\(\int \frac {2 \log (\frac {1}{120} (x+240 e^{e^{10}} x))}{x} \, dx\) [4492]

Optimal result

Integrand size = 20, antiderivative size = 17 \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx=\log ^2\left (\frac {x}{120}+2 e^{e^{10}} x\right ) \]

[Out]

ln(2*x*exp(exp(10))+1/120*x)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {12, 2468, 2338} \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx=\log ^2\left (\frac {1}{120} \left (1+240 e^{e^{10}}\right ) x\right ) \]

[In]

Int[(2*Log[(x + 240*E^E^10*x)/120])/x,x]

[Out]

Log[((1 + 240*E^E^10)*x)/120]^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]

Rule 2468

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^q*(a + b*Log[c*
ExpandToSum[v, x]^n])^p, x] /; FreeQ[{a, b, c, n, p, q}, x] && BinomialQ[u, x] && LinearQ[v, x] &&  !(Binomial
MatchQ[u, x] && LinearMatchQ[v, x])

Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {\log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx \\ & = 2 \int \frac {\log \left (\frac {1}{120} \left (1+240 e^{e^{10}}\right ) x\right )}{x} \, dx \\ & = \log ^2\left (\frac {1}{120} \left (1+240 e^{e^{10}}\right ) x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx=\log ^2\left (\left (\frac {1}{120}+2 e^{e^{10}}\right ) x\right ) \]

[In]

Integrate[(2*Log[(x + 240*E^E^10*x)/120])/x,x]

[Out]

Log[(1/120 + 2*E^E^10)*x]^2

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\ln \left (2 x \,{\mathrm e}^{{\mathrm e}^{10}}+\frac {x}{120}\right )^{2}\)	\(14\)
default	\(\ln \left (2 x \,{\mathrm e}^{{\mathrm e}^{10}}+\frac {x}{120}\right )^{2}\)	\(14\)
norman	\(\ln \left (2 x \,{\mathrm e}^{{\mathrm e}^{10}}+\frac {x}{120}\right )^{2}\)	\(14\)
risch	\(\ln \left (2 x \,{\mathrm e}^{{\mathrm e}^{10}}+\frac {x}{120}\right )^{2}\)	\(14\)
parts	\(2 \ln \left (2 x \,{\mathrm e}^{{\mathrm e}^{10}}+\frac {x}{120}\right ) \ln \left (x \right )-\frac {\left (480 \,{\mathrm e}^{{\mathrm e}^{10}}+2\right ) \ln \left (x \right )^{2}}{2 \left (240 \,{\mathrm e}^{{\mathrm e}^{10}}+1\right )}\)	\(39\)

[In]

int(2*ln(2*x*exp(exp(10))+1/120*x)/x,x,method=_RETURNVERBOSE)

[Out]

ln(2*x*exp(exp(10))+1/120*x)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx=\log \left (2 \, x e^{\left (e^{10}\right )} + \frac {1}{120} \, x\right )^{2} \]

[In]

integrate(2*log(2*x*exp(exp(10))+1/120*x)/x,x, algorithm="fricas")

[Out]

log(2*x*e^(e^10) + 1/120*x)^2

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx=\log {\left (\frac {x}{120} + 2 x e^{e^{10}} \right )}^{2} \]

[In]

integrate(2*ln(2*x*exp(exp(10))+1/120*x)/x,x)

[Out]

log(x/120 + 2*x*exp(exp(10)))**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 28 vs. \(2 (13) = 26\).

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.65 \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx=-2 \, {\left (\log \left (5\right ) + \log \left (3\right ) + 3 \, \log \left (2\right ) - \log \left (240 \, e^{\left (e^{10}\right )} + 1\right )\right )} \log \left (x\right ) + \log \left (x\right )^{2} \]

[In]

integrate(2*log(2*x*exp(exp(10))+1/120*x)/x,x, algorithm="maxima")

[Out]

-2*(log(5) + log(3) + 3*log(2) - log(240*e^(e^10) + 1))*log(x) + log(x)^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx=\log \left (2 \, x e^{\left (e^{10}\right )} + \frac {1}{120} \, x\right )^{2} \]

[In]

integrate(2*log(2*x*exp(exp(10))+1/120*x)/x,x, algorithm="giac")

[Out]

log(2*x*e^(e^10) + 1/120*x)^2

Mupad [B] (verification not implemented)

Time = 15.31 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.76 \[ \int \frac {2 \log \left (\frac {1}{120} \left (x+240 e^{e^{10}} x\right )\right )}{x} \, dx={\ln \left (\frac {x}{120}+2\,x\,{\mathrm {e}}^{{\mathrm {e}}^{10}}\right )}^2 \]

[In]

int((2*log(x/120 + 2*x*exp(exp(10))))/x,x)

[Out]

log(x/120 + 2*x*exp(exp(10)))^2