\(\int (b+a x) \log (x) \, dx\) [65]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 8, antiderivative size = 28 \[ \int (b+a x) \log (x) \, dx=-b x-\frac {a x^2}{4}+b x \log (x)+\frac {1}{2} a x^2 \log (x) \]

[Out]

-b*x-1/4*a*x^2+b*x*ln(x)+1/2*a*x^2*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2350} \[ \int (b+a x) \log (x) \, dx=-\frac {a x^2}{4}+\frac {1}{2} a x^2 \log (x)-b x+b x \log (x) \]

[In]

Int[(b + a*x)*Log[x],x]

[Out]

-(b*x) - (a*x^2)/4 + b*x*Log[x] + (a*x^2*Log[x])/2

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
 e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b
, c, d, e, n, r}, x] && IGtQ[q, 0]

Rubi steps \begin{align*} \text {integral}& = b x \log (x)+\frac {1}{2} a x^2 \log (x)-\int \left (b+\frac {a x}{2}\right ) \, dx \\ & = -b x-\frac {a x^2}{4}+b x \log (x)+\frac {1}{2} a x^2 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int (b+a x) \log (x) \, dx=-b x-\frac {a x^2}{4}+b x \log (x)+\frac {1}{2} a x^2 \log (x) \]

[In]

Integrate[(b + a*x)*Log[x],x]

[Out]

-(b*x) - (a*x^2)/4 + b*x*Log[x] + (a*x^2*Log[x])/2

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89

method result size
norman \(-b x -\frac {a \,x^{2}}{4}+b x \ln \left (x \right )+\frac {a \,x^{2} \ln \left (x \right )}{2}\) \(25\)
risch \(\left (\frac {1}{2} a \,x^{2}+b x \right ) \ln \left (x \right )-\frac {a \,x^{2}}{4}-b x\) \(25\)
parallelrisch \(-b x -\frac {a \,x^{2}}{4}+b x \ln \left (x \right )+\frac {a \,x^{2} \ln \left (x \right )}{2}\) \(25\)
parts \(-b x -\frac {a \,x^{2}}{4}+b x \ln \left (x \right )+\frac {a \,x^{2} \ln \left (x \right )}{2}\) \(25\)
default \(a \left (-\frac {x^{2}}{4}+\frac {x^{2} \ln \left (x \right )}{2}\right )+b \left (-x +x \ln \left (x \right )\right )\) \(27\)

[In]

int((a*x+b)*ln(x),x,method=_RETURNVERBOSE)

[Out]

-b*x-1/4*a*x^2+b*x*ln(x)+1/2*a*x^2*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int (b+a x) \log (x) \, dx=-\frac {1}{4} \, a x^{2} - b x + \frac {1}{2} \, {\left (a x^{2} + 2 \, b x\right )} \log \left (x\right ) \]

[In]

integrate((a*x+b)*log(x),x, algorithm="fricas")

[Out]

-1/4*a*x^2 - b*x + 1/2*(a*x^2 + 2*b*x)*log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int (b+a x) \log (x) \, dx=- \frac {a x^{2}}{4} - b x + \left (\frac {a x^{2}}{2} + b x\right ) \log {\left (x \right )} \]

[In]

integrate((a*x+b)*ln(x),x)

[Out]

-a*x**2/4 - b*x + (a*x**2/2 + b*x)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int (b+a x) \log (x) \, dx=-\frac {1}{4} \, a x^{2} - b x + \frac {1}{2} \, {\left (a x^{2} + 2 \, b x\right )} \log \left (x\right ) \]

[In]

integrate((a*x+b)*log(x),x, algorithm="maxima")

[Out]

-1/4*a*x^2 - b*x + 1/2*(a*x^2 + 2*b*x)*log(x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int (b+a x) \log (x) \, dx=\frac {1}{2} \, a x^{2} \log \left (x\right ) - \frac {1}{4} \, a x^{2} + b x \log \left (x\right ) - b x \]

[In]

integrate((a*x+b)*log(x),x, algorithm="giac")

[Out]

1/2*a*x^2*log(x) - 1/4*a*x^2 + b*x*log(x) - b*x

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int (b+a x) \log (x) \, dx=-\frac {x\,\left (4\,b+a\,x-4\,b\,\ln \left (x\right )-2\,a\,x\,\ln \left (x\right )\right )}{4} \]

[In]

int(log(x)*(b + a*x),x)

[Out]

-(x*(4*b + a*x - 4*b*log(x) - 2*a*x*log(x)))/4