\(\int -\frac {15 \log (5)}{2 x^2} \, dx\) [7162]

Optimal result

Integrand size = 9, antiderivative size = 18 \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=5 \left (e^4-\frac {(-3+x) \log (5)}{2 x}\right ) \]

[Out]

-5/2*(-3+x)*ln(5)/x+5*exp(4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 30} \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=\frac {15 \log (5)}{2 x} \]

[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{align*} \text {integral}& = -\left (\frac {1}{2} (15 \log (5)) \int \frac {1}{x^2} \, dx\right ) \\ & = \frac {15 \log (5)}{2 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.50 \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=\frac {15 \log (5)}{2 x} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.44

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

[In]

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

[Out]

15/2*ln(5)/x

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.39 \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=\frac {15 \, \log \left (5\right )}{2 \, x} \]

[In]

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

[Out]

15/2*log(5)/x

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.39 \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=\frac {15 \log {\left (5 \right )}}{2 x} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.39 \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=\frac {15 \, \log \left (5\right )}{2 \, x} \]

[In]

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

[Out]

15/2*log(5)/x

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.39 \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=\frac {15 \, \log \left (5\right )}{2 \, x} \]

[In]

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

[Out]

15/2*log(5)/x

Mupad [B] (verification not implemented)

Time = 13.57 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.39 \[ \int -\frac {15 \log (5)}{2 x^2} \, dx=\frac {15\,\ln \left (5\right )}{2\,x} \]

[In]

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

[Out]

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