∫ −1+(−1−x+e2x(−5x−10x2)) log(x) log(log(x))_________________________________

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Rubi [A] time = 0.67, antiderivative size = 37, normalized size of antiderivative = 1.54, number of steps used = 9, number of rules used = 6, integrand size = 41, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.146, Rules used = {6742, 2176, 2194, 6688, 2302, 29} \begin {gather*} -x+\frac {5 e^{2 x}}{2}-\frac {5}{2} e^{2 x} (2 x+1)-\log (x)-\log (\log (\log (x))) \end {gather*}

Int[(-1 + (-1 - x + E^(2*x)*(-5*x - 10*x^2))*Log[x]*Log[Log[x]])/(x*Log[x]*Log[Log[x]]),x]

(5*E^(2*x))/2 - x - (5*E^(2*x)*(1 + 2*x))/2 - Log[x] - Log[Log[Log[x]]]

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m

*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x

)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] && !$UseGamma === True

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre

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

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

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

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Mathematica [A] time = 0.05, size = 22, normalized size = 0.92 \begin {gather*} -x-5 e^{2 x} x-\log (x)-\log (\log (\log (x))) \end {gather*}

Integrate[(-1 + (-1 - x + E^(2*x)*(-5*x - 10*x^2))*Log[x]*Log[Log[x]])/(x*Log[x]*Log[Log[x]]),x]

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fricas [A] time = 0.60, size = 21, normalized size = 0.88 \begin {gather*} -5 \, x e^{\left (2 \, x\right )} - x - \log \relax (x) - \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

integrate((((-10*x^2-5*x)*exp(x)^2-x-1)*log(x)*log(log(x))-1)/x/log(x)/log(log(x)),x, algorithm="fricas")

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giac [A] time = 0.17, size = 21, normalized size = 0.88 \begin {gather*} -5 \, x e^{\left (2 \, x\right )} - x - \log \relax (x) - \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

integrate((((-10*x^2-5*x)*exp(x)^2-x-1)*log(x)*log(log(x))-1)/x/log(x)/log(log(x)),x, algorithm="giac")

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method	result	size

default	$-5 x \,{\mathrm e}^{2 x}-x -\ln \relax (x )-\ln \left (\ln \left (\ln \relax (x )\right )\right )$	$22$
risch	$-5 x \,{\mathrm e}^{2 x}-x -\ln \relax (x )-\ln \left (\ln \left (\ln \relax (x )\right )\right )$	$22$

int((((-10*x^2-5*x)*exp(x)^2-x-1)*ln(x)*ln(ln(x))-1)/x/ln(x)/ln(ln(x)),x,method=_RETURNVERBOSE)

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maxima [A] time = 0.37, size = 31, normalized size = 1.29 \begin {gather*} -\frac {5}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - x - \frac {5}{2} \, e^{\left (2 \, x\right )} - \log \relax (x) - \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

integrate((((-10*x^2-5*x)*exp(x)^2-x-1)*log(x)*log(log(x))-1)/x/log(x)/log(log(x)),x, algorithm="maxima")

-5/2*(2*x - 1)*e^(2*x) - x - 5/2*e^(2*x) - log(x) - log(log(log(x)))

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mupad [B] time = 2.28, size = 21, normalized size = 0.88 \begin {gather*} -x-\ln \left (\ln \left (\ln \relax (x)\right )\right )-\ln \relax (x)-5\,x\,{\mathrm {e}}^{2\,x} \end {gather*}

int(-(log(log(x))*log(x)*(x + exp(2*x)*(5*x + 10*x^2) + 1) + 1)/(x*log(log(x))*log(x)),x)

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sympy [A] time = 0.38, size = 20, normalized size = 0.83 \begin {gather*} - 5 x e^{2 x} - x - \log {\relax (x )} - \log {\left (\log {\left (\log {\relax (x )} \right )} \right )} \end {gather*}

integrate((((-10*x**2-5*x)*exp(x)**2-x-1)*ln(x)*ln(ln(x))-1)/x/ln(x)/ln(ln(x)),x)

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3.37.98 \(\int \frac {-1+(-1-x+e^{2 x} (-5 x-10 x^2)) \log (x) \log (\log (x))}{x \log (x) \log (\log (x))} \, dx\)