\(\int \frac {2 x+(5-6 x+x^2) \log (-5+x)}{(-20 x+4 x^2) \log (-5+x)} \, dx\) [929]

Optimal result

Integrand size = 35, antiderivative size = 24 \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {1}{4} \left (86+x-\log ^2(5)-\log (x)+\log \left (\log ^2(-5+x)\right )\right ) \]

[Out]

1/4*x+43/2+1/4*ln(ln(-5+x)^2)-1/4*ln(5)^2-1/4*ln(x)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {1607, 6874, 45, 2437, 2339, 29} \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {x}{4}-\frac {\log (x)}{4}+\frac {1}{2} \log (\log (x-5)) \]

[In]

Int[(2*x + (5 - 6*x + x^2)*Log[-5 + x])/((-20*x + 4*x^2)*Log[-5 + x]),x]

[Out]

x/4 - Log[x]/4 + Log[Log[-5 + x]]/2

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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2437

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[(f*(x/d))^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{x (-20+4 x) \log (-5+x)} \, dx \\ & = \int \left (\frac {-1+x}{4 x}+\frac {1}{2 (-5+x) \log (-5+x)}\right ) \, dx \\ & = \frac {1}{4} \int \frac {-1+x}{x} \, dx+\frac {1}{2} \int \frac {1}{(-5+x) \log (-5+x)} \, dx \\ & = \frac {1}{4} \int \left (1-\frac {1}{x}\right ) \, dx+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-5+x\right ) \\ & = \frac {x}{4}-\frac {\log (x)}{4}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-5+x)\right ) \\ & = \frac {x}{4}-\frac {\log (x)}{4}+\frac {1}{2} \log (\log (-5+x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {1}{4} (x-\log (x)+2 \log (\log (-5+x))) \]

[In]

Integrate[(2*x + (5 - 6*x + x^2)*Log[-5 + x])/((-20*x + 4*x^2)*Log[-5 + x]),x]

[Out]

(x - Log[x] + 2*Log[Log[-5 + x]])/4

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67

method	result	size
norman	\(\frac {x}{4}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (\ln \left (-5+x \right )\right )}{2}\)	\(16\)
risch	\(\frac {x}{4}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (\ln \left (-5+x \right )\right )}{2}\)	\(16\)
parts	\(\frac {x}{4}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (\ln \left (-5+x \right )\right )}{2}\)	\(16\)
derivativedivides	\(-\frac {5}{4}+\frac {x}{4}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (\ln \left (-5+x \right )\right )}{2}\)	\(17\)
default	\(-\frac {5}{4}+\frac {x}{4}-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (\ln \left (-5+x \right )\right )}{2}\)	\(17\)
parallelrisch	\(-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (\ln \left (-5+x \right )\right )}{2}+\frac {x}{4}+\frac {5}{2}\)	\(17\)

[In]

int(((x^2-6*x+5)*ln(-5+x)+2*x)/(4*x^2-20*x)/ln(-5+x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {1}{4} \, x - \frac {1}{4} \, \log \left (x\right ) + \frac {1}{2} \, \log \left (\log \left (x - 5\right )\right ) \]

[In]

integrate(((x^2-6*x+5)*log(-5+x)+2*x)/(4*x^2-20*x)/log(-5+x),x, algorithm="fricas")

[Out]

1/4*x - 1/4*log(x) + 1/2*log(log(x - 5))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {x}{4} - \frac {\log {\left (x \right )}}{4} + \frac {\log {\left (\log {\left (x - 5 \right )} \right )}}{2} \]

[In]

integrate(((x**2-6*x+5)*ln(-5+x)+2*x)/(4*x**2-20*x)/ln(-5+x),x)

[Out]

x/4 - log(x)/4 + log(log(x - 5))/2

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {1}{4} \, x - \frac {1}{4} \, \log \left (x\right ) + \frac {1}{2} \, \log \left (\log \left (x - 5\right )\right ) \]

[In]

integrate(((x^2-6*x+5)*log(-5+x)+2*x)/(4*x^2-20*x)/log(-5+x),x, algorithm="maxima")

[Out]

1/4*x - 1/4*log(x) + 1/2*log(log(x - 5))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {1}{4} \, x - \frac {1}{4} \, \log \left (x\right ) + \frac {1}{2} \, \log \left (\log \left (x - 5\right )\right ) \]

[In]

integrate(((x^2-6*x+5)*log(-5+x)+2*x)/(4*x^2-20*x)/log(-5+x),x, algorithm="giac")

[Out]

1/4*x - 1/4*log(x) + 1/2*log(log(x - 5))

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {2 x+\left (5-6 x+x^2\right ) \log (-5+x)}{\left (-20 x+4 x^2\right ) \log (-5+x)} \, dx=\frac {x}{4}+\frac {\ln \left (\ln \left (x-5\right )\right )}{2}-\frac {\ln \left (x\right )}{4} \]

[In]

int(-(2*x + log(x - 5)*(x^2 - 6*x + 5))/(log(x - 5)*(20*x - 4*x^2)),x)

[Out]

x/4 + log(log(x - 5))/2 - log(x)/4