3.46.13 \(\int \frac {-2-8 x-x^2 \log (x)}{x^2} \, dx\)

Optimal. Leaf size=23 \[ \frac {2}{x}+x-x \log (x)-4 \log \left (x^2 (3+\log (5))\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 0.70, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {14, 43, 2295} \begin {gather*} x+\frac {2}{x}+x (-\log (x))-8 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 - 8*x - x^2*Log[x])/x^2,x]

[Out]

2/x + x - 8*Log[x] - x*Log[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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 16, normalized size = 0.70 \begin {gather*} \frac {2}{x}+x-8 \log (x)-x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 - 8*x - x^2*Log[x])/x^2,x]

[Out]

2/x + x - 8*Log[x] - x*Log[x]

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fricas [A]  time = 0.57, size = 20, normalized size = 0.87 \begin {gather*} \frac {x^{2} - {\left (x^{2} + 8 \, x\right )} \log \relax (x) + 2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*log(x)-8*x-2)/x^2,x, algorithm="fricas")

[Out]

(x^2 - (x^2 + 8*x)*log(x) + 2)/x

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giac [A]  time = 0.12, size = 16, normalized size = 0.70 \begin {gather*} -x \log \relax (x) + x + \frac {2}{x} - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*log(x)-8*x-2)/x^2,x, algorithm="giac")

[Out]

-x*log(x) + x + 2/x - 8*log(x)

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maple [A]  time = 0.02, size = 17, normalized size = 0.74

method	result	size
default	\(x -x \ln \relax (x )-8 \ln \relax (x )+\frac {2}{x}\)	\(17\)
norman	\(\frac {2+x^{2}-8 x \ln \relax (x )-x^{2} \ln \relax (x )}{x}\)	\(22\)
risch	\(-x \ln \relax (x )-\frac {8 x \ln \relax (x )-x^{2}-2}{x}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2*ln(x)-8*x-2)/x^2,x,method=_RETURNVERBOSE)

[Out]

x-x*ln(x)-8*ln(x)+2/x

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maxima [A]  time = 0.37, size = 16, normalized size = 0.70 \begin {gather*} -x \log \relax (x) + x + \frac {2}{x} - 8 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2*log(x)-8*x-2)/x^2,x, algorithm="maxima")

[Out]

-x*log(x) + x + 2/x - 8*log(x)

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mupad [B]  time = 3.18, size = 17, normalized size = 0.74 \begin {gather*} \frac {2}{x}-x\,\left (\ln \relax (x)-1\right )-8\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*x + x^2*log(x) + 2)/x^2,x)

[Out]

2/x - x*(log(x) - 1) - 8*log(x)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.61 \begin {gather*} - x \log {\relax (x )} + x - 8 \log {\relax (x )} + \frac {2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2*ln(x)-8*x-2)/x**2,x)

[Out]

-x*log(x) + x - 8*log(x) + 2/x

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