\(\int \frac {115 \log (2)}{4 x^2} \, dx\) [6125]

Optimal result

Integrand size = 9, antiderivative size = 19 \[ \int \frac {115 \log (2)}{4 x^2} \, dx=\log (2) \left (\frac {5}{4} \left (16-\frac {23}{x}\right )-\log (4)\right ) \]

[Out]

ln(2)*(20-115/4/x-2*ln(2))

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.47, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 30} \[ \int \frac {115 \log (2)}{4 x^2} \, dx=-\frac {115 \log (2)}{4 x} \]

[In]

Int[(115*Log[2])/(4*x^2),x]

[Out]

(-115*Log[2])/(4*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}& = \frac {1}{4} (115 \log (2)) \int \frac {1}{x^2} \, dx \\ & = -\frac {115 \log (2)}{4 x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.47 \[ \int \frac {115 \log (2)}{4 x^2} \, dx=-\frac {115 \log (2)}{4 x} \]

[In]

Integrate[(115*Log[2])/(4*x^2),x]

[Out]

(-115*Log[2])/(4*x)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42

method	result	size
gosper	\(-\frac {115 \ln \left (2\right )}{4 x}\)	\(8\)
default	\(-\frac {115 \ln \left (2\right )}{4 x}\)	\(8\)
norman	\(-\frac {115 \ln \left (2\right )}{4 x}\)	\(8\)
risch	\(-\frac {115 \ln \left (2\right )}{4 x}\)	\(8\)
parallelrisch	\(-\frac {115 \ln \left (2\right )}{4 x}\)	\(8\)

[In]

int(115/4*ln(2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-115/4*ln(2)/x

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.37 \[ \int \frac {115 \log (2)}{4 x^2} \, dx=-\frac {115 \, \log \left (2\right )}{4 \, x} \]

[In]

integrate(115/4*log(2)/x^2,x, algorithm="fricas")

[Out]

-115/4*log(2)/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.42 \[ \int \frac {115 \log (2)}{4 x^2} \, dx=- \frac {115 \log {\left (2 \right )}}{4 x} \]

[In]

integrate(115/4*ln(2)/x**2,x)

[Out]

-115*log(2)/(4*x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.37 \[ \int \frac {115 \log (2)}{4 x^2} \, dx=-\frac {115 \, \log \left (2\right )}{4 \, x} \]

[In]

integrate(115/4*log(2)/x^2,x, algorithm="maxima")

[Out]

-115/4*log(2)/x

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.37 \[ \int \frac {115 \log (2)}{4 x^2} \, dx=-\frac {115 \, \log \left (2\right )}{4 \, x} \]

[In]

integrate(115/4*log(2)/x^2,x, algorithm="giac")

[Out]

-115/4*log(2)/x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 7, normalized size of antiderivative = 0.37 \[ \int \frac {115 \log (2)}{4 x^2} \, dx=-\frac {115\,\ln \left (2\right )}{4\,x} \]

[In]

int((115*log(2))/(4*x^2),x)

[Out]

-(115*log(2))/(4*x)