\(\int x^3 \log (c x) \, dx\) [1]

Optimal result

Integrand size = 8, antiderivative size = 19 \[ \int x^3 \log (c x) \, dx=-\frac {x^4}{16}+\frac {1}{4} x^4 \log (c x) \]

[Out]

-1/16*x^4+1/4*x^4*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} \[ \int x^3 \log (c x) \, dx=\frac {1}{4} x^4 \log (c x)-\frac {x^4}{16} \]

[In]

Int[x^3*Log[c*x],x]

[Out]

-1/16*x^4 + (x^4*Log[c*x])/4

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^4}{16}+\frac {1}{4} x^4 \log (c x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int x^3 \log (c x) \, dx=-\frac {x^4}{16}+\frac {1}{4} x^4 \log (c x) \]

[In]

Integrate[x^3*Log[c*x],x]

[Out]

-1/16*x^4 + (x^4*Log[c*x])/4

Maple [A] (verified)

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

method	result	size
norman	\(-\frac {x^{4}}{16}+\frac {x^{4} \ln \left (x c \right )}{4}\)	\(16\)
risch	\(-\frac {x^{4}}{16}+\frac {x^{4} \ln \left (x c \right )}{4}\)	\(16\)
parallelrisch	\(-\frac {x^{4}}{16}+\frac {x^{4} \ln \left (x c \right )}{4}\)	\(16\)
parts	\(-\frac {x^{4}}{16}+\frac {x^{4} \ln \left (x c \right )}{4}\)	\(16\)
derivativedivides	\(\frac {\frac {x^{4} c^{4} \ln \left (x c \right )}{4}-\frac {x^{4} c^{4}}{16}}{c^{4}}\)	\(26\)
default	\(\frac {\frac {x^{4} c^{4} \ln \left (x c \right )}{4}-\frac {x^{4} c^{4}}{16}}{c^{4}}\)	\(26\)

[In]

int(x^3*ln(x*c),x,method=_RETURNVERBOSE)

[Out]

-1/16*x^4+1/4*x^4*ln(x*c)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^3 \log (c x) \, dx=\frac {1}{4} \, x^{4} \log \left (c x\right ) - \frac {1}{16} \, x^{4} \]

[In]

integrate(x^3*log(c*x),x, algorithm="fricas")

[Out]

1/4*x^4*log(c*x) - 1/16*x^4

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.74 \[ \int x^3 \log (c x) \, dx=\frac {x^{4} \log {\left (c x \right )}}{4} - \frac {x^{4}}{16} \]

[In]

integrate(x**3*ln(c*x),x)

[Out]

x**4*log(c*x)/4 - x**4/16

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^3 \log (c x) \, dx=\frac {1}{4} \, x^{4} \log \left (c x\right ) - \frac {1}{16} \, x^{4} \]

[In]

integrate(x^3*log(c*x),x, algorithm="maxima")

[Out]

1/4*x^4*log(c*x) - 1/16*x^4

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int x^3 \log (c x) \, dx=\frac {1}{4} \, x^{4} \log \left (c x\right ) - \frac {1}{16} \, x^{4} \]

[In]

integrate(x^3*log(c*x),x, algorithm="giac")

[Out]

1/4*x^4*log(c*x) - 1/16*x^4

Mupad [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.58 \[ \int x^3 \log (c x) \, dx=\frac {x^4\,\left (\ln \left (c\,x\right )-\frac {1}{4}\right )}{4} \]

[In]

int(x^3*log(c*x),x)

[Out]

(x^4*(log(c*x) - 1/4))/4