)*((1-1/x)^2+3/x)*x^3

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Maxima [A]  time = 0.940232, size = 38, normalized size = 1.06 \begin{align*} \frac{1}{3} \, x^{3} \log \left (\frac{x - 1}{x}\right ) - \frac{1}{6} \, x^{2} - \frac{1}{3} \, x - \frac{1}{3} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log((x - 1)/x) - 1/6*x^2 - 1/3*x - 1/3*log(x - 1)

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Fricas [A]  time = 1.88659, size = 80, normalized size = 2.22 \begin{align*} \frac{1}{3} \, x^{3} \log \left (\frac{x - 1}{x}\right ) - \frac{1}{6} \, x^{2} - \frac{1}{3} \, x - \frac{1}{3} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log((x - 1)/x) - 1/6*x^2 - 1/3*x - 1/3*log(x - 1)

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Sympy [A]  time = 0.117856, size = 26, normalized size = 0.72 \begin{align*} \frac{x^{3} \log{\left (\frac{x - 1}{x} \right )}}{3} - \frac{x^{2}}{6} - \frac{x}{3} - \frac{\log{\left (x - 1 \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*ln((-1+x)/x),x)

[Out]

x**3*log((x - 1)/x)/3 - x**2/6 - x/3 - log(x - 1)/3

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Giac [A]  time = 1.05709, size = 39, normalized size = 1.08 \begin{align*} \frac{1}{3} \, x^{3} \log \left (\frac{x - 1}{x}\right ) - \frac{1}{6} \, x^{2} - \frac{1}{3} \, x - \frac{1}{3} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^3*log((x - 1)/x) - 1/6*x^2 - 1/3*x - 1/3*log(abs(x - 1))