3.12.52 \(\int \frac {e^5 x^2-2 e^{10} x^3+(3 e^5 x^2-4 e^{10} x^3) \log (\frac {3 x}{e})}{1-4 e^5 x+4 e^{10} x^2} \, dx\)

Optimal. Leaf size=20 \[ \frac {x^3 \log \left (\frac {3 x}{e}\right )}{\frac {1}{e^5}-2 x} \]

________________________________________________________________________________________

Rubi [B]  time = 0.39, antiderivative size = 78, normalized size of antiderivative = 3.90, number of steps used = 13, number of rules used = 10, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {27, 6688, 12, 6742, 77, 2357, 2295, 2304, 2314, 31} \begin {gather*} \frac {x^2}{2}-\frac {1}{2} x^2 \log (3 x)+\frac {x}{4 e^5}-\frac {1}{8 e^{10} \left (1-2 e^5 x\right )}+\frac {x \log (3 x)}{4 e^5 \left (1-2 e^5 x\right )}-\frac {x \log (3 x)}{4 e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*x^2 - 2*E^10*x^3 + (3*E^5*x^2 - 4*E^10*x^3)*Log[(3*x)/E])/(1 - 4*E^5*x + 4*E^10*x^2),x]

[Out]

x/(4*E^5) + x^2/2 - 1/(8*E^10*(1 - 2*E^5*x)) - (x*Log[3*x])/(4*E^5) - (x^2*Log[3*x])/2 + (x*Log[3*x])/(4*E^5*(
1 - 2*E^5*x))

Rule 12

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 6688

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

Rule 6742

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 0.09, size = 53, normalized size = 2.65 \begin {gather*} \frac {\left (1-2 e^5 x\right ) \log (x)-\left (1-2 e^5 x+8 e^{15} x^3\right ) (-1+\log (3 x))}{8 e^{10} \left (-1+2 e^5 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*x^2 - 2*E^10*x^3 + (3*E^5*x^2 - 4*E^10*x^3)*Log[(3*x)/E])/(1 - 4*E^5*x + 4*E^10*x^2),x]

[Out]

((1 - 2*E^5*x)*Log[x] - (1 - 2*E^5*x + 8*E^15*x^3)*(-1 + Log[3*x]))/(8*E^10*(-1 + 2*E^5*x))

________________________________________________________________________________________

fricas [A]  time = 0.80, size = 22, normalized size = 1.10 \begin {gather*} -\frac {x^{3} e^{5} \log \left (3 \, x e^{\left (-1\right )}\right )}{2 \, x e^{5} - 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3*exp(5)^2+3*x^2*exp(5))*log(3*x/exp(1))-2*x^3*exp(5)^2+x^2*exp(5))/(4*x^2*exp(5)^2-4*x*exp(5
)+1),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.42, size = 63, normalized size = 3.15 \begin {gather*} -\frac {8 \, x^{3} e^{15} \log \left (3 \, x\right ) - 8 \, x^{3} e^{15} - 2 \, x e^{5} \log \left (3 \, x\right ) + 2 \, x e^{5} \log \relax (x) + 2 \, x e^{5} + \log \left (3 \, x\right ) - \log \relax (x) - 1}{8 \, {\left (2 \, x e^{15} - e^{10}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3*exp(5)^2+3*x^2*exp(5))*log(3*x/exp(1))-2*x^3*exp(5)^2+x^2*exp(5))/(4*x^2*exp(5)^2-4*x*exp(5
)+1),x, algorithm="giac")

[Out]

-1/8*(8*x^3*e^15*log(3*x) - 8*x^3*e^15 - 2*x*e^5*log(3*x) + 2*x*e^5*log(x) + 2*x*e^5 + log(3*x) - log(x) - 1)/
(2*x*e^15 - e^10)

________________________________________________________________________________________

maple [A]  time = 0.26, size = 25, normalized size = 1.25

method	result	size
norman	\(-\frac {x^{3} {\mathrm e}^{5} \ln \left (3 \,{\mathrm e}^{-1} x \right )}{2 x \,{\mathrm e}^{5}-1}\)	\(25\)
risch	\(-\frac {\left (8 \,{\mathrm e}^{15} x^{3}-2 x \,{\mathrm e}^{5}+1\right ) {\mathrm e}^{-10} \ln \left (3 \,{\mathrm e}^{-1} x \right )}{8 \left (2 x \,{\mathrm e}^{5}-1\right )}-\frac {{\mathrm e}^{-10} \ln \relax (x )}{8}\)	\(41\)
derivativedivides	\(\frac {{\mathrm e} \left (-\frac {9 \,{\mathrm e}^{5} x^{2}}{4 \,{\mathrm e}^{6}}-\frac {9 \,{\mathrm e} \,{\mathrm e}^{5} x}{4 \left ({\mathrm e}^{6}\right )^{2}}-\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \ln \left (\frac {6 x \,{\mathrm e}^{6}}{{\mathrm e}}-3\right )}{8 \,{\mathrm e}^{6}}-\frac {9 \,{\mathrm e}^{5} {\mathrm e}^{6} x^{2} \ln \left (\frac {3 x}{{\mathrm e}}\right )}{2 \,{\mathrm e}^{12}}+\frac {9 \,{\mathrm e}^{5} {\mathrm e}^{6} x^{2}}{4 \,{\mathrm e}^{12}}-\frac {9 \,{\mathrm e} \,{\mathrm e}^{5} \left ({\mathrm e}^{6}\right )^{2} \ln \left (\frac {3 x}{{\mathrm e}}\right ) x}{\left ({\mathrm e}^{12}\right )^{2}}+\frac {9 \,{\mathrm e} \,{\mathrm e}^{5} \left ({\mathrm e}^{6}\right )^{2} x}{\left ({\mathrm e}^{12}\right )^{2}}+\frac {27 \,{\mathrm e} \,{\mathrm e}^{5} \ln \left (\frac {3 x}{{\mathrm e}}\right ) x}{4 \,{\mathrm e}^{12}}-\frac {27 \,{\mathrm e} \,{\mathrm e}^{5} x}{4 \,{\mathrm e}^{12}}+\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \ln \left (\frac {3 x}{{\mathrm e}}\right ) \ln \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}-\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \ln \left (\frac {3 x}{{\mathrm e}}\right ) \ln \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}+\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \dilog \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}-\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \dilog \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{9}\)	\(444\)
default	\(\frac {{\mathrm e} \left (-\frac {9 \,{\mathrm e}^{5} x^{2}}{4 \,{\mathrm e}^{6}}-\frac {9 \,{\mathrm e} \,{\mathrm e}^{5} x}{4 \left ({\mathrm e}^{6}\right )^{2}}-\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \ln \left (\frac {6 x \,{\mathrm e}^{6}}{{\mathrm e}}-3\right )}{8 \,{\mathrm e}^{6}}-\frac {9 \,{\mathrm e}^{5} {\mathrm e}^{6} x^{2} \ln \left (\frac {3 x}{{\mathrm e}}\right )}{2 \,{\mathrm e}^{12}}+\frac {9 \,{\mathrm e}^{5} {\mathrm e}^{6} x^{2}}{4 \,{\mathrm e}^{12}}-\frac {9 \,{\mathrm e} \,{\mathrm e}^{5} \left ({\mathrm e}^{6}\right )^{2} \ln \left (\frac {3 x}{{\mathrm e}}\right ) x}{\left ({\mathrm e}^{12}\right )^{2}}+\frac {9 \,{\mathrm e} \,{\mathrm e}^{5} \left ({\mathrm e}^{6}\right )^{2} x}{\left ({\mathrm e}^{12}\right )^{2}}+\frac {27 \,{\mathrm e} \,{\mathrm e}^{5} \ln \left (\frac {3 x}{{\mathrm e}}\right ) x}{4 \,{\mathrm e}^{12}}-\frac {27 \,{\mathrm e} \,{\mathrm e}^{5} x}{4 \,{\mathrm e}^{12}}+\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \ln \left (\frac {3 x}{{\mathrm e}}\right ) \ln \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}-\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \ln \left (\frac {3 x}{{\mathrm e}}\right ) \ln \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}+\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \dilog \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}+3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}-\frac {9 \left ({\mathrm e}\right )^{2} {\mathrm e}^{5} {\mathrm e}^{-12} \dilog \left (\frac {-\frac {6 x \,{\mathrm e}^{12}}{{\mathrm e}}+3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}{3 \,{\mathrm e}^{6}-3 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{16 \sqrt {\left ({\mathrm e}^{6}\right )^{2}-{\mathrm e}^{12}}}\right )}{9}\)	\(444\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^3*exp(5)^2+3*x^2*exp(5))*ln(3*x/exp(1))-2*x^3*exp(5)^2+x^2*exp(5))/(4*x^2*exp(5)^2-4*x*exp(5)+1),x,
method=_RETURNVERBOSE)

[Out]

-x^3*exp(5)*ln(3*x/exp(1))/(2*x*exp(5)-1)

________________________________________________________________________________________

maxima [B]  time = 0.78, size = 150, normalized size = 7.50 \begin {gather*} -\frac {1}{8} \, {\left (2 \, {\left (x^{2} e^{5} + 2 \, x\right )} e^{\left (-15\right )} + 3 \, e^{\left (-20\right )} \log \left (2 \, x e^{5} - 1\right ) - \frac {1}{2 \, x e^{25} - e^{20}}\right )} e^{10} + \frac {1}{8} \, {\left (2 \, x e^{\left (-10\right )} + 2 \, e^{\left (-15\right )} \log \left (2 \, x e^{5} - 1\right ) - \frac {1}{2 \, x e^{20} - e^{15}}\right )} e^{5} + \frac {1}{8} \, e^{\left (-10\right )} \log \left (2 \, x e^{5} - 1\right ) - \frac {4 \, x^{3} {\left (2 \, \log \relax (3) - 3\right )} e^{15} + 8 \, x^{3} e^{15} \log \relax (x) - 2 \, x^{2} e^{10} - 2 \, x {\left (\log \relax (3) - 2\right )} e^{5} + \log \relax (3) - 1}{8 \, {\left (2 \, x e^{15} - e^{10}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^3*exp(5)^2+3*x^2*exp(5))*log(3*x/exp(1))-2*x^3*exp(5)^2+x^2*exp(5))/(4*x^2*exp(5)^2-4*x*exp(5
)+1),x, algorithm="maxima")

[Out]

-1/8*(2*(x^2*e^5 + 2*x)*e^(-15) + 3*e^(-20)*log(2*x*e^5 - 1) - 1/(2*x*e^25 - e^20))*e^10 + 1/8*(2*x*e^(-10) +
2*e^(-15)*log(2*x*e^5 - 1) - 1/(2*x*e^20 - e^15))*e^5 + 1/8*e^(-10)*log(2*x*e^5 - 1) - 1/8*(4*x^3*(2*log(3) -
3)*e^15 + 8*x^3*e^15*log(x) - 2*x^2*e^10 - 2*x*(log(3) - 2)*e^5 + log(3) - 1)/(2*x*e^15 - e^10)

________________________________________________________________________________________

mupad [B]  time = 1.12, size = 22, normalized size = 1.10 \begin {gather*} -\frac {x^3\,{\mathrm {e}}^5\,\left (\ln \left (3\,x\right )-1\right )}{2\,x\,{\mathrm {e}}^5-1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2*exp(5) - 2*x^3*exp(10) + log(3*x*exp(-1))*(3*x^2*exp(5) - 4*x^3*exp(10)))/(4*x^2*exp(10) - 4*x*exp(5)
+ 1),x)

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 0.22, size = 44, normalized size = 2.20 \begin {gather*} - \frac {\log {\relax (x )}}{8 e^{10}} + \frac {\left (- 8 x^{3} e^{15} + 2 x e^{5} - 1\right ) \log {\left (\frac {3 x}{e} \right )}}{16 x e^{15} - 8 e^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**3*exp(5)**2+3*x**2*exp(5))*ln(3*x/exp(1))-2*x**3*exp(5)**2+x**2*exp(5))/(4*x**2*exp(5)**2-4*
x*exp(5)+1),x)

[Out]

-exp(-10)*log(x)/8 + (-8*x**3*exp(15) + 2*x*exp(5) - 1)*log(3*x*exp(-1))/(16*x*exp(15) - 8*exp(10))

________________________________________________________________________________________