\(\int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx\) [3260]

Optimal result

Integrand size = 36, antiderivative size = 22 \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=\frac {2+4 x+(1-\log (x))^2}{\frac {1}{5}+\log (x)} \]

[Out]

((1-ln(x))^2+2+4*x)/(1/5+ln(x))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {6873, 6874, 2395, 2334, 2336, 2209, 2339, 30} \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=\frac {20 x}{5 \log (x)+1}+\log (x)+\frac {86}{5 (5 \log (x)+1)} \]

[In]

Int[(-85 - 80*x + (10 + 100*x)*Log[x] + 25*Log[x]^2)/(x + 10*x*Log[x] + 25*x*Log[x]^2),x]

[Out]

Log[x] + 86/(5*(1 + 5*Log[x])) + (20*x)/(1 + 5*Log[x])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2336

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

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 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

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}& = \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x (1+5 \log (x))^2} \, dx \\ & = \int \left (\frac {1}{x}-\frac {2 (43+50 x)}{x (1+5 \log (x))^2}+\frac {20}{1+5 \log (x)}\right ) \, dx \\ & = \log (x)-2 \int \frac {43+50 x}{x (1+5 \log (x))^2} \, dx+20 \int \frac {1}{1+5 \log (x)} \, dx \\ & = \log (x)-2 \int \left (\frac {50}{(1+5 \log (x))^2}+\frac {43}{x (1+5 \log (x))^2}\right ) \, dx+20 \text {Subst}\left (\int \frac {e^x}{1+5 x} \, dx,x,\log (x)\right ) \\ & = \frac {4 \text {Ei}\left (\frac {1}{5} (1+5 \log (x))\right )}{\sqrt [5]{e}}+\log (x)-86 \int \frac {1}{x (1+5 \log (x))^2} \, dx-100 \int \frac {1}{(1+5 \log (x))^2} \, dx \\ & = \frac {4 \text {Ei}\left (\frac {1}{5} (1+5 \log (x))\right )}{\sqrt [5]{e}}+\log (x)+\frac {20 x}{1+5 \log (x)}-\frac {86}{5} \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,1+5 \log (x)\right )-20 \int \frac {1}{1+5 \log (x)} \, dx \\ & = \frac {4 \text {Ei}\left (\frac {1}{5} (1+5 \log (x))\right )}{\sqrt [5]{e}}+\log (x)+\frac {86}{5 (1+5 \log (x))}+\frac {20 x}{1+5 \log (x)}-20 \text {Subst}\left (\int \frac {e^x}{1+5 x} \, dx,x,\log (x)\right ) \\ & = \log (x)+\frac {86}{5 (1+5 \log (x))}+\frac {20 x}{1+5 \log (x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=5 \left (\frac {\log (x)}{5}+\frac {86+100 x}{25+125 \log (x)}\right ) \]

[In]

Integrate[(-85 - 80*x + (10 + 100*x)*Log[x] + 25*Log[x]^2)/(x + 10*x*Log[x] + 25*x*Log[x]^2),x]

[Out]

5*(Log[x]/5 + (86 + 100*x)/(25 + 125*Log[x]))

Maple [A] (verified)

Time = 2.44 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
risch	\(\ln \left (x \right )+\frac {20 x +\frac {86}{5}}{1+5 \ln \left (x \right )}\)	\(19\)
norman	\(\frac {5 \ln \left (x \right )^{2}+20 x +17}{1+5 \ln \left (x \right )}\)	\(21\)
parallelrisch	\(\frac {85+25 \ln \left (x \right )^{2}+100 x}{5+25 \ln \left (x \right )}\)	\(22\)
default	\(\frac {86}{5 \left (1+5 \ln \left (x \right )\right )}+\ln \left (x \right )+\frac {4 x}{\frac {1}{5}+\ln \left (x \right )}\)	\(23\)

[In]

int((25*ln(x)^2+(100*x+10)*ln(x)-80*x-85)/(25*x*ln(x)^2+10*x*ln(x)+x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+2/5*(50*x+43)/(1+5*ln(x))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=\frac {25 \, \log \left (x\right )^{2} + 100 \, x + 5 \, \log \left (x\right ) + 86}{5 \, {\left (5 \, \log \left (x\right ) + 1\right )}} \]

[In]

integrate((25*log(x)^2+(100*x+10)*log(x)-80*x-85)/(25*x*log(x)^2+10*x*log(x)+x),x, algorithm="fricas")

[Out]

1/5*(25*log(x)^2 + 100*x + 5*log(x) + 86)/(5*log(x) + 1)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=\frac {100 x + 86}{25 \log {\left (x \right )} + 5} + \log {\left (x \right )} \]

[In]

integrate((25*ln(x)**2+(100*x+10)*ln(x)-80*x-85)/(25*x*ln(x)**2+10*x*ln(x)+x),x)

[Out]

(100*x + 86)/(25*log(x) + 5) + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=\frac {100 \, x + 1}{5 \, {\left (5 \, \log \left (x\right ) + 1\right )}} + \frac {17}{5 \, \log \left (x\right ) + 1} + \log \left (x\right ) \]

[In]

integrate((25*log(x)^2+(100*x+10)*log(x)-80*x-85)/(25*x*log(x)^2+10*x*log(x)+x),x, algorithm="maxima")

[Out]

1/5*(100*x + 1)/(5*log(x) + 1) + 17/(5*log(x) + 1) + log(x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=\frac {2 \, {\left (50 \, x + 43\right )}}{5 \, {\left (5 \, \log \left (x\right ) + 1\right )}} + \log \left (x\right ) \]

[In]

integrate((25*log(x)^2+(100*x+10)*log(x)-80*x-85)/(25*x*log(x)^2+10*x*log(x)+x),x, algorithm="giac")

[Out]

2/5*(50*x + 43)/(5*log(x) + 1) + log(x)

Mupad [B] (verification not implemented)

Time = 10.98 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {-85-80 x+(10+100 x) \log (x)+25 \log ^2(x)}{x+10 x \log (x)+25 x \log ^2(x)} \, dx=\ln \left (x\right )+\frac {20\,x+\frac {86}{5}}{5\,\ln \left (x\right )+1} \]

[In]

int(-(80*x - 25*log(x)^2 - log(x)*(100*x + 10) + 85)/(x + 25*x*log(x)^2 + 10*x*log(x)),x)

[Out]

log(x) + (20*x + 86/5)/(5*log(x) + 1)