\(\int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+(-375 x+205 x^2-50 x^3+3 x^4) \log (x)}{(-1500 x+525 x^2+45 x^3-33 x^4+3 x^5) \log (x)} \, dx\) [3803]

Optimal result

Integrand size = 73, antiderivative size = 23 \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=7+\frac {x^2}{6 (5-x)^2}+\log ((4+x) \log (x)) \]

[Out]

7+1/6*x^2/(5-x)^2+ln((4+x)*ln(x))

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {6820, 1634, 2339, 29} \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=-\frac {5}{3 (5-x)}+\frac {25}{6 (5-x)^2}+\log (x+4)+\log (\log (x)) \]

[In]

Int[(-1500 + 525*x + 45*x^2 - 33*x^3 + 3*x^4 + (-375*x + 205*x^2 - 50*x^3 + 3*x^4)*Log[x])/((-1500*x + 525*x^2
+ 45*x^3 - 33*x^4 + 3*x^5)*Log[x]),x]

[Out]

25/(6*(5 - x)^2) - 5/(3*(5 - x)) + Log[4 + x] + Log[Log[x]]

Rule 29

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

Rule 1634

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 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 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-375+205 x-50 x^2+3 x^3}{3 (-5+x)^3 (4+x)}+\frac {1}{x \log (x)}\right ) \, dx \\ & = \frac {1}{3} \int \frac {-375+205 x-50 x^2+3 x^3}{(-5+x)^3 (4+x)} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = \frac {1}{3} \int \left (-\frac {25}{(-5+x)^3}-\frac {5}{(-5+x)^2}+\frac {3}{4+x}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \frac {25}{6 (5-x)^2}-\frac {5}{3 (5-x)}+\log (4+x)+\log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=\frac {25}{6 (-5+x)^2}+\frac {5}{3 (-5+x)}+\log (4+x)+\log (\log (x)) \]

[In]

Integrate[(-1500 + 525*x + 45*x^2 - 33*x^3 + 3*x^4 + (-375*x + 205*x^2 - 50*x^3 + 3*x^4)*Log[x])/((-1500*x + 5
25*x^2 + 45*x^3 - 33*x^4 + 3*x^5)*Log[x]),x]

[Out]

25/(6*(-5 + x)^2) + 5/(3*(-5 + x)) + Log[4 + x] + Log[Log[x]]

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87

method	result	size
norman	\(\frac {\frac {5 x}{3}-\frac {25}{6}}{\left (-5+x \right )^{2}}+\ln \left (\ln \left (x \right )\right )+\ln \left (4+x \right )\)	\(20\)
default	\(\ln \left (4+x \right )+\frac {25}{6 \left (-5+x \right )^{2}}+\frac {5}{3 \left (-5+x \right )}+\ln \left (\ln \left (x \right )\right )\)	\(23\)
parts	\(\ln \left (4+x \right )+\frac {25}{6 \left (-5+x \right )^{2}}+\frac {5}{3 \left (-5+x \right )}+\ln \left (\ln \left (x \right )\right )\)	\(23\)
risch	\(\frac {6 x^{2} \ln \left (4+x \right )-60 \ln \left (4+x \right ) x +150 \ln \left (4+x \right )+10 x -25}{6 x^{2}-60 x +150}+\ln \left (\ln \left (x \right )\right )\)	\(44\)
parallelrisch	\(\frac {6 x^{2} \ln \left (\ln \left (x \right )\right )+6 x^{2} \ln \left (4+x \right )-25-60 x \ln \left (\ln \left (x \right )\right )-60 \ln \left (4+x \right ) x +150 \ln \left (\ln \left (x \right )\right )+150 \ln \left (4+x \right )+10 x}{6 x^{2}-60 x +150}\)	\(59\)

[In]

int(((3*x^4-50*x^3+205*x^2-375*x)*ln(x)+3*x^4-33*x^3+45*x^2+525*x-1500)/(3*x^5-33*x^4+45*x^3+525*x^2-1500*x)/l
n(x),x,method=_RETURNVERBOSE)

[Out]

(5/3*x-25/6)/(-5+x)^2+ln(ln(x))+ln(4+x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.91 \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=\frac {6 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (x + 4\right ) + 6 \, {\left (x^{2} - 10 \, x + 25\right )} \log \left (\log \left (x\right )\right ) + 10 \, x - 25}{6 \, {\left (x^{2} - 10 \, x + 25\right )}} \]

[In]

integrate(((3*x^4-50*x^3+205*x^2-375*x)*log(x)+3*x^4-33*x^3+45*x^2+525*x-1500)/(3*x^5-33*x^4+45*x^3+525*x^2-15
00*x)/log(x),x, algorithm="fricas")

[Out]

1/6*(6*(x^2 - 10*x + 25)*log(x + 4) + 6*(x^2 - 10*x + 25)*log(log(x)) + 10*x - 25)/(x^2 - 10*x + 25)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=\frac {10 x - 25}{6 x^{2} - 60 x + 150} + \log {\left (x + 4 \right )} + \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((3*x**4-50*x**3+205*x**2-375*x)*ln(x)+3*x**4-33*x**3+45*x**2+525*x-1500)/(3*x**5-33*x**4+45*x**3+52
5*x**2-1500*x)/ln(x),x)

[Out]

(10*x - 25)/(6*x**2 - 60*x + 150) + log(x + 4) + log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=\frac {5 \, {\left (2 \, x - 5\right )}}{6 \, {\left (x^{2} - 10 \, x + 25\right )}} + \log \left (x + 4\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((3*x^4-50*x^3+205*x^2-375*x)*log(x)+3*x^4-33*x^3+45*x^2+525*x-1500)/(3*x^5-33*x^4+45*x^3+525*x^2-15
00*x)/log(x),x, algorithm="maxima")

[Out]

5/6*(2*x - 5)/(x^2 - 10*x + 25) + log(x + 4) + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=\frac {5 \, {\left (2 \, x - 5\right )}}{6 \, {\left (x^{2} - 10 \, x + 25\right )}} + \log \left (x + 4\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

integrate(((3*x^4-50*x^3+205*x^2-375*x)*log(x)+3*x^4-33*x^3+45*x^2+525*x-1500)/(3*x^5-33*x^4+45*x^3+525*x^2-15
00*x)/log(x),x, algorithm="giac")

[Out]

5/6*(2*x - 5)/(x^2 - 10*x + 25) + log(x + 4) + log(log(x))

Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.83 \[ \int \frac {-1500+525 x+45 x^2-33 x^3+3 x^4+\left (-375 x+205 x^2-50 x^3+3 x^4\right ) \log (x)}{\left (-1500 x+525 x^2+45 x^3-33 x^4+3 x^5\right ) \log (x)} \, dx=\ln \left (x+4\right )+\ln \left (\ln \left (x\right )\right )+\frac {\frac {5\,x}{3}-\frac {25}{6}}{{\left (x-5\right )}^2} \]

[In]

int((525*x - log(x)*(375*x - 205*x^2 + 50*x^3 - 3*x^4) + 45*x^2 - 33*x^3 + 3*x^4 - 1500)/(log(x)*(525*x^2 - 15
00*x + 45*x^3 - 33*x^4 + 3*x^5)),x)

[Out]

log(x + 4) + log(log(x)) + ((5*x)/3 - 25/6)/(x - 5)^2