3.608 \(\int x^m \log (x) \, dx\)

Optimal. Leaf size=26 \[ \frac {x^{m+1} \log (x)}{m+1}-\frac {x^{m+1}}{(m+1)^2} \]

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2304} \[ \frac {x^{m+1} \log (x)}{m+1}-\frac {x^{m+1}}{(m+1)^2} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Log[x],x]

[Out]

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

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.73 \[ \frac {x^{m+1} ((m+1) \log (x)-1)}{(m+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*Log[x],x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^m \log (x) \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[x^m*Log[x],x]

[Out]

Could not integrate

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fricas [A]  time = 1.15, size = 25, normalized size = 0.96 \[ \frac {{\left ({\left (m + 1\right )} x \log \relax (x) - x\right )} x^{m}}{m^{2} + 2 \, m + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{m} \log \relax (x)\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(x),x, algorithm="giac")

[Out]

integrate(x^m*log(x), x)

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maple [A]  time = 0.03, size = 19, normalized size = 0.73




method result size



risch \(\frac {x \left (m \ln \relax (x )+\ln \relax (x )-1\right ) x^{m}}{\left (1+m \right )^{2}}\) \(19\)
norman \(\frac {x \ln \relax (x ) {\mathrm e}^{m \ln \relax (x )}}{1+m}-\frac {x \,{\mathrm e}^{m \ln \relax (x )}}{m^{2}+2 m +1}\) \(34\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*ln(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.42, size = 26, normalized size = 1.00 \[ \frac {x^{m + 1} \log \relax (x)}{m + 1} - \frac {x^{m + 1}}{{\left (m + 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*log(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.40, size = 32, normalized size = 1.23 \[ \left \{\begin {array}{cl} \frac {{\ln \relax (x)}^2}{2} & \text {\ if\ \ }m=-1\\ \frac {x^{m+1}\,\left (\ln \relax (x)\,\left (m+1\right )-1\right )}{{\left (m+1\right )}^2} & \text {\ if\ \ }m\neq -1 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*log(x),x)

[Out]

piecewise(m == -1, log(x)^2/2, m ~= -1, (x^(m + 1)*(log(x)*(m + 1) - 1))/(m + 1)^2)

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sympy [A]  time = 0.76, size = 56, normalized size = 2.15 \[ \begin {cases} \frac {m x x^{m} \log {\relax (x )}}{m^{2} + 2 m + 1} + \frac {x x^{m} \log {\relax (x )}}{m^{2} + 2 m + 1} - \frac {x x^{m}}{m^{2} + 2 m + 1} & \text {for}\: m \neq -1 \\\frac {\log {\relax (x )}^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*ln(x),x)

[Out]

Piecewise((m*x*x**m*log(x)/(m**2 + 2*m + 1) + x*x**m*log(x)/(m**2 + 2*m + 1) - x*x**m/(m**2 + 2*m + 1), Ne(m,
-1)), (log(x)**2/2, True))

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