3.15.26 \(\int (180+110 x+(30+20 x) \log (x)) \, dx\)

Optimal. Leaf size=20 \[ 4 \left (-e^3+\frac {5}{2} x (3+x) (5+\log (x))\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.35, number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2313, 9} \begin {gather*} 55 x^2+10 \left (x^2+3 x\right ) \log (x)+180 x-5 (x+3)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

180*x + 55*x^2 - 5*(3 + x)^2 + 10*(3*x + 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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), 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 {gather*} \begin {aligned} \text {integral} &=180 x+55 x^2+\int (30+20 x) \log (x) \, dx\\ &=180 x+55 x^2+10 \left (3 x+x^2\right ) \log (x)-\int 10 (3+x) \, dx\\ &=180 x+55 x^2-5 (3+x)^2+10 \left (3 x+x^2\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.05 \begin {gather*} 150 x+50 x^2+30 x \log (x)+10 x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.70, size = 20, normalized size = 1.00 \begin {gather*} 50 \, x^{2} + 10 \, {\left (x^{2} + 3 \, x\right )} \log \relax (x) + 150 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.54, size = 21, normalized size = 1.05 \begin {gather*} 10 \, x^{2} \log \relax (x) + 50 \, x^{2} + 30 \, x \log \relax (x) + 150 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 22, normalized size = 1.10

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.45, size = 20, normalized size = 1.00 \begin {gather*} 50 \, x^{2} + 10 \, {\left (x^{2} + 3 \, x\right )} \log \relax (x) + 150 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.01, size = 10, normalized size = 0.50 \begin {gather*} 10\,x\,\left (\ln \relax (x)+5\right )\,\left (x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*x*(log(x) + 5)*(x + 3)

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sympy [A]  time = 0.09, size = 19, normalized size = 0.95 \begin {gather*} 50 x^{2} + 150 x + \left (10 x^{2} + 30 x\right ) \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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