3.56 \(\int x^p \log (x) \, dx\)

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

[Out]

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

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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} \[ \frac {x^{p+1} \log (x)}{p+1}-\frac {x^{p+1}}{(p+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 25, normalized size = 0.96 \[ \frac {{\left ({\left (p + 1\right )} x \log \relax (x) - x\right )} x^{p}}{p^{2} + 2 \, p + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^p*log(x), x)

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maple [A]  time = 0.02, size = 34, normalized size = 1.31 \[ \frac {x \,{\mathrm e}^{p \ln \relax (x )} \ln \relax (x )}{p +1}-\frac {x \,{\mathrm e}^{p \ln \relax (x )}}{p^{2}+2 p +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^p*ln(x),x)

[Out]

1/(p+1)*x*ln(x)*exp(ln(x)*p)-1/(p^2+2*p+1)*x*exp(ln(x)*p)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.24, size = 32, normalized size = 1.23 \[ \left \{\begin {array}{cl} \frac {{\ln \relax (x)}^2}{2} & \text {\ if\ \ }p=-1\\ \frac {x^{p+1}\,\left (\ln \relax (x)\,\left (p+1\right )-1\right )}{{\left (p+1\right )}^2} & \text {\ if\ \ }p\neq -1 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.74, size = 56, normalized size = 2.15 \[ \begin {cases} \frac {p x x^{p} \log {\relax (x )}}{p^{2} + 2 p + 1} + \frac {x x^{p} \log {\relax (x )}}{p^{2} + 2 p + 1} - \frac {x x^{p}}{p^{2} + 2 p + 1} & \text {for}\: p \neq -1 \\\frac {\log {\relax (x )}^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**p*ln(x),x)

[Out]

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

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