\(\int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx\) [1328]

Optimal result

Integrand size = 28, antiderivative size = 27 \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=5 \left (-4+x-2 \left (5-x+\log \left (\left (-\frac {9 x}{2}+\frac {\log (x)}{4}\right )^2\right )\right )\right ) \]

[Out]

15*x-70-10*ln((1/4*ln(x)-9/2*x)^2)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2641, 6873, 12, 6874, 6816} \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=15 x-20 \log (18 x-\log (x)) \]

[In]

Int[(-20 + 360*x - 270*x^2 + 15*x*Log[x])/(-18*x^2 + x*Log[x]),x]

[Out]

15*x - 20*Log[18*x - Log[x]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2641

Int[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.))^(p_.), x_Symbol] :> Int[u*x^(p*r)*
(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 {-20+360 x-270 x^2+15 x \log (x)}{x (-18 x+\log (x))} \, dx \\ & = \int \frac {5 \left (4-72 x+54 x^2-3 x \log (x)\right )}{x (18 x-\log (x))} \, dx \\ & = 5 \int \frac {4-72 x+54 x^2-3 x \log (x)}{x (18 x-\log (x))} \, dx \\ & = 5 \int \left (3-\frac {4 (-1+18 x)}{x (18 x-\log (x))}\right ) \, dx \\ & = 15 x-20 \int \frac {-1+18 x}{x (18 x-\log (x))} \, dx \\ & = 15 x-20 \log (18 x-\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.63 \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=5 (3 x-4 \log (18 x-\log (x))) \]

[In]

Integrate[(-20 + 360*x - 270*x^2 + 15*x*Log[x])/(-18*x^2 + x*Log[x]),x]

[Out]

5*(3*x - 4*Log[18*x - Log[x]])

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52

method	result	size
default	\(15 x -20 \ln \left (\ln \left (x \right )-18 x \right )\)	\(14\)
risch	\(15 x -20 \ln \left (\ln \left (x \right )-18 x \right )\)	\(14\)
parallelrisch	\(-20 \ln \left (-\frac {\ln \left (x \right )}{18}+x \right )+15 x\)	\(14\)
norman	\(15 x -20 \ln \left (18 x -\ln \left (x \right )\right )\)	\(16\)

[In]

int((15*x*ln(x)-270*x^2+360*x-20)/(x*ln(x)-18*x^2),x,method=_RETURNVERBOSE)

[Out]

15*x-20*ln(ln(x)-18*x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.48 \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=15 \, x - 20 \, \log \left (-18 \, x + \log \left (x\right )\right ) \]

[In]

integrate((15*x*log(x)-270*x^2+360*x-20)/(x*log(x)-18*x^2),x, algorithm="fricas")

[Out]

15*x - 20*log(-18*x + log(x))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.44 \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=15 x - 20 \log {\left (- 18 x + \log {\left (x \right )} \right )} \]

[In]

integrate((15*x*ln(x)-270*x**2+360*x-20)/(x*ln(x)-18*x**2),x)

[Out]

15*x - 20*log(-18*x + log(x))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.48 \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=15 \, x - 20 \, \log \left (-18 \, x + \log \left (x\right )\right ) \]

[In]

integrate((15*x*log(x)-270*x^2+360*x-20)/(x*log(x)-18*x^2),x, algorithm="maxima")

[Out]

15*x - 20*log(-18*x + log(x))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=15 \, x - 20 \, \log \left (18 \, x - \log \left (x\right )\right ) \]

[In]

integrate((15*x*log(x)-270*x^2+360*x-20)/(x*log(x)-18*x^2),x, algorithm="giac")

[Out]

15*x - 20*log(18*x - log(x))

Mupad [B] (verification not implemented)

Time = 8.44 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.56 \[ \int \frac {-20+360 x-270 x^2+15 x \log (x)}{-18 x^2+x \log (x)} \, dx=15\,x-20\,\ln \left (18\,x-\ln \left (x\right )\right ) \]

[In]

int((360*x + 15*x*log(x) - 270*x^2 - 20)/(x*log(x) - 18*x^2),x)

[Out]

15*x - 20*log(18*x - log(x))