\(\int x^2 \log (c x) \, dx\) [2]

Optimal result

Integrand size = 8, antiderivative size = 19 \[ \int x^2 \log (c x) \, dx=-\frac {x^3}{9}+\frac {1}{3} x^3 \log (c x) \]

[Out]

-1/9*x^3+1/3*x^3*ln(c*x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2341} \[ \int x^2 \log (c x) \, dx=\frac {1}{3} x^3 \log (c x)-\frac {x^3}{9} \]

[In]

Int[x^2*Log[c*x],x]

[Out]

-1/9*x^3 + (x^3*Log[c*x])/3

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^3}{9}+\frac {1}{3} x^3 \log (c x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int x^2 \log (c x) \, dx=-\frac {x^3}{9}+\frac {1}{3} x^3 \log (c x) \]

[In]

Integrate[x^2*Log[c*x],x]

[Out]

-1/9*x^3 + (x^3*Log[c*x])/3

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
norman	\(-\frac {x^{3}}{9}+\frac {x^{3} \ln \left (x c \right )}{3}\)	\(16\)
risch	\(-\frac {x^{3}}{9}+\frac {x^{3} \ln \left (x c \right )}{3}\)	\(16\)
parallelrisch	\(-\frac {x^{3}}{9}+\frac {x^{3} \ln \left (x c \right )}{3}\)	\(16\)
parts	\(-\frac {x^{3}}{9}+\frac {x^{3} \ln \left (x c \right )}{3}\)	\(16\)
derivativedivides	\(\frac {\frac {x^{3} c^{3} \ln \left (x c \right )}{3}-\frac {x^{3} c^{3}}{9}}{c^{3}}\)	\(26\)
default	\(\frac {\frac {x^{3} c^{3} \ln \left (x c \right )}{3}-\frac {x^{3} c^{3}}{9}}{c^{3}}\)	\(26\)

[In]

int(x^2*ln(x*c),x,method=_RETURNVERBOSE)

[Out]

-1/9*x^3+1/3*x^3*ln(x*c)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^2 \log (c x) \, dx=\frac {1}{3} \, x^{3} \log \left (c x\right ) - \frac {1}{9} \, x^{3} \]

[In]

integrate(x^2*log(c*x),x, algorithm="fricas")

[Out]

1/3*x^3*log(c*x) - 1/9*x^3

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int x^2 \log (c x) \, dx=\frac {x^{3} \log {\left (c x \right )}}{3} - \frac {x^{3}}{9} \]

[In]

integrate(x**2*ln(c*x),x)

[Out]

x**3*log(c*x)/3 - x**3/9

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^2 \log (c x) \, dx=\frac {1}{3} \, x^{3} \log \left (c x\right ) - \frac {1}{9} \, x^{3} \]

[In]

integrate(x^2*log(c*x),x, algorithm="maxima")

[Out]

1/3*x^3*log(c*x) - 1/9*x^3

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^2 \log (c x) \, dx=\frac {1}{3} \, x^{3} \log \left (c x\right ) - \frac {1}{9} \, x^{3} \]

[In]

integrate(x^2*log(c*x),x, algorithm="giac")

[Out]

1/3*x^3*log(c*x) - 1/9*x^3

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int x^2 \log (c x) \, dx=\frac {x^3\,\left (\ln \left (c\,x\right )-\frac {1}{3}\right )}{3} \]

[In]

int(x^2*log(c*x),x)

[Out]

(x^3*(log(c*x) - 1/3))/3