3.103.24 \(\int \frac {1-\log (10 x \log (3))}{25 x^2} \, dx\)

Optimal. Leaf size=13 \[ \frac {\log (10 x \log (3))}{25 x} \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2303} \begin {gather*} \frac {\log (10 x \log (3))}{25 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - Log[10*x*Log[3]])/(25*x^2),x]

[Out]

Log[10*x*Log[3]]/(25*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 2303

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(b*(d*x)^(m + 1)*Log[c*x^n])/(
d*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1] && EqQ[a*(m + 1) - b*n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{25} \int \frac {1-\log (10 x \log (3))}{x^2} \, dx\\ &=\frac {\log (10 x \log (3))}{25 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \begin {gather*} \frac {\log (10 x \log (3))}{25 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - Log[10*x*Log[3]])/(25*x^2),x]

[Out]

Log[10*x*Log[3]]/(25*x)

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fricas [A]  time = 0.69, size = 11, normalized size = 0.85 \begin {gather*} \frac {\log \left (10 \, x \log \relax (3)\right )}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-log(10*x*log(3))+1)/x^2,x, algorithm="fricas")

[Out]

1/25*log(10*x*log(3))/x

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giac [A]  time = 0.14, size = 11, normalized size = 0.85 \begin {gather*} \frac {\log \left (10 \, x \log \relax (3)\right )}{25 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-log(10*x*log(3))+1)/x^2,x, algorithm="giac")

[Out]

1/25*log(10*x*log(3))/x

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maple [A]  time = 0.03, size = 12, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\ln \left (10 x \ln \relax (3)\right )}{25 x}\)	\(12\)
default	\(\frac {\ln \left (10 x \ln \relax (3)\right )}{25 x}\)	\(12\)
norman	\(\frac {\ln \left (10 x \ln \relax (3)\right )}{25 x}\)	\(12\)
risch	\(\frac {\ln \left (10 x \ln \relax (3)\right )}{25 x}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/25*(-ln(10*x*ln(3))+1)/x^2,x,method=_RETURNVERBOSE)

[Out]

1/25*ln(10*x*ln(3))/x

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maxima [B]  time = 0.46, size = 30, normalized size = 2.31 \begin {gather*} \frac {1}{25} \, {\left (\frac {\log \left (10 \, x \log \relax (3)\right ) + 1}{x \log \relax (3)} - \frac {1}{x \log \relax (3)}\right )} \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-log(10*x*log(3))+1)/x^2,x, algorithm="maxima")

[Out]

1/25*((log(10*x*log(3)) + 1)/(x*log(3)) - 1/(x*log(3)))*log(3)

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mupad [B]  time = 6.29, size = 13, normalized size = 1.00 \begin {gather*} \frac {\ln \left (10\right )+\ln \left (\ln \relax (3)\right )+\ln \relax (x)}{25\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(10*x*log(3))/25 - 1/25)/x^2,x)

[Out]

(log(10) + log(log(3)) + log(x))/(25*x)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.77 \begin {gather*} \frac {\log {\left (10 x \log {\relax (3 )} \right )}}{25 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/25*(-ln(10*x*ln(3))+1)/x**2,x)

[Out]

log(10*x*log(3))/(25*x)

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