\(\int (1-13 x-2 x \log (x)) \, dx\) [6285]

Optimal result

Integrand size = 10, antiderivative size = 15 \[ \int (1-13 x-2 x \log (x)) \, dx=-14+x-6 x^2-x^2 \log (x) \]

[Out]

-14+x-x^2*ln(x)-6*x^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2341} \[ \int (1-13 x-2 x \log (x)) \, dx=-6 x^2+x^2 (-\log (x))+x \]

[In]

Int[1 - 13*x - 2*x*Log[x],x]

[Out]

x - 6*x^2 - x^2*Log[x]

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}& = x-\frac {13 x^2}{2}-2 \int x \log (x) \, dx \\ & = x-6 x^2-x^2 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int (1-13 x-2 x \log (x)) \, dx=x-6 x^2-x^2 \log (x) \]

[In]

Integrate[1 - 13*x - 2*x*Log[x],x]

[Out]

x - 6*x^2 - x^2*Log[x]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00

method	result	size
default	\(x -6 x^{2}-x^{2} \ln \left (x \right )\)	\(15\)
norman	\(x -6 x^{2}-x^{2} \ln \left (x \right )\)	\(15\)
risch	\(x -6 x^{2}-x^{2} \ln \left (x \right )\)	\(15\)
parallelrisch	\(x -6 x^{2}-x^{2} \ln \left (x \right )\)	\(15\)
parts	\(x -6 x^{2}-x^{2} \ln \left (x \right )\)	\(15\)

[In]

int(-2*x*ln(x)-13*x+1,x,method=_RETURNVERBOSE)

[Out]

x-6*x^2-x^2*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int (1-13 x-2 x \log (x)) \, dx=-x^{2} \log \left (x\right ) - 6 \, x^{2} + x \]

[In]

integrate(-2*x*log(x)-13*x+1,x, algorithm="fricas")

[Out]

-x^2*log(x) - 6*x^2 + x

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int (1-13 x-2 x \log (x)) \, dx=- x^{2} \log {\left (x \right )} - 6 x^{2} + x \]

[In]

integrate(-2*x*ln(x)-13*x+1,x)

[Out]

-x**2*log(x) - 6*x**2 + x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int (1-13 x-2 x \log (x)) \, dx=-x^{2} \log \left (x\right ) - 6 \, x^{2} + x \]

[In]

integrate(-2*x*log(x)-13*x+1,x, algorithm="maxima")

[Out]

-x^2*log(x) - 6*x^2 + x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int (1-13 x-2 x \log (x)) \, dx=-x^{2} \log \left (x\right ) - 6 \, x^{2} + x \]

[In]

integrate(-2*x*log(x)-13*x+1,x, algorithm="giac")

[Out]

-x^2*log(x) - 6*x^2 + x

Mupad [B] (verification not implemented)

Time = 11.12 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int (1-13 x-2 x \log (x)) \, dx=-x\,\left (6\,x+x\,\ln \left (x\right )-1\right ) \]

[In]

int(1 - 2*x*log(x) - 13*x,x)

[Out]

-x*(6*x + x*log(x) - 1)