3.4.12 \(\int \frac {5-\log (5)}{10 x^2} \, dx\)

Optimal. Leaf size=14 \[ \frac {-5+2 x+\log (5)}{10 x} \]

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Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 0.93, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {12, 30} \begin {gather*} -\frac {5-\log (5)}{10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - Log[5])/(10*x^2),x]

[Out]

-1/10*(5 - Log[5])/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} (5-\log (5)) \int \frac {1}{x^2} \, dx\\ &=-\frac {5-\log (5)}{10 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 11, normalized size = 0.79 \begin {gather*} \frac {-5+\log (5)}{10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - Log[5])/(10*x^2),x]

[Out]

(-5 + Log[5])/(10*x)

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fricas [A]  time = 0.57, size = 9, normalized size = 0.64 \begin {gather*} \frac {\log \relax (5) - 5}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-log(5)+5)/x^2,x, algorithm="fricas")

[Out]

1/10*(log(5) - 5)/x

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giac [A]  time = 0.32, size = 9, normalized size = 0.64 \begin {gather*} \frac {\log \relax (5) - 5}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-log(5)+5)/x^2,x, algorithm="giac")

[Out]

1/10*(log(5) - 5)/x

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maple [A]  time = 0.02, size = 10, normalized size = 0.71

method	result	size
gosper	\(\frac {\ln \relax (5)-5}{10 x}\)	\(10\)
norman	\(\frac {\frac {\ln \relax (5)}{10}-\frac {1}{2}}{x}\)	\(11\)
default	\(-\frac {-\frac {\ln \relax (5)}{10}+\frac {1}{2}}{x}\)	\(12\)
risch	\(\frac {\ln \relax (5)}{10 x}-\frac {1}{2 x}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*(-ln(5)+5)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/10*(ln(5)-5)/x

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maxima [A]  time = 0.77, size = 9, normalized size = 0.64 \begin {gather*} \frac {\log \relax (5) - 5}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-log(5)+5)/x^2,x, algorithm="maxima")

[Out]

1/10*(log(5) - 5)/x

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mupad [B]  time = 0.03, size = 10, normalized size = 0.71 \begin {gather*} \frac {\frac {\ln \relax (5)}{10}-\frac {1}{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5)/10 - 1/2)/x^2,x)

[Out]

(log(5)/10 - 1/2)/x

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sympy [A]  time = 0.06, size = 10, normalized size = 0.71 \begin {gather*} - \frac {\frac {1}{2} - \frac {\log {\relax (5 )}}{10}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(-ln(5)+5)/x**2,x)

[Out]

-(1/2 - log(5)/10)/x

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