\(\int (180+110 x+(30+20 x) \log (x)) \, dx\) [1426]

   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 = 13, antiderivative size = 20 \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=4 \left (-e^3+\frac {5}{2} x (3+x) (5+\log (x))\right ) \]

[Out]

10*(5+ln(x))*x*(3+x)-4*exp(3)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2350, 9} \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=55 x^2+10 x^2 \log (x)+180 x-5 (x+3)^2+30 x \log (x) \]

[In]

Int[180 + 110*x + (30 + 20*x)*Log[x],x]

[Out]

180*x + 55*x^2 - 5*(3 + x)^2 + 30*x*Log[x] + 10*x^2*Log[x]

Rule 9

Int[(a_)*((b_) + (c_.)*(x_)), x_Symbol] :> Simp[a*((b + c*x)^2/(2*c)), x] /; FreeQ[{a, b, c}, x]

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}& = 180 x+55 x^2+\int (30+20 x) \log (x) \, dx \\ & = 180 x+55 x^2+30 x \log (x)+10 x^2 \log (x)-\int 10 (3+x) \, dx \\ & = 180 x+55 x^2-5 (3+x)^2+30 x \log (x)+10 x^2 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=150 x+50 x^2+30 x \log (x)+10 x^2 \log (x) \]

[In]

Integrate[180 + 110*x + (30 + 20*x)*Log[x],x]

[Out]

150*x + 50*x^2 + 30*x*Log[x] + 10*x^2*Log[x]

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10

method result size
default \(150 x +10 x^{2} \ln \left (x \right )+50 x^{2}+30 x \ln \left (x \right )\) \(22\)
norman \(150 x +10 x^{2} \ln \left (x \right )+50 x^{2}+30 x \ln \left (x \right )\) \(22\)
risch \(\left (10 x^{2}+30 x \right ) \ln \left (x \right )+50 x^{2}+150 x\) \(22\)
parallelrisch \(150 x +10 x^{2} \ln \left (x \right )+50 x^{2}+30 x \ln \left (x \right )\) \(22\)
parts \(150 x +10 x^{2} \ln \left (x \right )+50 x^{2}+30 x \ln \left (x \right )\) \(22\)

[In]

int((20*x+30)*ln(x)+110*x+180,x,method=_RETURNVERBOSE)

[Out]

150*x+10*x^2*ln(x)+50*x^2+30*x*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=50 \, x^{2} + 10 \, {\left (x^{2} + 3 \, x\right )} \log \left (x\right ) + 150 \, x \]

[In]

integrate((20*x+30)*log(x)+110*x+180,x, algorithm="fricas")

[Out]

50*x^2 + 10*(x^2 + 3*x)*log(x) + 150*x

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=50 x^{2} + 150 x + \left (10 x^{2} + 30 x\right ) \log {\left (x \right )} \]

[In]

integrate((20*x+30)*ln(x)+110*x+180,x)

[Out]

50*x**2 + 150*x + (10*x**2 + 30*x)*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=50 \, x^{2} + 10 \, {\left (x^{2} + 3 \, x\right )} \log \left (x\right ) + 150 \, x \]

[In]

integrate((20*x+30)*log(x)+110*x+180,x, algorithm="maxima")

[Out]

50*x^2 + 10*(x^2 + 3*x)*log(x) + 150*x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=10 \, x^{2} \log \left (x\right ) + 50 \, x^{2} + 30 \, x \log \left (x\right ) + 150 \, x \]

[In]

integrate((20*x+30)*log(x)+110*x+180,x, algorithm="giac")

[Out]

10*x^2*log(x) + 50*x^2 + 30*x*log(x) + 150*x

Mupad [B] (verification not implemented)

Time = 8.69 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.50 \[ \int (180+110 x+(30+20 x) \log (x)) \, dx=10\,x\,\left (\ln \left (x\right )+5\right )\,\left (x+3\right ) \]

[In]

int(110*x + log(x)*(20*x + 30) + 180,x)

[Out]

10*x*(log(x) + 5)*(x + 3)