\(\int x^m \log (x) \, dx\) [608]

Optimal result

Integrand size = 6, antiderivative size = 26 \[ \int x^m \log (x) \, dx=-\frac {x^{1+m}}{(1+m)^2}+\frac {x^{1+m} \log (x)}{1+m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2341} \[ \int x^m \log (x) \, dx=\frac {x^{m+1} \log (x)}{m+1}-\frac {x^{m+1}}{(m+1)^2} \]

[In]

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

[Out]

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

Rule 2341

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*} \text {integral}& = -\frac {x^{1+m}}{(1+m)^2}+\frac {x^{1+m} \log (x)}{1+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int x^m \log (x) \, dx=\frac {x^{1+m} (-1+(1+m) \log (x))}{(1+m)^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int x^m \log (x) \, dx=\frac {{\left ({\left (m + 1\right )} x \log \left (x\right ) - x\right )} x^{m}}{m^{2} + 2 \, m + 1} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (20) = 40\).

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int x^m \log (x) \, dx=\begin {cases} \frac {m x x^{m} \log {\left (x \right )}}{m^{2} + 2 m + 1} + \frac {x x^{m} \log {\left (x \right )}}{m^{2} + 2 m + 1} - \frac {x x^{m}}{m^{2} + 2 m + 1} & \text {for}\: m \neq -1 \\\frac {\log {\left (x \right )}^{2}}{2} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int x^m \log (x) \, dx=\frac {x^{m + 1} \log \left (x\right )}{m + 1} - \frac {x^{m + 1}}{{\left (m + 1\right )}^{2}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int x^m \log (x) \, dx=\int { x^{m} \log \left (x\right ) \,d x } \]

[In]

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

[Out]

integrate(x^m*log(x), x)

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int x^m \log (x) \, dx=\left \{\begin {array}{cl} \frac {{\ln \left (x\right )}^2}{2} & \text {\ if\ \ }m=-1\\ \frac {x^{m+1}\,\left (\ln \left (x\right )\,\left (m+1\right )-1\right )}{{\left (m+1\right )}^2} & \text {\ if\ \ }m\neq -1 \end {array}\right . \]

[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)