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

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

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Rubi [A]  time = 0.03, 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 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])

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 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 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]

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*}

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Mathematica [A]  time = 0.01, size = 38, normalized size = 1.06 \[ \frac {1}{3} x^3 \log \left (\frac {x-1}{x}\right )-\frac {x^2}{6}-\frac {x}{3}-\frac {1}{3} \log (1-x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[(-1 + x)/x],x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^2 \log \left (\frac {-1+x}{x}\right ) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x^2*Log[(-1 + x)/x],x]

[Out]

Could not integrate

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fricas [A]  time = 0.93, size = 28, normalized size = 0.78 \[ \frac {1}{3} \, x^{3} \log \left (\frac {x - 1}{x}\right ) - \frac {1}{6} \, x^{2} - \frac {1}{3} \, x - \frac {1}{3} \, \log \left (x - 1\right ) \]

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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giac [B]  time = 1.05, size = 70, normalized size = 1.94 \[ \frac {\frac {2 \, {\left (x - 1\right )}}{x} - 3}{6 \, {\left (\frac {x - 1}{x} - 1\right )}^{2}} - \frac {\log \left (\frac {x - 1}{x}\right )}{3 \, {\left (\frac {x - 1}{x} - 1\right )}^{3}} - \frac {1}{3} \, \log \left (\frac {{\left | x - 1 \right |}}{{\left | x \right |}}\right ) + \frac {1}{3} \, \log \left ({\left | \frac {x - 1}{x} - 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 29, normalized size = 0.81

method	result	size
risch	\(-\frac {x}{3}-\frac {x^{2}}{6}-\frac {\ln \left (-1+x \right )}{3}+\frac {x^{3} \ln \left (\frac {-1+x}{x}\right )}{3}\)	\(29\)
derivativedivides	\(\frac {\ln \left (-\frac {1}{x}\right )}{3}-\frac {x}{3}-\frac {x^{2}}{6}+\frac {\ln \left (1-\frac {1}{x}\right ) \left (1-\frac {1}{x}\right ) \left (\left (1-\frac {1}{x}\right )^{2}+\frac {3}{x}\right ) x^{3}}{3}\)	\(53\)
default	\(\frac {\ln \left (-\frac {1}{x}\right )}{3}-\frac {x}{3}-\frac {x^{2}}{6}+\frac {\ln \left (1-\frac {1}{x}\right ) \left (1-\frac {1}{x}\right ) \left (\left (1-\frac {1}{x}\right )^{2}+\frac {3}{x}\right ) x^{3}}{3}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.45, size = 28, normalized size = 0.78 \[ \frac {1}{3} \, x^{3} \log \left (\frac {x - 1}{x}\right ) - \frac {1}{6} \, x^{2} - \frac {1}{3} \, x - \frac {1}{3} \, \log \left (x - 1\right ) \]

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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mupad [B]  time = 0.35, size = 40, normalized size = 1.11 \[ \frac {x^3\,\ln \left (\frac {x-1}{x}\right )}{3}-\frac {\ln \left (x\,\left (x-1\right )\right )}{6}-\frac {\ln \left (\frac {x-1}{x}\right )}{6}-\frac {x}{3}-\frac {x^2}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*log((x - 1)/x),x)

[Out]

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

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sympy [A]  time = 0.13, size = 26, normalized size = 0.72 \[ \frac {x^{3} \log {\left (\frac {x - 1}{x} \right )}}{3} - \frac {x^{2}}{6} - \frac {x}{3} - \frac {\log {\left (x - 1 \right )}}{3} \]

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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