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

Optimal. Leaf size=13 \[ \frac {1}{9} (3+x)^2 \log (2 x) \]

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Rubi [B]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 3.00, number of steps used = 7, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {12, 14, 2313, 9} \begin {gather*} \frac {x^2}{18}+\frac {1}{9} \left (x^2+6 x\right ) \log (2 x)+\frac {2 x}{3}-\frac {1}{18} (x+6)^2+\log (x) \end {gather*}

Antiderivative was successfully verified.

[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] + ((6*x + 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 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} &=\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 {1}{9} \left (6 x+x^2\right ) \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 {1}{9} \left (6 x+x^2\right ) \log (2 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 1.77 \begin {gather*} \log (x)+\frac {2}{3} x \log (2 x)+\frac {1}{9} x^2 \log (2 x) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.61, size = 14, normalized size = 1.08 \begin {gather*} \frac {1}{9} \, {\left (x^{2} + 6 \, x + 9\right )} \log \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.18, size = 16, normalized size = 1.23 \begin {gather*} \frac {1}{9} \, {\left (x^{2} + 6 \, x\right )} \log \left (2 \, x\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.05, size = 17, normalized size = 1.31

method	result	size
risch	\(\frac {\left (x^{2}+6 x \right ) \ln \left (2 x \right )}{9}+\ln \relax (x )\)	\(17\)
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	\(\frac {x^{2} \ln \left (2 x \right )}{9}+\frac {2 x \ln \left (2 x \right )}{3}+\ln \left (2 x \right )\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.39, size = 19, normalized size = 1.46 \begin {gather*} \frac {1}{9} \, x^{2} \log \left (2 \, x\right ) + \frac {2}{3} \, x \log \left (2 \, x\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 1.44, size = 11, normalized size = 0.85 \begin {gather*} \frac {\ln \left (2\,x\right )\,{\left (x+3\right )}^2}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.11, size = 17, normalized size = 1.31 \begin {gather*} \left (\frac {x^{2}}{9} + \frac {2 x}{3}\right ) \log {\left (2 x \right )} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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