\(\int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+(24-14 x+10 x^2-2 x^3) \log (3)+(3-x+x^2) \log ^2(3)+(16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)) \log (4)+(24-14 x+10 x^2-2 x^3+(6-2 x+2 x^2) \log (3)+(8-2 x+2 \log (3)) \log (4)) \log (-3+x-x^2-\log (4))+(3-x+x^2+\log (4)) \log ^2(-3+x-x^2-\log (4))} \, dx\) [6283]

Optimal result

Integrand size = 169, antiderivative size = 26 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{-4+x-\log (3)-\log \left (-3+x-x^2-\log (4)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6820, 12, 6818} \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=-\frac {5}{\log \left (-3 \left (x^2-x+3+\log (4)\right )\right )-x+4} \]

[In]

Int[(-20 + 15*x - 5*x^2 - 5*Log[4])/(48 - 40*x + 27*x^2 - 9*x^3 + x^4 + (24 - 14*x + 10*x^2 - 2*x^3)*Log[3] +
(3 - x + x^2)*Log[3]^2 + (16 - 8*x + x^2 + (8 - 2*x)*Log[3] + Log[3]^2)*Log[4] + (24 - 14*x + 10*x^2 - 2*x^3 +
(6 - 2*x + 2*x^2)*Log[3] + (8 - 2*x + 2*Log[3])*Log[4])*Log[-3 + x - x^2 - Log[4]] + (3 - x + x^2 + Log[4])*L
og[-3 + x - x^2 - Log[4]]^2),x]

[Out]

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

Rule 12

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

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 \frac {5 \left (-4+3 x-x^2-\log (4)\right )}{\left (3-x+x^2+\log (4)\right ) \left (4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )\right )^2} \, dx \\ & = 5 \int \frac {-4+3 x-x^2-\log (4)}{\left (3-x+x^2+\log (4)\right ) \left (4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )\right )^2} \, dx \\ & = -\frac {5}{4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=-\frac {5}{4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )} \]

[In]

Integrate[(-20 + 15*x - 5*x^2 - 5*Log[4])/(48 - 40*x + 27*x^2 - 9*x^3 + x^4 + (24 - 14*x + 10*x^2 - 2*x^3)*Log
[3] + (3 - x + x^2)*Log[3]^2 + (16 - 8*x + x^2 + (8 - 2*x)*Log[3] + Log[3]^2)*Log[4] + (24 - 14*x + 10*x^2 - 2
*x^3 + (6 - 2*x + 2*x^2)*Log[3] + (8 - 2*x + 2*Log[3])*Log[4])*Log[-3 + x - x^2 - Log[4]] + (3 - x + x^2 + Log
[4])*Log[-3 + x - x^2 - Log[4]]^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96

method	result	size
norman	\(-\frac {5}{\ln \left (3\right )+\ln \left (-2 \ln \left (2\right )-x^{2}+x -3\right )-x +4}\)	\(25\)
risch	\(-\frac {5}{\ln \left (3\right )+\ln \left (-2 \ln \left (2\right )-x^{2}+x -3\right )-x +4}\)	\(25\)
parallelrisch	\(-\frac {5}{\ln \left (3\right )+\ln \left (-2 \ln \left (2\right )-x^{2}+x -3\right )-x +4}\)	\(25\)

[In]

int((-10*ln(2)-5*x^2+15*x-20)/((2*ln(2)+x^2-x+3)*ln(-2*ln(2)-x^2+x-3)^2+(2*(2*ln(3)-2*x+8)*ln(2)+(2*x^2-2*x+6)
*ln(3)-2*x^3+10*x^2-14*x+24)*ln(-2*ln(2)-x^2+x-3)+2*(ln(3)^2+(-2*x+8)*ln(3)+x^2-8*x+16)*ln(2)+(x^2-x+3)*ln(3)^
2+(-2*x^3+10*x^2-14*x+24)*ln(3)+x^4-9*x^3+27*x^2-40*x+48),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{x - \log \left (3\right ) - \log \left (-x^{2} + x - 2 \, \log \left (2\right ) - 3\right ) - 4} \]

[In]

integrate((-10*log(2)-5*x^2+15*x-20)/((2*log(2)+x^2-x+3)*log(-2*log(2)-x^2+x-3)^2+(2*(2*log(3)-2*x+8)*log(2)+(
2*x^2-2*x+6)*log(3)-2*x^3+10*x^2-14*x+24)*log(-2*log(2)-x^2+x-3)+2*(log(3)^2+(-2*x+8)*log(3)+x^2-8*x+16)*log(2
)+(x^2-x+3)*log(3)^2+(-2*x^3+10*x^2-14*x+24)*log(3)+x^4-9*x^3+27*x^2-40*x+48),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=- \frac {5}{- x + \log {\left (- x^{2} + x - 3 - 2 \log {\left (2 \right )} \right )} + \log {\left (3 \right )} + 4} \]

[In]

integrate((-10*ln(2)-5*x**2+15*x-20)/((2*ln(2)+x**2-x+3)*ln(-2*ln(2)-x**2+x-3)**2+(2*(2*ln(3)-2*x+8)*ln(2)+(2*
x**2-2*x+6)*ln(3)-2*x**3+10*x**2-14*x+24)*ln(-2*ln(2)-x**2+x-3)+2*(ln(3)**2+(-2*x+8)*ln(3)+x**2-8*x+16)*ln(2)+
(x**2-x+3)*ln(3)**2+(-2*x**3+10*x**2-14*x+24)*ln(3)+x**4-9*x**3+27*x**2-40*x+48),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{x - \log \left (3\right ) - \log \left (-x^{2} + x - 2 \, \log \left (2\right ) - 3\right ) - 4} \]

[In]

integrate((-10*log(2)-5*x^2+15*x-20)/((2*log(2)+x^2-x+3)*log(-2*log(2)-x^2+x-3)^2+(2*(2*log(3)-2*x+8)*log(2)+(
2*x^2-2*x+6)*log(3)-2*x^3+10*x^2-14*x+24)*log(-2*log(2)-x^2+x-3)+2*(log(3)^2+(-2*x+8)*log(3)+x^2-8*x+16)*log(2
)+(x^2-x+3)*log(3)^2+(-2*x^3+10*x^2-14*x+24)*log(3)+x^4-9*x^3+27*x^2-40*x+48),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\frac {5}{x - \log \left (3\right ) - \log \left (-x^{2} + x - 2 \, \log \left (2\right ) - 3\right ) - 4} \]

[In]

integrate((-10*log(2)-5*x^2+15*x-20)/((2*log(2)+x^2-x+3)*log(-2*log(2)-x^2+x-3)^2+(2*(2*log(3)-2*x+8)*log(2)+(
2*x^2-2*x+6)*log(3)-2*x^3+10*x^2-14*x+24)*log(-2*log(2)-x^2+x-3)+2*(log(3)^2+(-2*x+8)*log(3)+x^2-8*x+16)*log(2
)+(x^2-x+3)*log(3)^2+(-2*x^3+10*x^2-14*x+24)*log(3)+x^4-9*x^3+27*x^2-40*x+48),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {-20+15 x-5 x^2-5 \log (4)}{48-40 x+27 x^2-9 x^3+x^4+\left (24-14 x+10 x^2-2 x^3\right ) \log (3)+\left (3-x+x^2\right ) \log ^2(3)+\left (16-8 x+x^2+(8-2 x) \log (3)+\log ^2(3)\right ) \log (4)+\left (24-14 x+10 x^2-2 x^3+\left (6-2 x+2 x^2\right ) \log (3)+(8-2 x+2 \log (3)) \log (4)\right ) \log \left (-3+x-x^2-\log (4)\right )+\left (3-x+x^2+\log (4)\right ) \log ^2\left (-3+x-x^2-\log (4)\right )} \, dx=\int -\frac {5\,x^2-15\,x+10\,\ln \left (2\right )+20}{{\ln \left (3\right )}^2\,\left (x^2-x+3\right )-40\,x+\ln \left (-x^2+x-2\,\ln \left (2\right )-3\right )\,\left (\ln \left (3\right )\,\left (2\,x^2-2\,x+6\right )-14\,x+2\,\ln \left (2\right )\,\left (2\,\ln \left (3\right )-2\,x+8\right )+10\,x^2-2\,x^3+24\right )-\ln \left (3\right )\,\left (2\,x^3-10\,x^2+14\,x-24\right )+2\,\ln \left (2\right )\,\left ({\ln \left (3\right )}^2-\ln \left (3\right )\,\left (2\,x-8\right )-8\,x+x^2+16\right )+27\,x^2-9\,x^3+x^4+{\ln \left (-x^2+x-2\,\ln \left (2\right )-3\right )}^2\,\left (x^2-x+2\,\ln \left (2\right )+3\right )+48} \,d x \]

[In]

int(-(10*log(2) - 15*x + 5*x^2 + 20)/(log(3)^2*(x^2 - x + 3) - 40*x + log(x - 2*log(2) - x^2 - 3)*(log(3)*(2*x
^2 - 2*x + 6) - 14*x + 2*log(2)*(2*log(3) - 2*x + 8) + 10*x^2 - 2*x^3 + 24) - log(3)*(14*x - 10*x^2 + 2*x^3 -
24) + 2*log(2)*(log(3)^2 - log(3)*(2*x - 8) - 8*x + x^2 + 16) + 27*x^2 - 9*x^3 + x^4 + log(x - 2*log(2) - x^2
- 3)^2*(2*log(2) - x + x^2 + 3) + 48),x)

[Out]

int(-(10*log(2) - 15*x + 5*x^2 + 20)/(log(3)^2*(x^2 - x + 3) - 40*x + log(x - 2*log(2) - x^2 - 3)*(log(3)*(2*x
^2 - 2*x + 6) - 14*x + 2*log(2)*(2*log(3) - 2*x + 8) + 10*x^2 - 2*x^3 + 24) - log(3)*(14*x - 10*x^2 + 2*x^3 -
24) + 2*log(2)*(log(3)^2 - log(3)*(2*x - 8) - 8*x + x^2 + 16) + 27*x^2 - 9*x^3 + x^4 + log(x - 2*log(2) - x^2
- 3)^2*(2*log(2) - x + x^2 + 3) + 48), x)