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

Optimal. Leaf size=26 \[ \frac {5}{-4+x-\log (3)-\log \left (-3+x-x^2-\log (4)\right )} \]

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Rubi [A]  time = 0.33, antiderivative size = 22, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, integrand size = 169, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {5}{\log \left (-3 \left (x^2-x+3+\log (4)\right )\right )-x+4} \end {gather*}

Antiderivative was successfully verified.

[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 6686

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 22, normalized size = 0.85 \begin {gather*} -\frac {5}{4-x+\log \left (-3 \left (3-x+x^2+\log (4)\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[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])])

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fricas [A]  time = 0.76, size = 26, normalized size = 1.00 \begin {gather*} \frac {5}{x - \log \relax (3) - \log \left (-x^{2} + x - 2 \, \log \relax (2) - 3\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.35, size = 26, normalized size = 1.00 \begin {gather*} \frac {5}{x - \log \relax (3) - \log \left (-x^{2} + x - 2 \, \log \relax (2) - 3\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.41, size = 25, normalized size = 0.96




method result size



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



Verification of antiderivative is not currently implemented for this CAS.

[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(-2*ln(2)-x^2+x-3)+ln(3)-x+4)

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maxima [A]  time = 0.52, size = 26, normalized size = 1.00 \begin {gather*} \frac {5}{x - \log \relax (3) - \log \left (-x^{2} + x - 2 \, \log \relax (2) - 3\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {5\,x^2-15\,x+10\,\ln \relax (2)+20}{{\ln \relax (3)}^2\,\left (x^2-x+3\right )-40\,x+\ln \left (-x^2+x-2\,\ln \relax (2)-3\right )\,\left (\ln \relax (3)\,\left (2\,x^2-2\,x+6\right )-14\,x+2\,\ln \relax (2)\,\left (2\,\ln \relax (3)-2\,x+8\right )+10\,x^2-2\,x^3+24\right )-\ln \relax (3)\,\left (2\,x^3-10\,x^2+14\,x-24\right )+2\,\ln \relax (2)\,\left ({\ln \relax (3)}^2-\ln \relax (3)\,\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 \relax (2)-3\right )}^2\,\left (x^2-x+2\,\ln \relax (2)+3\right )+48} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.24, size = 22, normalized size = 0.85 \begin {gather*} - \frac {5}{- x + \log {\left (- x^{2} + x - 3 - 2 \log {\relax (2 )} \right )} + \log {\relax (3 )} + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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