3.72.62 \(\int -\frac {15 \log (5)}{2 x^2} \, dx\)

Optimal. Leaf size=18 \[ 5 \left (e^4-\frac {(-3+x) \log (5)}{2 x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 9, normalized size of antiderivative = 0.50, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 30} \begin {gather*} \frac {15 \log (5)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*Log[5])/(2*x^2),x]

[Out]

(15*Log[5])/(2*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} &=-\left (\frac {1}{2} (15 \log (5)) \int \frac {1}{x^2} \, dx\right )\\ &=\frac {15 \log (5)}{2 x}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 9, normalized size = 0.50 \begin {gather*} \frac {15 \log (5)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*Log[5])/(2*x^2),x]

[Out]

(15*Log[5])/(2*x)

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 7, normalized size = 0.39 \begin {gather*} \frac {15 \, \log \relax (5)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15/2*log(5)/x^2,x, algorithm="fricas")

[Out]

15/2*log(5)/x

________________________________________________________________________________________

giac [A]  time = 0.22, size = 7, normalized size = 0.39 \begin {gather*} \frac {15 \, \log \relax (5)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15/2*log(5)/x^2,x, algorithm="giac")

[Out]

15/2*log(5)/x

________________________________________________________________________________________

maple [A]  time = 0.02, size = 8, normalized size = 0.44

method	result	size
gosper	\(\frac {15 \ln \relax (5)}{2 x}\)	\(8\)
default	\(\frac {15 \ln \relax (5)}{2 x}\)	\(8\)
norman	\(\frac {15 \ln \relax (5)}{2 x}\)	\(8\)
risch	\(\frac {15 \ln \relax (5)}{2 x}\)	\(8\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-15/2*ln(5)/x^2,x,method=_RETURNVERBOSE)

[Out]

15/2*ln(5)/x

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 7, normalized size = 0.39 \begin {gather*} \frac {15 \, \log \relax (5)}{2 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15/2*log(5)/x^2,x, algorithm="maxima")

[Out]

15/2*log(5)/x

________________________________________________________________________________________

mupad [B]  time = 0.02, size = 7, normalized size = 0.39 \begin {gather*} \frac {15\,\ln \relax (5)}{2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*log(5))/(2*x^2),x)

[Out]

(15*log(5))/(2*x)

________________________________________________________________________________________

sympy [A]  time = 0.06, size = 7, normalized size = 0.39 \begin {gather*} \frac {15 \log {\relax (5 )}}{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(-15/2*ln(5)/x**2,x)

[Out]

15*log(5)/(2*x)

________________________________________________________________________________________