3.40.79 \(\int \frac {-10-3 x^2+x^2 \log (x)}{6 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {2+4 \left (2-x^2\right )+x^2 \log (x)}{6 x} \]

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Rubi [A]  time = 0.01, antiderivative size = 20, normalized size of antiderivative = 0.83, number of steps used = 6, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {12, 14, 2295} \begin {gather*} -\frac {2 x}{3}+\frac {5}{3 x}+\frac {1}{6} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10 - 3*x^2 + x^2*Log[x])/(6*x^2),x]

[Out]

5/(3*x) - (2*x)/3 + (x*Log[x])/6

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 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} &=\frac {1}{6} \int \frac {-10-3 x^2+x^2 \log (x)}{x^2} \, dx\\ &=\frac {1}{6} \int \left (\frac {-10-3 x^2}{x^2}+\log (x)\right ) \, dx\\ &=\frac {1}{6} \int \frac {-10-3 x^2}{x^2} \, dx+\frac {1}{6} \int \log (x) \, dx\\ &=-\frac {x}{6}+\frac {1}{6} x \log (x)+\frac {1}{6} \int \left (-3-\frac {10}{x^2}\right ) \, dx\\ &=\frac {5}{3 x}-\frac {2 x}{3}+\frac {1}{6} x \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 0.83 \begin {gather*} \frac {5}{3 x}-\frac {2 x}{3}+\frac {1}{6} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 - 3*x^2 + x^2*Log[x])/(6*x^2),x]

[Out]

5/(3*x) - (2*x)/3 + (x*Log[x])/6

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fricas [A]  time = 0.74, size = 18, normalized size = 0.75 \begin {gather*} \frac {x^{2} \log \relax (x) - 4 \, x^{2} + 10}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(x^2*log(x)-3*x^2-10)/x^2,x, algorithm="fricas")

[Out]

1/6*(x^2*log(x) - 4*x^2 + 10)/x

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giac [A]  time = 0.13, size = 14, normalized size = 0.58 \begin {gather*} \frac {1}{6} \, x \log \relax (x) - \frac {2}{3} \, x + \frac {5}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(x^2*log(x)-3*x^2-10)/x^2,x, algorithm="giac")

[Out]

1/6*x*log(x) - 2/3*x + 5/3/x

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maple [A]  time = 0.01, size = 15, normalized size = 0.62

method	result	size
default	\(\frac {x \ln \relax (x )}{6}-\frac {2 x}{3}+\frac {5}{3 x}\)	\(15\)
norman	\(\frac {\frac {5}{3}-\frac {2 x^{2}}{3}+\frac {x^{2} \ln \relax (x )}{6}}{x}\)	\(19\)
risch	\(\frac {x \ln \relax (x )}{6}-\frac {2 x^{2}-5}{3 x}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/6*(x^2*ln(x)-3*x^2-10)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/6*x*ln(x)-2/3*x+5/3/x

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maxima [A]  time = 0.41, size = 14, normalized size = 0.58 \begin {gather*} \frac {1}{6} \, x \log \relax (x) - \frac {2}{3} \, x + \frac {5}{3 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(x^2*log(x)-3*x^2-10)/x^2,x, algorithm="maxima")

[Out]

1/6*x*log(x) - 2/3*x + 5/3/x

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mupad [B]  time = 2.60, size = 14, normalized size = 0.58 \begin {gather*} x\,\left (\frac {\ln \relax (x)}{6}-\frac {2}{3}\right )+\frac {5}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^2/2 - (x^2*log(x))/6 + 5/3)/x^2,x)

[Out]

x*(log(x)/6 - 2/3) + 5/(3*x)

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sympy [A]  time = 0.10, size = 15, normalized size = 0.62 \begin {gather*} \frac {x \log {\relax (x )}}{6} - \frac {2 x}{3} + \frac {5}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/6*(x**2*ln(x)-3*x**2-10)/x**2,x)

[Out]

x*log(x)/6 - 2*x/3 + 5/(3*x)

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