∫ e(1+x)−e log2(9)___________________________________ 1−x−x2+x3+(6−6x2) log2(9)+(9+9x) log4(9)+((2−2x2) log2(9)+(6+6x) log4(9)) log(1+x)+(1+x) log4(9) log2(1+x) dx

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Rubi [A] time = 0.26, antiderivative size = 24, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 3, integrand size = 93, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.032, Rules used = {6688, 12, 6686} \begin {gather*} \frac {e}{-x+\log ^2(9) \log (x+1)+1+3 \log ^2(9)} \end {gather*}

Int[(E*(1 + x) - E*Log[9]^2)/(1 - x - x^2 + x^3 + (6 - 6*x^2)*Log[9]^2 + (9 + 9*x)*Log[9]^4 + ((2 - 2*x^2)*Log

[9]^2 + (6 + 6*x)*Log[9]^4)*Log[1 + x] + (1 + x)*Log[9]^4*Log[1 + x]^2),x]

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

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

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e \left (1+x-\log ^2(9)\right )}{(1+x) \left (1-x+3 \log ^2(9)+\log ^2(9) \log (1+x)\right )^2} \, dx\\ &=e \int \frac {1+x-\log ^2(9)}{(1+x) \left (1-x+3 \log ^2(9)+\log ^2(9) \log (1+x)\right )^2} \, dx\\ &=\frac {e}{1-x+3 \log ^2(9)+\log ^2(9) \log (1+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.03, size = 24, normalized size = 1.20 \begin {gather*} \frac {e}{1-x+3 \log ^2(9)+\log ^2(9) \log (1+x)} \end {gather*}

Integrate[(E*(1 + x) - E*Log[9]^2)/(1 - x - x^2 + x^3 + (6 - 6*x^2)*Log[9]^2 + (9 + 9*x)*Log[9]^4 + ((2 - 2*x^

2)*Log[9]^2 + (6 + 6*x)*Log[9]^4)*Log[1 + x] + (1 + x)*Log[9]^4*Log[1 + x]^2),x]

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fricas [A] time = 0.73, size = 26, normalized size = 1.30 \begin {gather*} \frac {e}{4 \, \log \relax (3)^{2} \log \left (x + 1\right ) + 12 \, \log \relax (3)^{2} - x + 1} \end {gather*}

integrate((-4*exp(1)*log(3)^2+(x+1)*exp(1))/(16*(x+1)*log(3)^4*log(x+1)^2+(16*(6*x+6)*log(3)^4+4*(-2*x^2+2)*lo

g(3)^2)*log(x+1)+16*(9*x+9)*log(3)^4+4*(-6*x^2+6)*log(3)^2+x^3-x^2-x+1),x, algorithm="fricas")

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

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giac [A] time = 0.26, size = 26, normalized size = 1.30 \begin {gather*} \frac {e}{4 \, \log \relax (3)^{2} \log \left (x + 1\right ) + 12 \, \log \relax (3)^{2} - x + 1} \end {gather*}

integrate((-4*exp(1)*log(3)^2+(x+1)*exp(1))/(16*(x+1)*log(3)^4*log(x+1)^2+(16*(6*x+6)*log(3)^4+4*(-2*x^2+2)*lo

g(3)^2)*log(x+1)+16*(9*x+9)*log(3)^4+4*(-6*x^2+6)*log(3)^2+x^3-x^2-x+1),x, algorithm="giac")

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

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method	result	size

norman	\(\frac {{\mathrm e}}{4 \ln \relax (3)^{2} \ln \left (x +1\right )+12 \ln \relax (3)^{2}-x +1}\)	\(27\)
risch	\(\frac {{\mathrm e}}{4 \ln \relax (3)^{2} \ln \left (x +1\right )+12 \ln \relax (3)^{2}-x +1}\)	\(27\)

int((-4*exp(1)*ln(3)^2+(x+1)*exp(1))/(16*(x+1)*ln(3)^4*ln(x+1)^2+(16*(6*x+6)*ln(3)^4+4*(-2*x^2+2)*ln(3)^2)*ln(

x+1)+16*(9*x+9)*ln(3)^4+4*(-6*x^2+6)*ln(3)^2+x^3-x^2-x+1),x,method=_RETURNVERBOSE)

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maxima [A] time = 0.60, size = 26, normalized size = 1.30 \begin {gather*} \frac {e}{4 \, \log \relax (3)^{2} \log \left (x + 1\right ) + 12 \, \log \relax (3)^{2} - x + 1} \end {gather*}

integrate((-4*exp(1)*log(3)^2+(x+1)*exp(1))/(16*(x+1)*log(3)^4*log(x+1)^2+(16*(6*x+6)*log(3)^4+4*(-2*x^2+2)*lo

g(3)^2)*log(x+1)+16*(9*x+9)*log(3)^4+4*(-6*x^2+6)*log(3)^2+x^3-x^2-x+1),x, algorithm="maxima")

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

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mupad [F] time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {4\,\mathrm {e}\,{\ln \relax (3)}^2-\mathrm {e}\,\left (x+1\right )}{\ln \left (x+1\right )\,\left (16\,{\ln \relax (3)}^4\,\left (6\,x+6\right )-4\,{\ln \relax (3)}^2\,\left (2\,x^2-2\right )\right )-x+16\,{\ln \relax (3)}^4\,\left (9\,x+9\right )-4\,{\ln \relax (3)}^2\,\left (6\,x^2-6\right )-x^2+x^3+16\,{\ln \left (x+1\right )}^2\,{\ln \relax (3)}^4\,\left (x+1\right )+1} \,d x \end {gather*}

int(-(4*exp(1)*log(3)^2 - exp(1)*(x + 1))/(log(x + 1)*(16*log(3)^4*(6*x + 6) - 4*log(3)^2*(2*x^2 - 2)) - x + 1

6*log(3)^4*(9*x + 9) - 4*log(3)^2*(6*x^2 - 6) - x^2 + x^3 + 16*log(x + 1)^2*log(3)^4*(x + 1) + 1),x)

int(-(4*exp(1)*log(3)^2 - exp(1)*(x + 1))/(log(x + 1)*(16*log(3)^4*(6*x + 6) - 4*log(3)^2*(2*x^2 - 2)) - x + 1

6*log(3)^4*(9*x + 9) - 4*log(3)^2*(6*x^2 - 6) - x^2 + x^3 + 16*log(x + 1)^2*log(3)^4*(x + 1) + 1), x)

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sympy [A] time = 0.18, size = 24, normalized size = 1.20 \begin {gather*} \frac {e}{- x + 4 \log {\relax (3 )}^{2} \log {\left (x + 1 \right )} + 1 + 12 \log {\relax (3 )}^{2}} \end {gather*}

integrate((-4*exp(1)*ln(3)**2+(x+1)*exp(1))/(16*(x+1)*ln(3)**4*ln(x+1)**2+(16*(6*x+6)*ln(3)**4+4*(-2*x**2+2)*l

n(3)**2)*ln(x+1)+16*(9*x+9)*ln(3)**4+4*(-6*x**2+6)*ln(3)**2+x**3-x**2-x+1),x)

E/(-x + 4*log(3)**2*log(x + 1) + 1 + 12*log(3)**2)

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3.23.49 \(\int \frac {e (1+x)-e \log ^2(9)}{1-x-x^2+x^3+(6-6 x^2) \log ^2(9)+(9+9 x) \log ^4(9)+((2-2 x^2) \log ^2(9)+(6+6 x) \log ^4(9)) \log (1+x)+(1+x) \log ^4(9) \log ^2(1+x)} \, dx\)