\(\int \frac {(20-48 x+12 x^2) \log ^4(3)+(5 x-6 x^2+x^3) \log ^4(3) \log (5)+(5 x-6 x^2+x^3) \log ^4(3) \log (-5 x+6 x^2-x^3)}{(5 x-6 x^2+x^3) \log (5)+(5 x-6 x^2+x^3) \log (-5 x+6 x^2-x^3)} \, dx\) [5082]

Optimal result

Integrand size = 114, antiderivative size = 24 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=\log ^4(3) (x+4 \log (\log (5)+\log ((5-x) (-1+x) x))) \]

[Out]

(x+4*ln(ln((-1+x)*x*(5-x))+ln(5)))*ln(3)^4

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {6820, 12, 6860, 6816} \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=4 \log ^4(3) \log \left (\log \left (-x \left (x^2-6 x+5\right )\right )+\log (5)\right )+x \log ^4(3) \]

[In]

Int[((20 - 48*x + 12*x^2)*Log[3]^4 + (5*x - 6*x^2 + x^3)*Log[3]^4*Log[5] + (5*x - 6*x^2 + x^3)*Log[3]^4*Log[-5
*x + 6*x^2 - x^3])/((5*x - 6*x^2 + x^3)*Log[5] + (5*x - 6*x^2 + x^3)*Log[-5*x + 6*x^2 - x^3]),x]

[Out]

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

Rule 12

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

Rule 6816

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

Rule 6820

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

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\log ^4(3) \left (20-6 x^2 (-2+\log (5))+x^3 \log (5)+x (-48+5 \log (5))+x \left (5-6 x+x^2\right ) \log \left (-x \left (5-6 x+x^2\right )\right )\right )}{x \left (5-6 x+x^2\right ) \left (\log (5)+\log \left (-x \left (5-6 x+x^2\right )\right )\right )} \, dx \\ & = \log ^4(3) \int \frac {20-6 x^2 (-2+\log (5))+x^3 \log (5)+x (-48+5 \log (5))+x \left (5-6 x+x^2\right ) \log \left (-x \left (5-6 x+x^2\right )\right )}{x \left (5-6 x+x^2\right ) \left (\log (5)+\log \left (-x \left (5-6 x+x^2\right )\right )\right )} \, dx \\ & = \log ^4(3) \int \left (1+\frac {4 \left (5-12 x+3 x^2\right )}{(-5+x) (-1+x) x \left (\log (5)+\log \left (-x \left (5-6 x+x^2\right )\right )\right )}\right ) \, dx \\ & = x \log ^4(3)+\left (4 \log ^4(3)\right ) \int \frac {5-12 x+3 x^2}{(-5+x) (-1+x) x \left (\log (5)+\log \left (-x \left (5-6 x+x^2\right )\right )\right )} \, dx \\ & = x \log ^4(3)+4 \log ^4(3) \log \left (\log (5)+\log \left (-x \left (5-6 x+x^2\right )\right )\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=\log ^4(3) \left (x+4 \log \left (\log (5)+\log \left (-x \left (5-6 x+x^2\right )\right )\right )\right ) \]

[In]

Integrate[((20 - 48*x + 12*x^2)*Log[3]^4 + (5*x - 6*x^2 + x^3)*Log[3]^4*Log[5] + (5*x - 6*x^2 + x^3)*Log[3]^4*
Log[-5*x + 6*x^2 - x^3])/((5*x - 6*x^2 + x^3)*Log[5] + (5*x - 6*x^2 + x^3)*Log[-5*x + 6*x^2 - x^3]),x]

[Out]

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

Maple [A] (verified)

Time = 0.66 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38

method	result	size
norman	\(x \ln \left (3\right )^{4}+4 \ln \left (3\right )^{4} \ln \left (\ln \left (-x^{3}+6 x^{2}-5 x \right )+\ln \left (5\right )\right )\)	\(33\)
risch	\(x \ln \left (3\right )^{4}+4 \ln \left (3\right )^{4} \ln \left (\ln \left (-x^{3}+6 x^{2}-5 x \right )+\ln \left (5\right )\right )\)	\(33\)
parallelrisch	\(x \ln \left (3\right )^{4}+4 \ln \left (3\right )^{4} \ln \left (\ln \left (-x^{3}+6 x^{2}-5 x \right )+\ln \left (5\right )\right )\)	\(33\)

[In]

int(((x^3-6*x^2+5*x)*ln(3)^4*ln(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*ln(3)^4*ln(5)+(12*x^2-48*x+20)*ln(3)^4)/((x^3-
6*x^2+5*x)*ln(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*ln(5)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log \left (3\right )^{4} + 4 \, \log \left (3\right )^{4} \log \left (\log \left (5\right ) + \log \left (-x^{3} + 6 \, x^{2} - 5 \, x\right )\right ) \]

[In]

integrate(((x^3-6*x^2+5*x)*log(3)^4*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*log(3)^4*log(5)+(12*x^2-48*x+20)*log(3
)^4)/((x^3-6*x^2+5*x)*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*log(5)),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log {\left (3 \right )}^{4} + 4 \log {\left (3 \right )}^{4} \log {\left (\log {\left (- x^{3} + 6 x^{2} - 5 x \right )} + \log {\left (5 \right )} \right )} \]

[In]

integrate(((x**3-6*x**2+5*x)*ln(3)**4*ln(-x**3+6*x**2-5*x)+(x**3-6*x**2+5*x)*ln(3)**4*ln(5)+(12*x**2-48*x+20)*
ln(3)**4)/((x**3-6*x**2+5*x)*ln(-x**3+6*x**2-5*x)+(x**3-6*x**2+5*x)*ln(5)),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log \left (3\right )^{4} + 4 \, \log \left (3\right )^{4} \log \left (\log \left (5\right ) + \log \left (x - 1\right ) + \log \left (x\right ) + \log \left (-x + 5\right )\right ) \]

[In]

integrate(((x^3-6*x^2+5*x)*log(3)^4*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*log(3)^4*log(5)+(12*x^2-48*x+20)*log(3
)^4)/((x^3-6*x^2+5*x)*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*log(5)),x, algorithm="maxima")

[Out]

x*log(3)^4 + 4*log(3)^4*log(log(5) + log(x - 1) + log(x) + log(-x + 5))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x \log \left (3\right )^{4} + 4 \, \log \left (3\right )^{4} \log \left (\log \left (5\right ) + \log \left (-x^{3} + 6 \, x^{2} - 5 \, x\right )\right ) \]

[In]

integrate(((x^3-6*x^2+5*x)*log(3)^4*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*log(3)^4*log(5)+(12*x^2-48*x+20)*log(3
)^4)/((x^3-6*x^2+5*x)*log(-x^3+6*x^2-5*x)+(x^3-6*x^2+5*x)*log(5)),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 15.55 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\left (20-48 x+12 x^2\right ) \log ^4(3)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log (5)+\left (5 x-6 x^2+x^3\right ) \log ^4(3) \log \left (-5 x+6 x^2-x^3\right )}{\left (5 x-6 x^2+x^3\right ) \log (5)+\left (5 x-6 x^2+x^3\right ) \log \left (-5 x+6 x^2-x^3\right )} \, dx=x\,{\ln \left (3\right )}^4+4\,{\ln \left (3\right )}^4\,\ln \left (\ln \left (-5\,x^3+30\,x^2-25\,x\right )\right ) \]

[In]

int((log(3)^4*(12*x^2 - 48*x + 20) + log(6*x^2 - 5*x - x^3)*log(3)^4*(5*x - 6*x^2 + x^3) + log(3)^4*log(5)*(5*
x - 6*x^2 + x^3))/(log(5)*(5*x - 6*x^2 + x^3) + log(6*x^2 - 5*x - x^3)*(5*x - 6*x^2 + x^3)),x)

[Out]

x*log(3)^4 + 4*log(3)^4*log(log(30*x^2 - 25*x - 5*x^3))