3.70.58 \(\int \frac {-1-4 x+8 x^2+(-4 x+16 x^2) \log (x)}{9 x} \, dx\) [6958]

Optimal. Leaf size=16 \[ \frac {1}{9} (-1+4 x (-1+2 x)) \log (x) \]

[Out]

1/9*(4*x*(-1+2*x)-1)*ln(x)

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Rubi [A]
time = 0.02, antiderivative size = 23, normalized size of antiderivative = 1.44, number of steps used = 7, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2350} \begin {gather*} \frac {8}{9} x^2 \log (x)-\frac {4}{9} x \log (x)-\frac {\log (x)}{9} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-1 - 4*x + 8*x^2 + (-4*x + 16*x^2)*Log[x])/(9*x),x]

[Out]

-1/9*Log[x] - (4*x*Log[x])/9 + (8*x^2*Log[x])/9

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 {gather*} \begin {aligned} \text {integral} &=\frac {1}{9} \int \frac {-1-4 x+8 x^2+\left (-4 x+16 x^2\right ) \log (x)}{x} \, dx\\ &=\frac {1}{9} \int \left (\frac {-1-4 x+8 x^2}{x}+4 (-1+4 x) \log (x)\right ) \, dx\\ &=\frac {1}{9} \int \frac {-1-4 x+8 x^2}{x} \, dx+\frac {4}{9} \int (-1+4 x) \log (x) \, dx\\ &=-\frac {4}{9} \left (x-2 x^2\right ) \log (x)+\frac {1}{9} \int \left (-4-\frac {1}{x}+8 x\right ) \, dx-\frac {4}{9} \int (-1+2 x) \, dx\\ &=-\frac {\log (x)}{9}-\frac {4}{9} \left (x-2 x^2\right ) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.01, size = 23, normalized size = 1.44 \begin {gather*} -\frac {\log (x)}{9}-\frac {4}{9} x \log (x)+\frac {8}{9} x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-1 - 4*x + 8*x^2 + (-4*x + 16*x^2)*Log[x])/(9*x),x]

[Out]

-1/9*Log[x] - (4*x*Log[x])/9 + (8*x^2*Log[x])/9

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Maple [A]
time = 0.11, size = 18, normalized size = 1.12

method	result	size
default	\(\frac {8 x^{2} \ln \left (x \right )}{9}-\frac {4 x \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\)	\(18\)
norman	\(\frac {8 x^{2} \ln \left (x \right )}{9}-\frac {4 x \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\)	\(18\)
risch	\(\frac {\left (8 x^{2}-4 x \right ) \ln \left (x \right )}{9}-\frac {\ln \left (x \right )}{9}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/9*((16*x^2-4*x)*ln(x)+8*x^2-4*x-1)/x,x,method=_RETURNVERBOSE)

[Out]

8/9*x^2*ln(x)-4/9*x*ln(x)-1/9*ln(x)

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Maxima [A]
time = 0.28, size = 17, normalized size = 1.06 \begin {gather*} \frac {8}{9} \, x^{2} \log \left (x\right ) - \frac {4}{9} \, x \log \left (x\right ) - \frac {1}{9} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((16*x^2-4*x)*log(x)+8*x^2-4*x-1)/x,x, algorithm="maxima")

[Out]

8/9*x^2*log(x) - 4/9*x*log(x) - 1/9*log(x)

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Fricas [A]
time = 0.37, size = 14, normalized size = 0.88 \begin {gather*} \frac {1}{9} \, {\left (8 \, x^{2} - 4 \, x - 1\right )} \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((16*x^2-4*x)*log(x)+8*x^2-4*x-1)/x,x, algorithm="fricas")

[Out]

1/9*(8*x^2 - 4*x - 1)*log(x)

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Sympy [A]
time = 0.04, size = 19, normalized size = 1.19 \begin {gather*} \left (\frac {8 x^{2}}{9} - \frac {4 x}{9}\right ) \log {\left (x \right )} - \frac {\log {\left (x \right )}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((16*x**2-4*x)*ln(x)+8*x**2-4*x-1)/x,x)

[Out]

(8*x**2/9 - 4*x/9)*log(x) - log(x)/9

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Giac [A]
time = 0.42, size = 18, normalized size = 1.12 \begin {gather*} \frac {4}{9} \, {\left (2 \, x^{2} - x\right )} \log \left (x\right ) - \frac {1}{9} \, \log \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/9*((16*x^2-4*x)*log(x)+8*x^2-4*x-1)/x,x, algorithm="giac")

[Out]

4/9*(2*x^2 - x)*log(x) - 1/9*log(x)

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Mupad [B]
time = 4.25, size = 14, normalized size = 0.88 \begin {gather*} -\frac {\ln \left (x\right )\,\left (-8\,x^2+4\,x+1\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((4*x)/9 + (log(x)*(4*x - 16*x^2))/9 - (8*x^2)/9 + 1/9)/x,x)

[Out]

-(log(x)*(4*x - 8*x^2 + 1))/9

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