\(\int \frac {5-5 \log (\frac {1}{2} (5 x-e x))}{x \log (\frac {1}{2} (5 x-e x))} \, dx\) [1790]

Optimal result

Integrand size = 36, antiderivative size = 21 \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=5+5 \log \left (\frac {16 \log \left (\frac {1}{2} (5-e) x\right )}{x}\right ) \]

[Out]

5+5*ln(16*ln(1/2*(5-exp(1))*x)/x)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2494, 2412, 45} \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=5 \log \left (\log \left (\frac {1}{2} (5-e) x\right )\right )-5 \log (x) \]

[In]

Int[(5 - 5*Log[(5*x - E*x)/2])/(x*Log[(5*x - E*x)/2]),x]

[Out]

-5*Log[x] + 5*Log[Log[((5 - E)*x)/2]]

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 2412

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

Rule 2494

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p
, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] &&  !LinearMatchQ[v, x] &&  !(EqQ[n, 1] && MatchQ[c*v, (e_.
)*((f_) + (g_.)*x) /; FreeQ[{e, f, g}, x]])

Rubi steps \begin{align*} \text {integral}& = \int \frac {5-5 \log \left (\frac {1}{2} (5-e) x\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx \\ & = \int \frac {5-5 \log \left (\frac {1}{2} (5-e) x\right )}{x \log \left (\frac {1}{2} (5-e) x\right )} \, dx \\ & = \text {Subst}\left (\int \frac {5-5 x}{x} \, dx,x,\log \left (\frac {1}{2} (5-e) x\right )\right ) \\ & = \text {Subst}\left (\int \left (-5+\frac {5}{x}\right ) \, dx,x,\log \left (\frac {1}{2} (5-e) x\right )\right ) \\ & = -5 \log (x)+5 \log \left (\log \left (\frac {1}{2} (5-e) x\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=-5 \log (x)+5 \log \left (\log \left (-\frac {1}{2} (-5+e) x\right )\right ) \]

[In]

Integrate[(5 - 5*Log[(5*x - E*x)/2])/(x*Log[(5*x - E*x)/2]),x]

[Out]

-5*Log[x] + 5*Log[Log[-1/2*((-5 + E)*x)]]

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90

method	result	size
risch	\(-5 \ln \left (x \right )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\)	\(19\)
parts	\(-5 \ln \left (x \right )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\)	\(19\)
derivativedivides	\(-5 \ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\)	\(27\)
default	\(-5 \ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\)	\(27\)
norman	\(-5 \ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\)	\(27\)
parallelrisch	\(-5 \ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )+5 \ln \left (\ln \left (-\frac {x \,{\mathrm e}}{2}+\frac {5 x}{2}\right )\right )\)	\(27\)

[In]

int((-5*ln(-1/2*x*exp(1)+5/2*x)+5)/x/ln(-1/2*x*exp(1)+5/2*x),x,method=_RETURNVERBOSE)

[Out]

-5*ln(x)+5*ln(ln(-1/2*x*exp(1)+5/2*x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=-5 \, \log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right ) + 5 \, \log \left (\log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right )\right ) \]

[In]

integrate((-5*log(-1/2*x*exp(1)+5/2*x)+5)/x/log(-1/2*x*exp(1)+5/2*x),x, algorithm="fricas")

[Out]

-5*log(-1/2*x*e + 5/2*x) + 5*log(log(-1/2*x*e + 5/2*x))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=- 5 \log {\left (x \right )} + 5 \log {\left (\log {\left (- \frac {e x}{2} + \frac {5 x}{2} \right )} \right )} \]

[In]

integrate((-5*ln(-1/2*x*exp(1)+5/2*x)+5)/x/ln(-1/2*x*exp(1)+5/2*x),x)

[Out]

-5*log(x) + 5*log(log(-E*x/2 + 5*x/2))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=-5 \, \log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right ) + 5 \, \log \left (\log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right )\right ) \]

[In]

integrate((-5*log(-1/2*x*exp(1)+5/2*x)+5)/x/log(-1/2*x*exp(1)+5/2*x),x, algorithm="maxima")

[Out]

-5*log(-1/2*x*e + 5/2*x) + 5*log(log(-1/2*x*e + 5/2*x))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=-5 \, \log \left (x\right ) + 5 \, \log \left (\log \left (-\frac {1}{2} \, x e + \frac {5}{2} \, x\right )\right ) \]

[In]

integrate((-5*log(-1/2*x*exp(1)+5/2*x)+5)/x/log(-1/2*x*exp(1)+5/2*x),x, algorithm="giac")

[Out]

-5*log(x) + 5*log(log(-1/2*x*e + 5/2*x))

Mupad [B] (verification not implemented)

Time = 8.74 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {5-5 \log \left (\frac {1}{2} (5 x-e x)\right )}{x \log \left (\frac {1}{2} (5 x-e x)\right )} \, dx=5\,\ln \left (\ln \left (\frac {5\,x}{2}-\frac {x\,\mathrm {e}}{2}\right )\right )-5\,\ln \left (x\right ) \]

[In]

int(-(5*log((5*x)/2 - (x*exp(1))/2) - 5)/(x*log((5*x)/2 - (x*exp(1))/2)),x)

[Out]

5*log(log((5*x)/2 - (x*exp(1))/2)) - 5*log(x)