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

Optimal. Leaf size=24 \[ \frac {1}{4} \left (86+x-\log ^2(5)-\log (x)+\log \left (\log ^2(-5+x)\right )\right ) \]

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Rubi [A]  time = 0.32, antiderivative size = 21, normalized size of antiderivative = 0.88, number of steps used = 8, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {1593, 6742, 43, 2390, 2302, 29} \begin {gather*} \frac {x}{4}-\frac {\log (x)}{4}+\frac {1}{2} \log (\log (x-5)) \end {gather*}

Antiderivative was successfully verified.

[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 43

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 1593

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 2302

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 2390

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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \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} \operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-5+x\right )\\ &=\frac {x}{4}-\frac {\log (x)}{4}+\frac {1}{2} \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 17, normalized size = 0.71 \begin {gather*} \frac {1}{4} (x-\log (x)+2 \log (\log (-5+x))) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.82, size = 15, normalized size = 0.62 \begin {gather*} \frac {1}{4} \, x - \frac {1}{4} \, \log \relax (x) + \frac {1}{2} \, \log \left (\log \left (x - 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.38, size = 15, normalized size = 0.62 \begin {gather*} \frac {1}{4} \, x - \frac {1}{4} \, \log \relax (x) + \frac {1}{2} \, \log \left (\log \left (x - 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.14, size = 16, normalized size = 0.67

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.66, size = 15, normalized size = 0.62 \begin {gather*} \frac {1}{4} \, x - \frac {1}{4} \, \log \relax (x) + \frac {1}{2} \, \log \left (\log \left (x - 5\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.84, size = 15, normalized size = 0.62 \begin {gather*} \frac {x}{4}+\frac {\ln \left (\ln \left (x-5\right )\right )}{2}-\frac {\ln \relax (x)}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.14, size = 15, normalized size = 0.62 \begin {gather*} \frac {x}{4} - \frac {\log {\relax (x )}}{4} + \frac {\log {\left (\log {\left (x - 5 \right )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x/4 - log(x)/4 + log(log(x - 5))/2

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