\(\int \frac {50 \log (6) \log (2 x)}{x} \, dx\) [1606]

Optimal result

Integrand size = 11, antiderivative size = 10 \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25 \log (6) \log ^2(2 x) \]

[Out]

25*ln(2*x)^2*ln(6)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {12, 2338} \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25 \log (6) \log ^2(2 x) \]

[In]

Int[(50*Log[6]*Log[2*x])/x,x]

[Out]

25*Log[6]*Log[2*x]^2

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}& = (50 \log (6)) \int \frac {\log (2 x)}{x} \, dx \\ & = 25 \log (6) \log ^2(2 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25 \log (6) \log ^2(2 x) \]

[In]

Integrate[(50*Log[6]*Log[2*x])/x,x]

[Out]

25*Log[6]*Log[2*x]^2

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 11, normalized size of antiderivative = 1.10

method	result	size
derivativedivides	\(25 \ln \left (2 x \right )^{2} \ln \left (6\right )\)	\(11\)
default	\(25 \ln \left (2 x \right )^{2} \ln \left (6\right )\)	\(11\)
norman	\(25 \ln \left (2 x \right )^{2} \ln \left (6\right )\)	\(11\)
parts	\(50 \ln \left (2 x \right ) \ln \left (6\right ) \ln \left (x \right )-25 \ln \left (6\right ) \ln \left (x \right )^{2}\)	\(20\)
risch	\(25 \ln \left (2 x \right )^{2} \ln \left (2\right )+25 \ln \left (2 x \right )^{2} \ln \left (3\right )\)	\(22\)

[In]

int(50*ln(6)*ln(2*x)/x,x,method=_RETURNVERBOSE)

[Out]

25*ln(2*x)^2*ln(6)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25 \, \log \left (6\right ) \log \left (2 \, x\right )^{2} \]

[In]

integrate(50*log(6)*log(2*x)/x,x, algorithm="fricas")

[Out]

25*log(6)*log(2*x)^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25 \log {\left (6 \right )} \log {\left (2 x \right )}^{2} \]

[In]

integrate(50*ln(6)*ln(2*x)/x,x)

[Out]

25*log(6)*log(2*x)**2

Maxima [A] (verification not implemented)

none

Time = 0.17 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25 \, \log \left (6\right ) \log \left (2 \, x\right )^{2} \]

[In]

integrate(50*log(6)*log(2*x)/x,x, algorithm="maxima")

[Out]

25*log(6)*log(2*x)^2

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25 \, \log \left (6\right ) \log \left (2 \, x\right )^{2} \]

[In]

integrate(50*log(6)*log(2*x)/x,x, algorithm="giac")

[Out]

25*log(6)*log(2*x)^2

Mupad [B] (verification not implemented)

Time = 8.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.00 \[ \int \frac {50 \log (6) \log (2 x)}{x} \, dx=25\,{\ln \left (2\,x\right )}^2\,\ln \left (6\right ) \]

[In]

int((50*log(2*x)*log(6))/x,x)

[Out]

25*log(2*x)^2*log(6)