3.21.59 \(\int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} (2+2 x-2 x^2)+e^5 (1+x) \log (2)+(-e^5 x-2 e^{5+x} x) \log (x)}{80 x+20 e^{10+2 x} x+e^5 (-360 x+40 x^2)+e^{10} (405 x-90 x^2+5 x^3)+(40 e^5 x+e^{10} (-90 x+10 x^2)) \log (2)+5 e^{10} x \log ^2(2)+e^x (80 e^5 x+e^{10} (-180 x+20 x^2)+20 e^{10} x \log (2))} \, dx\)

Optimal. Leaf size=28 \[ \frac {x+\log (x)}{5 \left (4+e^5 \left (-5+2 \left (-2+e^x\right )+x+\log (2)\right )\right )} \]

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Rubi [F]  time = 4.88, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{80 x+20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+5 e^{10} x \log ^2(2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4 + E^5*(-9 - 8*x) + 4*x + E^(5 + x)*(2 + 2*x - 2*x^2) + E^5*(1 + x)*Log[2] + (-(E^5*x) - 2*E^(5 + x)*x)*
Log[x])/(80*x + 20*E^(10 + 2*x)*x + E^5*(-360*x + 40*x^2) + E^10*(405*x - 90*x^2 + 5*x^3) + (40*E^5*x + E^10*(
-90*x + 10*x^2))*Log[2] + 5*E^10*x*Log[2]^2 + E^x*(80*E^5*x + E^10*(-180*x + 20*x^2) + 20*E^10*x*Log[2])),x]

[Out]

(Log[x]*Defer[Int][(-4 - 2*E^(5 + x) - E^5*(-9 + x + Log[2]))^(-1), x])/5 + Defer[Int][x/(-4 - 2*E^(5 + x) - E
^5*(-9 + x + Log[2])), x]/5 + ((4 - E^5*(10 - Log[2]))*Log[x]*Defer[Int][(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[
2]))^(-2), x])/5 + ((4 - E^5*(10 - Log[2]))*Defer[Int][x/(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[2]))^2, x])/5 +
(E^5*Log[x]*Defer[Int][x/(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[2]))^2, x])/5 + (E^5*Defer[Int][x^2/(4 + 2*E^(5
+ x) + E^5*(-9 + x + Log[2]))^2, x])/5 + Defer[Int][(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[2]))^(-1), x]/5 + Def
er[Int][1/(x*(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[2]))), x]/5 - ((4 - E^5*(10 - Log[2]))*Defer[Int][Defer[Int]
[(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[2]))^(-2), x]/x, x])/5 - (E^5*Defer[Int][Defer[Int][x/(4 + 2*E^(5 + x) +
E^5*(-9 + x + Log[2]))^2, x]/x, x])/5 - Defer[Int][Defer[Int][-(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[2]))^(-1)
, x]/x, x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4+e^5 (-9-8 x)+4 x+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (1+x) \log (2)+\left (-e^5 x-2 e^{5+x} x\right ) \log (x)}{20 e^{10+2 x} x+e^5 \left (-360 x+40 x^2\right )+e^{10} \left (405 x-90 x^2+5 x^3\right )+\left (40 e^5 x+e^{10} \left (-90 x+10 x^2\right )\right ) \log (2)+e^x \left (80 e^5 x+e^{10} \left (-180 x+20 x^2\right )+20 e^{10} x \log (2)\right )+x \left (80+5 e^{10} \log ^2(2)\right )} \, dx\\ &=\int \frac {4 (1+x)+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (-9+x (-8+\log (2))+\log (2))-e^5 \left (1+2 e^x\right ) x \log (x)}{5 x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {4 (1+x)+e^{5+x} \left (2+2 x-2 x^2\right )+e^5 (-9+x (-8+\log (2))+\log (2))-e^5 \left (1+2 e^x\right ) x \log (x)}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx\\ &=\frac {1}{5} \int \left (\frac {\left (4-10 e^5+e^5 x+e^5 \log (2)\right ) (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}+\frac {1+x-x^2-x \log (x)}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )}\right ) \, dx\\ &=\frac {1}{5} \int \frac {\left (4-10 e^5+e^5 x+e^5 \log (2)\right ) (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \int \frac {1+x-x^2-x \log (x)}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )} \, dx\\ &=\frac {1}{5} \int \frac {1+x-x^2-x \log (x)}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx+\frac {1}{5} \int \left (\frac {10 e^5 \left (1-\frac {4+e^5 \log (2)}{10 e^5}\right ) (-x-\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}+\frac {e^5 x (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}\right ) \, dx\\ &=\frac {1}{5} \int \left (\frac {x}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}+\frac {1}{2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}+\frac {1}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )}+\frac {\log (x)}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )}\right ) \, dx+\frac {1}{5} e^5 \int \frac {x (x+\log (x))}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \frac {-x-\log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {x}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {1}{2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {1}{x \left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )} \, dx+\frac {1}{5} \int \frac {\log (x)}{-2 e^{5+x}-e^5 x-4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )} \, dx+\frac {1}{5} e^5 \int \frac {x (x+\log (x))}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \frac {-x-\log (x)}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {x}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx-\frac {1}{5} \int \frac {\int -\frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx}{x} \, dx+\frac {1}{5} e^5 \int \left (\frac {x^2}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}+\frac {x \log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}\right ) \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \left (-\frac {x}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}-\frac {\log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2}\right ) \, dx+\frac {1}{5} \log (x) \int \frac {1}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx\\ &=\frac {1}{5} \int \frac {x}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {\int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx}{x} \, dx+\frac {1}{5} e^5 \int \frac {x^2}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} e^5 \int \frac {x \log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \left (4-e^5 (10-\log (2))\right ) \int \frac {x}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \left (4-e^5 (10-\log (2))\right ) \int \frac {\log (x)}{\left (2 e^{5+x}+e^5 x+4 \left (1+\frac {1}{4} e^5 (-9+\log (2))\right )\right )^2} \, dx+\frac {1}{5} \log (x) \int \frac {1}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx\\ &=\frac {1}{5} \int \frac {x}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \int \frac {1}{x \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \, dx+\frac {1}{5} \int \frac {\int \frac {1}{4+2 e^{5+x}+e^5 (-9+x+\log (2))} \, dx}{x} \, dx+\frac {1}{5} e^5 \int \frac {x^2}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx-\frac {1}{5} e^5 \int \frac {\int \frac {x}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx}{x} \, dx+\frac {1}{5} \left (4-e^5 (10-\log (2))\right ) \int \frac {x}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx+\frac {1}{5} \left (-4+e^5 (10-\log (2))\right ) \int \frac {\int \frac {1}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx}{x} \, dx+\frac {1}{5} \log (x) \int \frac {1}{-4-2 e^{5+x}-e^5 (-9+x+\log (2))} \, dx+\frac {1}{5} \left (e^5 \log (x)\right ) \int \frac {x}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx+\frac {1}{5} \left (\left (4-e^5 (10-\log (2))\right ) \log (x)\right ) \int \frac {1}{\left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.94, size = 28, normalized size = 1.00 \begin {gather*} \frac {x+\log (x)}{5 \left (4+2 e^{5+x}+e^5 (-9+x+\log (2))\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + E^5*(-9 - 8*x) + 4*x + E^(5 + x)*(2 + 2*x - 2*x^2) + E^5*(1 + x)*Log[2] + (-(E^5*x) - 2*E^(5 +
x)*x)*Log[x])/(80*x + 20*E^(10 + 2*x)*x + E^5*(-360*x + 40*x^2) + E^10*(405*x - 90*x^2 + 5*x^3) + (40*E^5*x +
E^10*(-90*x + 10*x^2))*Log[2] + 5*E^10*x*Log[2]^2 + E^x*(80*E^5*x + E^10*(-180*x + 20*x^2) + 20*E^10*x*Log[2])
),x]

[Out]

(x + Log[x])/(5*(4 + 2*E^(5 + x) + E^5*(-9 + x + Log[2])))

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fricas [A]  time = 0.59, size = 27, normalized size = 0.96 \begin {gather*} \frac {x + \log \relax (x)}{5 \, {\left ({\left (x - 9\right )} e^{5} + e^{5} \log \relax (2) + 2 \, e^{\left (x + 5\right )} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*log(x)+(-2*x^2+2*x+2)*exp(5)*exp(x)+(x+1)*exp(5)*log(2)+(-8*x-9)*exp(
5)+4*x+4)/(20*x*exp(5)^2*exp(x)^2+(20*x*exp(5)^2*log(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)
^2*log(2)^2+((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*log(2)+(5*x^3-90*x^2+405*x)*exp(5)^2+(40*x^2-360*x)*exp(5)+80
*x),x, algorithm="fricas")

[Out]

1/5*(x + log(x))/((x - 9)*e^5 + e^5*log(2) + 2*e^(x + 5) + 4)

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giac [A]  time = 1.09, size = 29, normalized size = 1.04 \begin {gather*} \frac {x + \log \relax (x)}{5 \, {\left (x e^{5} + e^{5} \log \relax (2) - 9 \, e^{5} + 2 \, e^{\left (x + 5\right )} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*log(x)+(-2*x^2+2*x+2)*exp(5)*exp(x)+(x+1)*exp(5)*log(2)+(-8*x-9)*exp(
5)+4*x+4)/(20*x*exp(5)^2*exp(x)^2+(20*x*exp(5)^2*log(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)
^2*log(2)^2+((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*log(2)+(5*x^3-90*x^2+405*x)*exp(5)^2+(40*x^2-360*x)*exp(5)+80
*x),x, algorithm="giac")

[Out]

1/5*(x + log(x))/(x*e^5 + e^5*log(2) - 9*e^5 + 2*e^(x + 5) + 4)

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maple [B]  time = 0.06, size = 55, normalized size = 1.96

method	result	size
risch	\(\frac {\ln \relax (x )}{10 \,{\mathrm e}^{5+x}+5 \,{\mathrm e}^{5} \ln \relax (2)+5 x \,{\mathrm e}^{5}-45 \,{\mathrm e}^{5}+20}+\frac {x}{10 \,{\mathrm e}^{5+x}+5 \,{\mathrm e}^{5} \ln \relax (2)+5 x \,{\mathrm e}^{5}-45 \,{\mathrm e}^{5}+20}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x*exp(5)*exp(x)-x*exp(5))*ln(x)+(-2*x^2+2*x+2)*exp(5)*exp(x)+(x+1)*exp(5)*ln(2)+(-8*x-9)*exp(5)+4*x+4
)/(20*x*exp(5)^2*exp(x)^2+(20*x*exp(5)^2*ln(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)^2*ln(2)^
2+((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*ln(2)+(5*x^3-90*x^2+405*x)*exp(5)^2+(40*x^2-360*x)*exp(5)+80*x),x,metho
d=_RETURNVERBOSE)

[Out]

1/5/(2*exp(5+x)+exp(5)*ln(2)+x*exp(5)-9*exp(5)+4)*ln(x)+1/5*x/(2*exp(5+x)+exp(5)*ln(2)+x*exp(5)-9*exp(5)+4)

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maxima [A]  time = 0.62, size = 27, normalized size = 0.96 \begin {gather*} \frac {x + \log \relax (x)}{5 \, {\left (x e^{5} + {\left (\log \relax (2) - 9\right )} e^{5} + 2 \, e^{\left (x + 5\right )} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*log(x)+(-2*x^2+2*x+2)*exp(5)*exp(x)+(x+1)*exp(5)*log(2)+(-8*x-9)*exp(
5)+4*x+4)/(20*x*exp(5)^2*exp(x)^2+(20*x*exp(5)^2*log(2)+(20*x^2-180*x)*exp(5)^2+80*x*exp(5))*exp(x)+5*x*exp(5)
^2*log(2)^2+((10*x^2-90*x)*exp(5)^2+40*x*exp(5))*log(2)+(5*x^3-90*x^2+405*x)*exp(5)^2+(40*x^2-360*x)*exp(5)+80
*x),x, algorithm="maxima")

[Out]

1/5*(x + log(x))/(x*e^5 + (log(2) - 9)*e^5 + 2*e^(x + 5) + 4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {4\,x+{\mathrm {e}}^{x+5}\,\left (-2\,x^2+2\,x+2\right )-\ln \relax (x)\,\left (2\,x\,{\mathrm {e}}^{x+5}+x\,{\mathrm {e}}^5\right )-{\mathrm {e}}^5\,\left (8\,x+9\right )+{\mathrm {e}}^5\,\ln \relax (2)\,\left (x+1\right )+4}{80\,x-{\mathrm {e}}^5\,\left (360\,x-40\,x^2\right )+{\mathrm {e}}^{10}\,\left (5\,x^3-90\,x^2+405\,x\right )+20\,x\,{\mathrm {e}}^{2\,x+10}-\ln \relax (2)\,\left ({\mathrm {e}}^{10}\,\left (90\,x-10\,x^2\right )-40\,x\,{\mathrm {e}}^5\right )+{\mathrm {e}}^x\,\left (80\,x\,{\mathrm {e}}^5-{\mathrm {e}}^{10}\,\left (180\,x-20\,x^2\right )+20\,x\,{\mathrm {e}}^{10}\,\ln \relax (2)\right )+5\,x\,{\mathrm {e}}^{10}\,{\ln \relax (2)}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x - log(x)*(x*exp(5) + 2*x*exp(5)*exp(x)) - exp(5)*(8*x + 9) + exp(5)*log(2)*(x + 1) + exp(5)*exp(x)*(2
*x - 2*x^2 + 2) + 4)/(80*x - exp(5)*(360*x - 40*x^2) + exp(10)*(405*x - 90*x^2 + 5*x^3) - log(2)*(exp(10)*(90*
x - 10*x^2) - 40*x*exp(5)) + exp(x)*(80*x*exp(5) - exp(10)*(180*x - 20*x^2) + 20*x*exp(10)*log(2)) + 5*x*exp(1
0)*log(2)^2 + 20*x*exp(2*x)*exp(10)),x)

[Out]

int((4*x + exp(x + 5)*(2*x - 2*x^2 + 2) - log(x)*(2*x*exp(x + 5) + x*exp(5)) - exp(5)*(8*x + 9) + exp(5)*log(2
)*(x + 1) + 4)/(80*x - exp(5)*(360*x - 40*x^2) + exp(10)*(405*x - 90*x^2 + 5*x^3) + 20*x*exp(2*x + 10) - log(2
)*(exp(10)*(90*x - 10*x^2) - 40*x*exp(5)) + exp(x)*(80*x*exp(5) - exp(10)*(180*x - 20*x^2) + 20*x*exp(10)*log(
2)) + 5*x*exp(10)*log(2)^2), x)

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sympy [A]  time = 0.38, size = 34, normalized size = 1.21 \begin {gather*} \frac {x + \log {\relax (x )}}{5 x e^{5} + 10 e^{5} e^{x} - 45 e^{5} + 20 + 5 e^{5} \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x*exp(5)*exp(x)-x*exp(5))*ln(x)+(-2*x**2+2*x+2)*exp(5)*exp(x)+(x+1)*exp(5)*ln(2)+(-8*x-9)*exp(5
)+4*x+4)/(20*x*exp(5)**2*exp(x)**2+(20*x*exp(5)**2*ln(2)+(20*x**2-180*x)*exp(5)**2+80*x*exp(5))*exp(x)+5*x*exp
(5)**2*ln(2)**2+((10*x**2-90*x)*exp(5)**2+40*x*exp(5))*ln(2)+(5*x**3-90*x**2+405*x)*exp(5)**2+(40*x**2-360*x)*
exp(5)+80*x),x)

[Out]

(x + log(x))/(5*x*exp(5) + 10*exp(5)*exp(x) - 45*exp(5) + 20 + 5*exp(5)*log(2))

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