∫ −2 log(x)+(2+(2+x) log(x)) log(x+9e8x)____________________________

Optimal. Leaf size=25 \[ x-\log \left (\frac {36 \log ^2\left (x+9 e^8 x\right )}{x^2 \log ^2(x)}\right ) \]

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Rubi [A] time = 0.41, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 11, number of rules used = 5, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.119, Rules used = {2444, 6742, 43, 2302, 29} \begin {gather*} x+2 \log (x)+2 \log (\log (x))-2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right ) \end {gather*}

Int[(-2*Log[x] + (2 + (2 + x)*Log[x])*Log[x + 9*E^8*x])/(x*Log[x]*Log[x + 9*E^8*x]),x]

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d

*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L

Int[((a_.) + Log[(c_.)*(v_)^(n_.)]*(b_.))^(p_.)*(u_.), x_Symbol] :> Int[u*(a + b*Log[c*ExpandToSum[v, x]^n])^p

, x] /; FreeQ[{a, b, c, n, p}, x] && LinearQ[v, x] && !LinearMatchQ[v, x] && !(EqQ[n, 1] && MatchQ[c*v, (e_.

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log \left (x+9 e^8 x\right )}{x \log (x) \log \left (\left (1+9 e^8\right ) x\right )} \, dx\\ &=\int \left (\frac {2+2 \log (x)+x \log (x)}{x \log (x)}-\frac {2}{x \log \left (\left (1+9 e^8\right ) x\right )}\right ) \, dx\\ &=-\left (2 \int \frac {1}{x \log \left (\left (1+9 e^8\right ) x\right )} \, dx\right )+\int \frac {2+2 \log (x)+x \log (x)}{x \log (x)} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\left (1+9 e^8\right ) x\right )\right )\right )+\int \left (\frac {2+x}{x}+\frac {2}{x \log (x)}\right ) \, dx\\ &=-2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right )+2 \int \frac {1}{x \log (x)} \, dx+\int \frac {2+x}{x} \, dx\\ &=-2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right )+2 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )+\int \left (1+\frac {2}{x}\right ) \, dx\\ &=x+2 \log (x)+2 \log (\log (x))-2 \log \left (\log \left (\left (1+9 e^8\right ) x\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.03, size = 23, normalized size = 0.92 \begin {gather*} x+2 \log (x)+2 \log (\log (x))-2 \log \left (\log \left (x+9 e^8 x\right )\right ) \end {gather*}

Integrate[(-2*Log[x] + (2 + (2 + x)*Log[x])*Log[x + 9*E^8*x])/(x*Log[x]*Log[x + 9*E^8*x]),x]

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fricas [A] time = 0.77, size = 24, normalized size = 0.96 \begin {gather*} x + 2 \, \log \relax (x) - 2 \, \log \left (\log \relax (x) + \log \left (9 \, e^{8} + 1\right )\right ) + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="fricas")

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giac [A] time = 0.22, size = 28, normalized size = 1.12 \begin {gather*} x + 2 \, \log \relax (x) - 2 \, \log \left (-\log \relax (x) - \log \left (9 \, e^{8} + 1\right )\right ) + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="giac")

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method	result	size

risch	\(2 \ln \relax (x )+x\)	\(7\)
default	\(2 \ln \relax (x )+x +2 \ln \left (\ln \relax (x )\right )-2 \ln \left (\ln \left (\left (9 \,{\mathrm e}^{8}+1\right ) x \right )\right )\)	\(26\)
norman	\(x +2 \ln \left (9 x \,{\mathrm e}^{8}+x \right )+2 \ln \left (\ln \relax (x )\right )-2 \ln \left (\ln \left (9 x \,{\mathrm e}^{8}+x \right )\right )\)	\(33\)

int((((2+x)*ln(x)+2)*ln(9*x*exp(4)^2+x)-2*ln(x))/x/ln(x)/ln(9*x*exp(4)^2+x),x,method=_RETURNVERBOSE)

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maxima [A] time = 0.35, size = 22, normalized size = 0.88 \begin {gather*} x + 2 \, \log \relax (x) - 2 \, \log \left (\log \left (9 \, x e^{8} + x\right )\right ) + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

integrate((((2+x)*log(x)+2)*log(9*x*exp(4)^2+x)-2*log(x))/x/log(x)/log(9*x*exp(4)^2+x),x, algorithm="maxima")

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mupad [B] time = 4.60, size = 23, normalized size = 0.92 \begin {gather*} x+2\,\ln \left (\ln \relax (x)\right )-2\,\ln \left (\ln \left (x\,\left (9\,{\mathrm {e}}^8+1\right )\right )\right )+2\,\ln \relax (x) \end {gather*}

int(-(2*log(x) - log(x + 9*x*exp(8))*(log(x)*(x + 2) + 2))/(x*log(x + 9*x*exp(8))*log(x)),x)

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sympy [A] time = 0.35, size = 27, normalized size = 1.08 \begin {gather*} x + 2 \log {\relax (x )} - 2 \log {\left (\log {\relax (x )} + \log {\left (1 + 9 e^{8} \right )} \right )} + 2 \log {\left (\log {\relax (x )} \right )} \end {gather*}

integrate((((2+x)*ln(x)+2)*ln(9*x*exp(4)**2+x)-2*ln(x))/x/ln(x)/ln(9*x*exp(4)**2+x),x)

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3.73.66 \(\int \frac {-2 \log (x)+(2+(2+x) \log (x)) \log (x+9 e^8 x)}{x \log (x) \log (x+9 e^8 x)} \, dx\)