∫ −30+10x+e3(−8+4x)+(−40+20x) log(x)+(−5−2e3−10 log(x)) log( 1_ 10(5+2e3+10 log(x))) _____________________________________________________________________________________________________________________

Optimal. Leaf size=31 \[ x \left (-4+x-\log \left (-1+\frac {e^3}{5}-x+\frac {1}{2} (3+2 x)+\log (x)\right )\right ) \]

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Rubi [A] time = 0.41, antiderivative size = 26, normalized size of antiderivative = 0.84, number of steps used = 12, number of rules used = 7, integrand size = 64, $\frac {\text {number of rules}}{\text {integrand size}}$ = 0.109, Rules used = {6741, 6742, 6688, 2299, 2178, 2549, 12} \begin {gather*} x^2-4 x-x \log \left (\frac {1}{10} \left (10 \log (x)+2 e^3+5\right )\right ) \end {gather*}

Int[(-30 + 10*x + E^3*(-8 + 4*x) + (-40 + 20*x)*Log[x] + (-5 - 2*E^3 - 10*Log[x])*Log[(5 + 2*E^3 + 10*Log[x])/

-4*x + x^2 - x*Log[(5 + 2*E^3 + 10*Log[x])/10]

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

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral

Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] && !$UseGamma === True

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

, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[1/n]

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[x*Simplify[D[u, x]/u], x], x] /; ProductQ[

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

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

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

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx\\ &=\int \left (\frac {2 \left (-15 \left (1+\frac {4 e^3}{15}\right )+5 \left (1+\frac {2 e^3}{5}\right ) x-20 \log (x)+10 x \log (x)\right )}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)}-\log \left (\frac {1}{10} \left (5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)\right )\right )\right ) \, dx\\ &=2 \int \frac {-15 \left (1+\frac {4 e^3}{15}\right )+5 \left (1+\frac {2 e^3}{5}\right ) x-20 \log (x)+10 x \log (x)}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx-\int \log \left (\frac {1}{10} \left (5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)\right )\right ) \, dx\\ &=-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+2 \int \frac {5 (-3+x)+2 e^3 (-2+x)+10 (-2+x) \log (x)}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx+\int \frac {10}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx\\ &=-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+2 \int \left (-2+x+\frac {5}{-5 \left (1+\frac {2 e^3}{5}\right )-10 \log (x)}\right ) \, dx+10 \int \frac {1}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx\\ &=-4 x+x^2-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+10 \int \frac {1}{-5 \left (1+\frac {2 e^3}{5}\right )-10 \log (x)} \, dx+10 \operatorname {Subst}\left (\int \frac {e^x}{5 \left (1+\frac {2 e^3}{5}\right )+10 x} \, dx,x,\log (x)\right )\\ &=-4 x+x^2+e^{-\frac {1}{2}-\frac {e^3}{5}} \text {Ei}\left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+10 \operatorname {Subst}\left (\int \frac {e^x}{-5 \left (1+\frac {2 e^3}{5}\right )-10 x} \, dx,x,\log (x)\right )\\ &=-4 x+x^2-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A] time = 0.18, size = 24, normalized size = 0.77 \begin {gather*} -4 x+x^2-x \log \left (\frac {1}{2}+\frac {e^3}{5}+\log (x)\right ) \end {gather*}

Integrate[(-30 + 10*x + E^3*(-8 + 4*x) + (-40 + 20*x)*Log[x] + (-5 - 2*E^3 - 10*Log[x])*Log[(5 + 2*E^3 + 10*Lo

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fricas [A] time = 1.16, size = 19, normalized size = 0.61 \begin {gather*} x^{2} - x \log \left (\frac {1}{5} \, e^{3} + \log \relax (x) + \frac {1}{2}\right ) - 4 \, x \end {gather*}

integrate(((-10*log(x)-2*exp(3)-5)*log(log(x)+1/5*exp(3)+1/2)+(20*x-40)*log(x)+(4*x-8)*exp(3)+10*x-30)/(10*log

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giac [A] time = 0.18, size = 25, normalized size = 0.81 \begin {gather*} x^{2} + x \log \left (10\right ) - x \log \left (2 \, e^{3} + 10 \, \log \relax (x) + 5\right ) - 4 \, x \end {gather*}

integrate(((-10*log(x)-2*exp(3)-5)*log(log(x)+1/5*exp(3)+1/2)+(20*x-40)*log(x)+(4*x-8)*exp(3)+10*x-30)/(10*log

x^2 + x*log(10) - x*log(2*e^3 + 10*log(x) + 5) - 4*x

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method	result	size

norman	$x^{2}-4 x -x \ln \left (\ln \relax (x )+\frac {{\mathrm e}^{3}}{5}+\frac {1}{2}\right )$	$20$
risch	$x^{2}-4 x -x \ln \left (\ln \relax (x )+\frac {{\mathrm e}^{3}}{5}+\frac {1}{2}\right )$	$20$

int(((-10*ln(x)-2*exp(3)-5)*ln(ln(x)+1/5*exp(3)+1/2)+(20*x-40)*ln(x)+(4*x-8)*exp(3)+10*x-30)/(10*ln(x)+2*exp(3

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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 4 \, e^{\left (-\frac {1}{5} \, e^{3} - \frac {1}{2}\right )} E_{1}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) \log \relax (x) - 2 \, e^{\left (-\frac {2}{5} \, e^{3} - 1\right )} E_{1}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) \log \relax (x) + x {\left (\log \relax (5) + \log \relax (2)\right )} - 4 \, e^{\left (-\frac {1}{5} \, e^{3} - \frac {1}{2}\right )} E_{2}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) + e^{\left (-\frac {2}{5} \, e^{3} - 1\right )} E_{2}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) + \frac {4}{5} \, e^{\left (-\frac {1}{5} \, e^{3} + \frac {5}{2}\right )} E_{1}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) + 3 \, e^{\left (-\frac {1}{5} \, e^{3} - \frac {1}{2}\right )} E_{1}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) - \frac {2}{5} \, e^{\left (-\frac {2}{5} \, e^{3} + 2\right )} E_{1}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) - e^{\left (-\frac {2}{5} \, e^{3} - 1\right )} E_{1}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) - x \log \left (2 \, e^{3} + 10 \, \log \relax (x) + 5\right ) + 10 \, \int \frac {1}{2 \, e^{3} + 10 \, \log \relax (x) + 5}\,{d x} \end {gather*}

integrate(((-10*log(x)-2*exp(3)-5)*log(log(x)+1/5*exp(3)+1/2)+(20*x-40)*log(x)+(4*x-8)*exp(3)+10*x-30)/(10*log

4*e^(-1/5*e^3 - 1/2)*exp_integral_e(1, -1/5*e^3 - log(x) - 1/2)*log(x) - 2*e^(-2/5*e^3 - 1)*exp_integral_e(1,

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

x) - 1/2) + e^(-2/5*e^3 - 1)*exp_integral_e(2, -2/5*e^3 - 2*log(x) - 1) + 4/5*e^(-1/5*e^3 + 5/2)*exp_integral_

e(1, -1/5*e^3 - log(x) - 1/2) + 3*e^(-1/5*e^3 - 1/2)*exp_integral_e(1, -1/5*e^3 - log(x) - 1/2) - 2/5*e^(-2/5*

e^3 + 2)*exp_integral_e(1, -2/5*e^3 - 2*log(x) - 1) - e^(-2/5*e^3 - 1)*exp_integral_e(1, -2/5*e^3 - 2*log(x) -

1) - x*log(2*e^3 + 10*log(x) + 5) + 10*integrate(1/(2*e^3 + 10*log(x) + 5), x)

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mupad [B] time = 4.66, size = 17, normalized size = 0.55 \begin {gather*} -x\,\left (\ln \left (\frac {{\mathrm {e}}^3}{5}+\ln \relax (x)+\frac {1}{2}\right )-x+4\right ) \end {gather*}

int((10*x - log(exp(3)/5 + log(x) + 1/2)*(2*exp(3) + 10*log(x) + 5) + log(x)*(20*x - 40) + exp(3)*(4*x - 8) -

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sympy [A] time = 0.39, size = 20, normalized size = 0.65 \begin {gather*} x^{2} - x \log {\left (\log {\relax (x )} + \frac {1}{2} + \frac {e^{3}}{5} \right )} - 4 x \end {gather*}

integrate(((-10*ln(x)-2*exp(3)-5)*ln(ln(x)+1/5*exp(3)+1/2)+(20*x-40)*ln(x)+(4*x-8)*exp(3)+10*x-30)/(10*ln(x)+2

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3.74.17 \(\int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+(-5-2 e^3-10 \log (x)) \log (\frac {1}{10} (5+2 e^3+10 \log (x)))}{5+2 e^3+10 \log (x)} \, dx\)