\(\int \frac {-20-91 x+x^2+8 x^3+(5+18 x-6 x^2) \log (x)}{25 x+50 x^2-x^3-2 x^4+(-5 x-9 x^2+2 x^3) \log (x)} \, dx\) [1482]

Optimal result

Integrand size = 65, antiderivative size = 27 \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=5+e^5-\log \left ((5-x) x \left (\frac {1}{2}+x\right ) (-5-x+\log (x))\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6873, 6860, 1626, 6816} \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=-\log (5-x)-\log (x)-\log (2 x+1)-\log (x-\log (x)+5) \]

[In]

Int[(-20 - 91*x + x^2 + 8*x^3 + (5 + 18*x - 6*x^2)*Log[x])/(25*x + 50*x^2 - x^3 - 2*x^4 + (-5*x - 9*x^2 + 2*x^
3)*Log[x]),x]

[Out]

-Log[5 - x] - Log[x] - Log[1 + 2*x] - Log[5 + x - Log[x]]

Rule 1626

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

Rule 6816

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

Rule 6860

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rule 6873

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{x \left (5+9 x-2 x^2\right ) (5+x-\log (x))} \, dx \\ & = \int \left (\frac {5+18 x-6 x^2}{(-5+x) x (1+2 x)}+\frac {1-x}{x (5+x-\log (x))}\right ) \, dx \\ & = \int \frac {5+18 x-6 x^2}{(-5+x) x (1+2 x)} \, dx+\int \frac {1-x}{x (5+x-\log (x))} \, dx \\ & = -\log (5+x-\log (x))+\int \left (\frac {1}{5-x}-\frac {1}{x}-\frac {2}{1+2 x}\right ) \, dx \\ & = -\log (5-x)-\log (x)-\log (1+2 x)-\log (5+x-\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=-\log (x)-\log \left (5+9 x-2 x^2\right )-\log (5+x-\log (x)) \]

[In]

Integrate[(-20 - 91*x + x^2 + 8*x^3 + (5 + 18*x - 6*x^2)*Log[x])/(25*x + 50*x^2 - x^3 - 2*x^4 + (-5*x - 9*x^2
+ 2*x^3)*Log[x]),x]

[Out]

-Log[x] - Log[5 + 9*x - 2*x^2] - Log[5 + x - Log[x]]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04

method	result	size
parallelrisch	\(-\ln \left (-5+x \right )-\ln \left (\frac {1}{2}+x \right )-\ln \left (x -\ln \left (x \right )+5\right )-\ln \left (x \right )\)	\(28\)
risch	\(-\ln \left (2 x^{3}-9 x^{2}-5 x \right )-\ln \left (\ln \left (x \right )-5-x \right )\)	\(29\)
default	\(-\ln \left (x \right )-\ln \left (-5+x \right )-\ln \left (1+2 x \right )-\ln \left (\ln \left (x \right )-5-x \right )\)	\(30\)
norman	\(-\ln \left (x \right )-\ln \left (-5+x \right )-\ln \left (1+2 x \right )-\ln \left (x -\ln \left (x \right )+5\right )\)	\(30\)

[In]

int(((-6*x^2+18*x+5)*ln(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*ln(x)-2*x^4-x^3+50*x^2+25*x),x,method=_RETURN
VERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=-\log \left (2 \, x^{3} - 9 \, x^{2} - 5 \, x\right ) - \log \left (-x + \log \left (x\right ) - 5\right ) \]

[In]

integrate(((-6*x^2+18*x+5)*log(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*log(x)-2*x^4-x^3+50*x^2+25*x),x, algor
ithm="fricas")

[Out]

-log(2*x^3 - 9*x^2 - 5*x) - log(-x + log(x) - 5)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=- \log {\left (- x + \log {\left (x \right )} - 5 \right )} - \log {\left (2 x^{3} - 9 x^{2} - 5 x \right )} \]

[In]

integrate(((-6*x**2+18*x+5)*ln(x)+8*x**3+x**2-91*x-20)/((2*x**3-9*x**2-5*x)*ln(x)-2*x**4-x**3+50*x**2+25*x),x)

[Out]

-log(-x + log(x) - 5) - log(2*x**3 - 9*x**2 - 5*x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=-\log \left (2 \, x + 1\right ) - \log \left (x - 5\right ) - \log \left (x\right ) - \log \left (-x + \log \left (x\right ) - 5\right ) \]

[In]

integrate(((-6*x^2+18*x+5)*log(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*log(x)-2*x^4-x^3+50*x^2+25*x),x, algor
ithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=-\log \left (2 \, x^{2} - 9 \, x - 5\right ) - \log \left (x\right ) - \log \left (-x + \log \left (x\right ) - 5\right ) \]

[In]

integrate(((-6*x^2+18*x+5)*log(x)+8*x^3+x^2-91*x-20)/((2*x^3-9*x^2-5*x)*log(x)-2*x^4-x^3+50*x^2+25*x),x, algor
ithm="giac")

[Out]

-log(2*x^2 - 9*x - 5) - log(x) - log(-x + log(x) - 5)

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96 \[ \int \frac {-20-91 x+x^2+8 x^3+\left (5+18 x-6 x^2\right ) \log (x)}{25 x+50 x^2-x^3-2 x^4+\left (-5 x-9 x^2+2 x^3\right ) \log (x)} \, dx=-\ln \left (x-\ln \left (x\right )+5\right )-\ln \left (x\,\left (-2\,x^2+9\,x+5\right )\right ) \]

[In]

int(-(log(x)*(18*x - 6*x^2 + 5) - 91*x + x^2 + 8*x^3 - 20)/(x^3 - 50*x^2 - 25*x + 2*x^4 + log(x)*(5*x + 9*x^2
- 2*x^3)),x)

[Out]

- log(x - log(x) + 5) - log(x*(9*x - 2*x^2 + 5))