\(\int \frac {4 \log (x)}{-x+x \log (2)} \, dx\) [5866]

Optimal result

Integrand size = 14, antiderivative size = 14 \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=1+\frac {2 \log ^2(x)}{-1+\log (2)} \]

[Out]

2*ln(x)^2/(ln(2)-1)+1

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6, 12, 2338} \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=-\frac {2 \log ^2(x)}{1-\log (2)} \]

[In]

Int[(4*Log[x])/(-x + x*Log[2]),x]

[Out]

(-2*Log[x]^2)/(1 - Log[2])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2338

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {4 \log (x)}{x (-1+\log (2))} \, dx \\ & = -\frac {4 \int \frac {\log (x)}{x} \, dx}{1-\log (2)} \\ & = -\frac {2 \log ^2(x)}{1-\log (2)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=-\frac {4 \log ^2(x)}{2-\log (4)} \]

[In]

Integrate[(4*Log[x])/(-x + x*Log[2]),x]

[Out]

(-4*Log[x]^2)/(2 - Log[4])

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
default	\(\frac {2 \ln \left (x \right )^{2}}{\ln \left (2\right )-1}\)	\(13\)
norman	\(\frac {2 \ln \left (x \right )^{2}}{\ln \left (2\right )-1}\)	\(13\)
risch	\(\frac {2 \ln \left (x \right )^{2}}{\ln \left (2\right )-1}\)	\(13\)
parts	\(\frac {2 \ln \left (x \right )^{2}}{\ln \left (2\right )-1}\)	\(13\)

[In]

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

[Out]

2*ln(x)^2/(ln(2)-1)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=\frac {2 \, \log \left (x\right )^{2}}{\log \left (2\right ) - 1} \]

[In]

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

[Out]

2*log(x)^2/(log(2) - 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71 \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=\frac {2 \log {\left (x \right )}^{2}}{-1 + \log {\left (2 \right )}} \]

[In]

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

[Out]

2*log(x)**2/(-1 + log(2))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=\frac {2 \, \log \left (x\right )^{2}}{\log \left (2\right ) - 1} \]

[In]

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

[Out]

2*log(x)^2/(log(2) - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=\frac {2 \, \log \left (x\right )^{2}}{\log \left (2\right ) - 1} \]

[In]

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

[Out]

2*log(x)^2/(log(2) - 1)

Mupad [B] (verification not implemented)

Time = 12.42 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {4 \log (x)}{-x+x \log (2)} \, dx=\frac {2\,{\ln \left (x\right )}^2}{\ln \left (2\right )-1} \]

[In]

int(-(4*log(x))/(x - x*log(2)),x)

[Out]

(2*log(x)^2)/(log(2) - 1)