3.21.66 \(\int \frac {75-5 x+18 x^2-5 x^3+(-15+3 x-4 x^2+x^3) \log (x)}{25 x^2-5 x^3+(-5 x^2+x^3) \log (x)} \, dx\)

Optimal. Leaf size=24 \[ 3-\frac {3}{x}+x+\log \left (\frac {5-x}{4 (-5+\log (x))^2}\right ) \]

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Rubi [A]  time = 0.50, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 5, integrand size = 57, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {6741, 6742, 1620, 2302, 29} \begin {gather*} x-\frac {3}{x}+\log (5-x)-2 \log (5-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(75 - 5*x + 18*x^2 - 5*x^3 + (-15 + 3*x - 4*x^2 + x^3)*Log[x])/(25*x^2 - 5*x^3 + (-5*x^2 + x^3)*Log[x]),x]

[Out]

-3/x + x + Log[5 - x] - 2*Log[5 - Log[x]]

Rule 29

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

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 6741

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

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 {75-5 x+18 x^2-5 x^3+\left (-15+3 x-4 x^2+x^3\right ) \log (x)}{(5-x) x^2 (5-\log (x))} \, dx\\ &=\int \left (\frac {-15+3 x-4 x^2+x^3}{(-5+x) x^2}-\frac {2}{x (-5+\log (x))}\right ) \, dx\\ &=-\left (2 \int \frac {1}{x (-5+\log (x))} \, dx\right )+\int \frac {-15+3 x-4 x^2+x^3}{(-5+x) x^2} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-5+\log (x)\right )\right )+\int \left (1+\frac {1}{-5+x}+\frac {3}{x^2}\right ) \, dx\\ &=-\frac {3}{x}+x+\log (5-x)-2 \log (5-\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 22, normalized size = 0.92 \begin {gather*} -\frac {3}{x}+x+\log (5-x)-2 \log (5-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(75 - 5*x + 18*x^2 - 5*x^3 + (-15 + 3*x - 4*x^2 + x^3)*Log[x])/(25*x^2 - 5*x^3 + (-5*x^2 + x^3)*Log[
x]),x]

[Out]

-3/x + x + Log[5 - x] - 2*Log[5 - Log[x]]

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fricas [A]  time = 0.62, size = 23, normalized size = 0.96 \begin {gather*} \frac {x^{2} + x \log \left (x - 5\right ) - 2 \, x \log \left (\log \relax (x) - 5\right ) - 3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+3*x-15)*log(x)-5*x^3+18*x^2-5*x+75)/((x^3-5*x^2)*log(x)-5*x^3+25*x^2),x, algorithm="fric
as")

[Out]

(x^2 + x*log(x - 5) - 2*x*log(log(x) - 5) - 3)/x

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giac [A]  time = 1.98, size = 18, normalized size = 0.75 \begin {gather*} x - \frac {3}{x} + \log \left (x - 5\right ) - 2 \, \log \left (\log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+3*x-15)*log(x)-5*x^3+18*x^2-5*x+75)/((x^3-5*x^2)*log(x)-5*x^3+25*x^2),x, algorithm="giac
")

[Out]

x - 3/x + log(x - 5) - 2*log(log(x) - 5)

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maple [A]  time = 0.04, size = 22, normalized size = 0.92

method	result	size
norman	\(\frac {x^{2}-3}{x}-2 \ln \left (\ln \relax (x )-5\right )+\ln \left (x -5\right )\)	\(22\)
risch	\(\frac {\ln \left (x -5\right ) x +x^{2}-3}{x}-2 \ln \left (\ln \relax (x )-5\right )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-4*x^2+3*x-15)*ln(x)-5*x^3+18*x^2-5*x+75)/((x^3-5*x^2)*ln(x)-5*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

(x^2-3)/x-2*ln(ln(x)-5)+ln(x-5)

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maxima [A]  time = 0.42, size = 21, normalized size = 0.88 \begin {gather*} \frac {x^{2} - 3}{x} + \log \left (x - 5\right ) - 2 \, \log \left (\log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-4*x^2+3*x-15)*log(x)-5*x^3+18*x^2-5*x+75)/((x^3-5*x^2)*log(x)-5*x^3+25*x^2),x, algorithm="maxi
ma")

[Out]

(x^2 - 3)/x + log(x - 5) - 2*log(log(x) - 5)

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mupad [B]  time = 1.25, size = 18, normalized size = 0.75 \begin {gather*} x+\ln \left (x-5\right )-2\,\ln \left (\ln \relax (x)-5\right )-\frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(3*x - 4*x^2 + x^3 - 15) - 5*x + 18*x^2 - 5*x^3 + 75)/(log(x)*(5*x^2 - x^3) - 25*x^2 + 5*x^3),x)

[Out]

x + log(x - 5) - 2*log(log(x) - 5) - 3/x

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sympy [A]  time = 0.15, size = 17, normalized size = 0.71 \begin {gather*} x + \log {\left (x - 5 \right )} - 2 \log {\left (\log {\relax (x )} - 5 \right )} - \frac {3}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-4*x**2+3*x-15)*ln(x)-5*x**3+18*x**2-5*x+75)/((x**3-5*x**2)*ln(x)-5*x**3+25*x**2),x)

[Out]

x + log(x - 5) - 2*log(log(x) - 5) - 3/x

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