\(\int \frac {9+6 x+x^2+(6 x+2 x^2) \log (2 x)}{9 x} \, dx\) [2375]

Optimal result

Integrand size = 29, antiderivative size = 13 \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\frac {1}{9} (3+x)^2 \log (2 x) \]

[Out]

1/9*ln(2*x)*(3+x)^2

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(44\) vs. \(2(13)=26\).

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 3.38, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {12, 14, 2350, 9} \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\frac {x^2}{18}+\frac {1}{9} x^2 \log (2 x)+\frac {2 x}{3}-\frac {1}{18} (x+6)^2+\frac {2}{3} x \log (2 x)+\log (x) \]

[In]

Int[(9 + 6*x + x^2 + (6*x + 2*x^2)*Log[2*x])/(9*x),x]

[Out]

(2*x)/3 + x^2/18 - (6 + x)^2/18 + Log[x] + (2*x*Log[2*x])/3 + (x^2*Log[2*x])/9

Rule 9

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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}& = \frac {1}{9} \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{x} \, dx \\ & = \frac {1}{9} \int \left (\frac {9+6 x+x^2}{x}+2 (3+x) \log (2 x)\right ) \, dx \\ & = \frac {1}{9} \int \frac {9+6 x+x^2}{x} \, dx+\frac {2}{9} \int (3+x) \log (2 x) \, dx \\ & = \frac {2}{3} x \log (2 x)+\frac {1}{9} x^2 \log (2 x)+\frac {1}{9} \int \left (6+\frac {9}{x}+x\right ) \, dx-\frac {2}{9} \int \frac {6+x}{2} \, dx \\ & = \frac {2 x}{3}+\frac {x^2}{18}-\frac {1}{18} (6+x)^2+\log (x)+\frac {2}{3} x \log (2 x)+\frac {1}{9} x^2 \log (2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.77 \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\log (x)+\frac {2}{3} x \log (2 x)+\frac {1}{9} x^2 \log (2 x) \]

[In]

Integrate[(9 + 6*x + x^2 + (6*x + 2*x^2)*Log[2*x])/(9*x),x]

[Out]

Log[x] + (2*x*Log[2*x])/3 + (x^2*Log[2*x])/9

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31

method	result	size
risch	\(\frac {\left (x^{2}+6 x \right ) \ln \left (2 x \right )}{9}+\ln \left (x \right )\)	\(17\)
parts	\(\frac {2 x \ln \left (2 x \right )}{3}+\frac {x^{2} \ln \left (2 x \right )}{9}+\ln \left (x \right )\)	\(20\)
derivativedivides	\(\frac {x^{2} \ln \left (2 x \right )}{9}+\frac {2 x \ln \left (2 x \right )}{3}+\ln \left (2 x \right )\)	\(22\)
default	\(\frac {x^{2} \ln \left (2 x \right )}{9}+\frac {2 x \ln \left (2 x \right )}{3}+\ln \left (2 x \right )\)	\(22\)
norman	\(\frac {x^{2} \ln \left (2 x \right )}{9}+\frac {2 x \ln \left (2 x \right )}{3}+\ln \left (2 x \right )\)	\(22\)
parallelrisch	\(\frac {x^{2} \ln \left (2 x \right )}{9}+\frac {2 x \ln \left (2 x \right )}{3}+\ln \left (2 x \right )\)	\(22\)

[In]

int(1/9*((2*x^2+6*x)*ln(2*x)+x^2+6*x+9)/x,x,method=_RETURNVERBOSE)

[Out]

1/9*(x^2+6*x)*ln(2*x)+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\frac {1}{9} \, {\left (x^{2} + 6 \, x + 9\right )} \log \left (2 \, x\right ) \]

[In]

integrate(1/9*((2*x^2+6*x)*log(2*x)+x^2+6*x+9)/x,x, algorithm="fricas")

[Out]

1/9*(x^2 + 6*x + 9)*log(2*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\left (\frac {x^{2}}{9} + \frac {2 x}{3}\right ) \log {\left (2 x \right )} + \log {\left (x \right )} \]

[In]

integrate(1/9*((2*x**2+6*x)*ln(2*x)+x**2+6*x+9)/x,x)

[Out]

(x**2/9 + 2*x/3)*log(2*x) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\frac {1}{9} \, x^{2} \log \left (2 \, x\right ) + \frac {2}{3} \, x \log \left (2 \, x\right ) + \log \left (x\right ) \]

[In]

integrate(1/9*((2*x^2+6*x)*log(2*x)+x^2+6*x+9)/x,x, algorithm="maxima")

[Out]

1/9*x^2*log(2*x) + 2/3*x*log(2*x) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\frac {1}{9} \, {\left (x^{2} + 6 \, x\right )} \log \left (2 \, x\right ) + \log \left (x\right ) \]

[In]

integrate(1/9*((2*x^2+6*x)*log(2*x)+x^2+6*x+9)/x,x, algorithm="giac")

[Out]

1/9*(x^2 + 6*x)*log(2*x) + log(x)

Mupad [B] (verification not implemented)

Time = 10.59 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {9+6 x+x^2+\left (6 x+2 x^2\right ) \log (2 x)}{9 x} \, dx=\frac {\ln \left (2\,x\right )\,{\left (x+3\right )}^2}{9} \]

[In]

int(((2*x)/3 + (log(2*x)*(6*x + 2*x^2))/9 + x^2/9 + 1)/x,x)

[Out]

(log(2*x)*(x + 3)^2)/9