∫ −8+24x−24x2+8x3+(−8+8x) log(5ex)−8 log2(5ex)____________________________________

Optimal. Leaf size=21 \[ 4 \left (4+2 x+\frac {\log ^2\left (5 e^x\right )}{(-1+x)^2}\right ) \]

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Rubi [A] time = 0.11, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {6688, 2168, 31} \begin {gather*} 8 x+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2} \end {gather*}

Int[(-8 + 24*x - 24*x^2 + 8*x^3 + (-8 + 8*x)*Log[5*E^x] - 8*Log[5*E^x]^2)/(-1 + 3*x - 3*x^2 + x^3),x]

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Int[(u_)^(m_)*(v_)^(n_.), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(u^(m + 1)*v^

n)/(a*(m + 1)), x] - Dist[(b*n)/(a*(m + 1)), Int[u^(m + 1)*v^(n - 1), x], x] /; NeQ[b*u - a*v, 0]] /; FreeQ[{m

, n}, x] && PiecewiseLinearQ[u, v, x] && NeQ[m, -1] && ((LtQ[m, -1] && GtQ[n, 0] && !(ILtQ[m + n, -2] && (Fra

ctionQ[m] || GeQ[2*n + m + 1, 0]))) || (IGtQ[n, 0] && IGtQ[m, 0] && LeQ[n, m]) || (IGtQ[n, 0] && !IntegerQ[m]

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (8+\frac {8 \log \left (5 e^x\right )}{(-1+x)^2}-\frac {8 \log ^2\left (5 e^x\right )}{(-1+x)^3}\right ) \, dx\\ &=8 x+8 \int \frac {\log \left (5 e^x\right )}{(-1+x)^2} \, dx-8 \int \frac {\log ^2\left (5 e^x\right )}{(-1+x)^3} \, dx\\ &=8 x+\frac {8 \log \left (5 e^x\right )}{1-x}+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2}+8 \int \frac {1}{-1+x} \, dx-8 \int \frac {\log \left (5 e^x\right )}{(-1+x)^2} \, dx\\ &=8 x+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2}+8 \log (1-x)-8 \int \frac {1}{-1+x} \, dx\\ &=8 x+\frac {4 \log ^2\left (5 e^x\right )}{(1-x)^2}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.04, size = 20, normalized size = 0.95 \begin {gather*} -12+8 x+\frac {4 \log ^2\left (5 e^x\right )}{(-1+x)^2} \end {gather*}

Integrate[(-8 + 24*x - 24*x^2 + 8*x^3 + (-8 + 8*x)*Log[5*E^x] - 8*Log[5*E^x]^2)/(-1 + 3*x - 3*x^2 + x^3),x]

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fricas [A] time = 0.62, size = 36, normalized size = 1.71 \begin {gather*} \frac {4 \, {\left (2 \, x^{3} - 4 \, x^{2} + 2 \, x \log \relax (5) + \log \relax (5)^{2} + 4 \, x - 1\right )}}{x^{2} - 2 \, x + 1} \end {gather*}

integrate((-8*log(5*exp(x))^2+(8*x-8)*log(5*exp(x))+8*x^3-24*x^2+24*x-8)/(x^3-3*x^2+3*x-1),x, algorithm="frica

4*(2*x^3 - 4*x^2 + 2*x*log(5) + log(5)^2 + 4*x - 1)/(x^2 - 2*x + 1)

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giac [A] time = 0.20, size = 25, normalized size = 1.19 \begin {gather*} 8 \, x + \frac {4 \, {\left (2 \, x \log \relax (5) + \log \relax (5)^{2} + 2 \, x - 1\right )}}{{\left (x - 1\right )}^{2}} \end {gather*}

integrate((-8*log(5*exp(x))^2+(8*x-8)*log(5*exp(x))+8*x^3-24*x^2+24*x-8)/(x^3-3*x^2+3*x-1),x, algorithm="giac"

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method	result	size

default	\(8 x +\frac {4 \ln \left (5 \,{\mathrm e}^{x}\right )^{2}}{\left (x -1\right )^{2}}\)	\(19\)
risch	\(\frac {4 \ln \left ({\mathrm e}^{x}\right )^{2}}{x^{2}-2 x +1}+\frac {8 \ln \relax (5) \ln \left ({\mathrm e}^{x}\right )}{x^{2}-2 x +1}+\frac {8 x^{3}+4 \ln \relax (5)^{2}-16 x^{2}+8 x}{x^{2}-2 x +1}\)	\(67\)

int((-8*ln(5*exp(x))^2+(8*x-8)*ln(5*exp(x))+8*x^3-24*x^2+24*x-8)/(x^3-3*x^2+3*x-1),x,method=_RETURNVERBOSE)

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maxima [B] time = 0.37, size = 151, normalized size = 7.19 \begin {gather*} 8 \, x - \frac {4 \, {\left (2 \, x - 1\right )} \log \left (5 \, e^{x}\right )}{x^{2} - 2 \, x + 1} + \frac {4 \, \log \left (5 \, e^{x}\right )^{2}}{x^{2} - 2 \, x + 1} - \frac {4 \, {\left (6 \, x - 5\right )}}{x^{2} - 2 \, x + 1} + \frac {12 \, {\left (4 \, x - 3\right )}}{x^{2} - 2 \, x + 1} - \frac {12 \, {\left (2 \, x - 1\right )}}{x^{2} - 2 \, x + 1} + \frac {4 \, \log \left (5 \, e^{x}\right )}{x^{2} - 2 \, x + 1} + \frac {8 \, \log \left (5 \, e^{x}\right )}{x - 1} + \frac {4}{x^{2} - 2 \, x + 1} - 8 \, \log \left (2 \, x - 2\right ) + 8 \, \log \left (x - 1\right ) \end {gather*}

integrate((-8*log(5*exp(x))^2+(8*x-8)*log(5*exp(x))+8*x^3-24*x^2+24*x-8)/(x^3-3*x^2+3*x-1),x, algorithm="maxim

8*x - 4*(2*x - 1)*log(5*e^x)/(x^2 - 2*x + 1) + 4*log(5*e^x)^2/(x^2 - 2*x + 1) - 4*(6*x - 5)/(x^2 - 2*x + 1) +

12*(4*x - 3)/(x^2 - 2*x + 1) - 12*(2*x - 1)/(x^2 - 2*x + 1) + 4*log(5*e^x)/(x^2 - 2*x + 1) + 8*log(5*e^x)/(x -

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mupad [B] time = 3.46, size = 29, normalized size = 1.38 \begin {gather*} 8\,x+\frac {8\,\ln \relax (5)+8}{x-1}+\frac {4\,{\left (\ln \relax (5)+1\right )}^2}{{\left (x-1\right )}^2} \end {gather*}

int((24*x - 8*log(5*exp(x))^2 + log(5*exp(x))*(8*x - 8) - 24*x^2 + 8*x^3 - 8)/(3*x - 3*x^2 + x^3 - 1),x)

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sympy [A] time = 0.26, size = 27, normalized size = 1.29 \begin {gather*} 8 x + \frac {x \left (8 + 8 \log {\relax (5 )}\right ) - 4 + 4 \log {\relax (5 )}^{2}}{x^{2} - 2 x + 1} \end {gather*}

integrate((-8*ln(5*exp(x))**2+(8*x-8)*ln(5*exp(x))+8*x**3-24*x**2+24*x-8)/(x**3-3*x**2+3*x-1),x)

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3.56.43 \(\int \frac {-8+24 x-24 x^2+8 x^3+(-8+8 x) \log (5 e^x)-8 \log ^2(5 e^x)}{-1+3 x-3 x^2+x^3} \, dx\)