3.64.79 \(\int \frac {-48+2 x^2+(43-3 x^2) \log (3)}{x^2 \log (3)} \, dx\)

Optimal. Leaf size=32 \[ \frac {1}{2}+\frac {5}{x}-x+\frac {(6-x) (-4+x) \left (2-\frac {2}{\log (3)}\right )}{x} \]

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Rubi [A]  time = 0.02, antiderivative size = 27, normalized size of antiderivative = 0.84, number of steps used = 3, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {12, 14} \begin {gather*} \frac {x (2-3 \log (3))}{\log (3)}+\frac {48-43 \log (3)}{x \log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-48 + 2*x^2 + (43 - 3*x^2)*Log[3])/(x^2*Log[3]),x]

[Out]

(48 - 43*Log[3])/(x*Log[3]) + (x*(2 - 3*Log[3]))/Log[3]

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]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {-48+2 x^2+\left (43-3 x^2\right ) \log (3)}{x^2} \, dx}{\log (3)}\\ &=\frac {\int \left (2 \left (1-\frac {3 \log (3)}{2}\right )+\frac {-48+43 \log (3)}{x^2}\right ) \, dx}{\log (3)}\\ &=\frac {48-43 \log (3)}{x \log (3)}+\frac {x (2-3 \log (3))}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.75 \begin {gather*} \frac {\frac {48-43 \log (3)}{x}+x (2-3 \log (3))}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-48 + 2*x^2 + (43 - 3*x^2)*Log[3])/(x^2*Log[3]),x]

[Out]

((48 - 43*Log[3])/x + x*(2 - 3*Log[3]))/Log[3]

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fricas [A]  time = 0.55, size = 26, normalized size = 0.81 \begin {gather*} \frac {2 \, x^{2} - {\left (3 \, x^{2} + 43\right )} \log \relax (3) + 48}{x \log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+43)*log(3)+2*x^2-48)/x^2/log(3),x, algorithm="fricas")

[Out]

(2*x^2 - (3*x^2 + 43)*log(3) + 48)/(x*log(3))

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giac [A]  time = 0.21, size = 25, normalized size = 0.78 \begin {gather*} -\frac {3 \, x \log \relax (3) - 2 \, x + \frac {43 \, \log \relax (3) - 48}{x}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+43)*log(3)+2*x^2-48)/x^2/log(3),x, algorithm="giac")

[Out]

-(3*x*log(3) - 2*x + (43*log(3) - 48)/x)/log(3)

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maple [A]  time = 0.07, size = 26, normalized size = 0.81

method	result	size
default	\(\frac {-3 x \ln \relax (3)+2 x -\frac {43 \ln \relax (3)-48}{x}}{\ln \relax (3)}\)	\(26\)
risch	\(-3 x +\frac {2 x}{\ln \relax (3)}-\frac {43}{x}+\frac {48}{x \ln \relax (3)}\)	\(26\)
gosper	\(-\frac {3 x^{2} \ln \relax (3)-2 x^{2}+43 \ln \relax (3)-48}{\ln \relax (3) x}\)	\(28\)
norman	\(\frac {-\frac {43 \ln \relax (3)-48}{\ln \relax (3)}-\frac {\left (3 \ln \relax (3)-2\right ) x^{2}}{\ln \relax (3)}}{x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-3*x^2+43)*ln(3)+2*x^2-48)/x^2/ln(3),x,method=_RETURNVERBOSE)

[Out]

1/ln(3)*(-3*x*ln(3)+2*x-(43*ln(3)-48)/x)

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maxima [A]  time = 0.41, size = 25, normalized size = 0.78 \begin {gather*} -\frac {x {\left (3 \, \log \relax (3) - 2\right )} + \frac {43 \, \log \relax (3) - 48}{x}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x^2+43)*log(3)+2*x^2-48)/x^2/log(3),x, algorithm="maxima")

[Out]

-(x*(3*log(3) - 2) + (43*log(3) - 48)/x)/log(3)

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mupad [B]  time = 0.06, size = 27, normalized size = 0.84 \begin {gather*} -\frac {43\,\ln \relax (3)-48}{x\,\ln \relax (3)}-\frac {x\,\left (\ln \left (27\right )-2\right )}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(3)*(3*x^2 - 43) - 2*x^2 + 48)/(x^2*log(3)),x)

[Out]

- (43*log(3) - 48)/(x*log(3)) - (x*(log(27) - 2))/log(3)

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sympy [A]  time = 0.10, size = 19, normalized size = 0.59 \begin {gather*} \frac {x \left (2 - 3 \log {\relax (3 )}\right ) + \frac {48 - 43 \log {\relax (3 )}}{x}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-3*x**2+43)*ln(3)+2*x**2-48)/x**2/ln(3),x)

[Out]

(x*(2 - 3*log(3)) + (48 - 43*log(3))/x)/log(3)

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