\(\int \frac {-e^2+(-2 e^2-e^{50} x) \log (x)}{e^2 x \log (x)} \, dx\) [9548]

Optimal result

Integrand size = 32, antiderivative size = 20 \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=\log \left (\frac {3 e^{-e^{48} x}}{2 x^2 \log (x)}\right ) \]

[Out]

ln(3/2/ln(x)/x^2/exp(exp(25)^2*x/exp(2)))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {12, 6873, 6874, 45, 2339, 29} \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=-e^{48} x-2 \log (x)-\log (\log (x)) \]

[In]

Int[(-E^2 + (-2*E^2 - E^50*x)*Log[x])/(E^2*x*Log[x]),x]

[Out]

-(E^48*x) - 2*Log[x] - Log[Log[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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{x \log (x)} \, dx}{e^2} \\ & = \frac {\int \frac {e^2 \left (-1-2 \log (x)-e^{48} x \log (x)\right )}{x \log (x)} \, dx}{e^2} \\ & = \int \frac {-1-2 \log (x)-e^{48} x \log (x)}{x \log (x)} \, dx \\ & = \int \left (\frac {-2-e^{48} x}{x}-\frac {1}{x \log (x)}\right ) \, dx \\ & = \int \frac {-2-e^{48} x}{x} \, dx-\int \frac {1}{x \log (x)} \, dx \\ & = \int \left (-e^{48}-\frac {2}{x}\right ) \, dx-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -e^{48} x-2 \log (x)-\log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80 \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=-e^{48} x-2 \log (x)-\log (\log (x)) \]

[In]

Integrate[(-E^2 + (-2*E^2 - E^50*x)*Log[x])/(E^2*x*Log[x]),x]

[Out]

-(E^48*x) - 2*Log[x] - Log[Log[x]]

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80

method	result	size
risch	\(-{\mathrm e}^{48} x -2 \ln \left (x \right )-\ln \left (\ln \left (x \right )\right )\)	\(16\)
parts	\(-\ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )-{\mathrm e}^{-2} x \,{\mathrm e}^{50}\)	\(20\)
norman	\(-\ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )-{\mathrm e}^{-2} x \,{\mathrm e}^{50}\)	\(22\)
default	\({\mathrm e}^{-2} \left (-x \,{\mathrm e}^{50}-2 \,{\mathrm e}^{2} \ln \left (x \right )-{\mathrm e}^{2} \ln \left (\ln \left (x \right )\right )\right )\)	\(27\)
parallelrisch	\({\mathrm e}^{-2} \left (-x \,{\mathrm e}^{50}-2 \,{\mathrm e}^{2} \ln \left (x \right )-{\mathrm e}^{2} \ln \left (\ln \left (x \right )\right )\right )\)	\(27\)

[In]

int(((-x*exp(25)^2-2*exp(2))*ln(x)-exp(2))/x/exp(2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

-exp(48)*x-2*ln(x)-ln(ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=-x e^{48} - 2 \, \log \left (x\right ) - \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((-x*exp(25)^2-2*exp(2))*log(x)-exp(2))/x/exp(2)/log(x),x, algorithm="fricas")

[Out]

-x*e^48 - 2*log(x) - log(log(x))

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=- x e^{48} - 2 \log {\left (x \right )} - \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((-x*exp(25)**2-2*exp(2))*ln(x)-exp(2))/x/exp(2)/ln(x),x)

[Out]

-x*exp(48) - 2*log(x) - log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=-{\left (x e^{50} + 2 \, e^{2} \log \left (x\right ) + e^{2} \log \left (\log \left (x\right )\right )\right )} e^{\left (-2\right )} \]

[In]

integrate(((-x*exp(25)^2-2*exp(2))*log(x)-exp(2))/x/exp(2)/log(x),x, algorithm="maxima")

[Out]

-(x*e^50 + 2*e^2*log(x) + e^2*log(log(x)))*e^(-2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=-{\left (x e^{50} + 2 \, e^{2} \log \left (x\right ) + e^{2} \log \left (\log \left (x\right )\right )\right )} e^{\left (-2\right )} \]

[In]

integrate(((-x*exp(25)^2-2*exp(2))*log(x)-exp(2))/x/exp(2)/log(x),x, algorithm="giac")

[Out]

-(x*e^50 + 2*e^2*log(x) + e^2*log(log(x)))*e^(-2)

Mupad [B] (verification not implemented)

Time = 14.71 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {-e^2+\left (-2 e^2-e^{50} x\right ) \log (x)}{e^2 x \log (x)} \, dx=-\ln \left (\ln \left (x\right )\right )-2\,\ln \left (x\right )-x\,{\mathrm {e}}^{48} \]

[In]

int(-(exp(-2)*(exp(2) + log(x)*(2*exp(2) + x*exp(50))))/(x*log(x)),x)

[Out]

- log(log(x)) - 2*log(x) - x*exp(48)