3.86.33 \(\int \frac {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} (16 x+8 x^2+x^3)} (-80-80 x-15 x^2+e^x (-7-16 x-3 x^2)+(16+16 x+3 x^2) \log (3))}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{\left (2+\frac {1}{3} (-2+x)\right )^2 x}+x}{-5-e^x+\log (3)} \]

________________________________________________________________________________________

Rubi [F]  time = 4.38, 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 {-45+e^x (-9+9 x)+9 \log (3)+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{225+9 e^{2 x}+e^x (90-18 \log (3))-90 \log (3)+9 \log ^2(3)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-45 + E^x*(-9 + 9*x) + 9*Log[3] + E^((16*x + 8*x^2 + x^3)/9)*(-80 - 80*x - 15*x^2 + E^x*(-7 - 16*x - 3*x^
2) + (16 + 16*x + 3*x^2)*Log[3]))/(225 + 9*E^(2*x) + E^x*(90 - 18*Log[3]) - 90*Log[3] + 9*Log[3]^2),x]

[Out]

(1 - x)^2/(2*(5 - Log[3])) + x/(5 - Log[3]) - x^2/(2*(5 - Log[3])) - x/(5 + E^x - Log[3]) + ((1 - x)*Log[1 + E
^x/(5 - Log[3])])/(5 - Log[3]) + (x*Log[1 + E^x/(5 - Log[3])])/(5 - Log[3]) - Log[5 + E^x - Log[3]]/(5 - Log[3
]) + (7*Defer[Int][E^((x*(4 + x)^2)/9)/(-E^x - 5*(1 - Log[3]/5)), x])/9 + (16*Defer[Int][(E^((x*(4 + x)^2)/9)*
x)/(-E^x - 5*(1 - Log[3]/5)), x])/9 + Defer[Int][(E^((x*(4 + x)^2)/9)*x^2)/(-E^x - 5*(1 - Log[3]/5)), x]/3 - (
5 - Log[3])*Defer[Int][E^((x*(4 + x)^2)/9)/(E^x + 5*(1 - Log[3]/5))^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x (-9+9 x)-45 \left (1-\frac {\log (3)}{5}\right )+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{9 \left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx\\ &=\frac {1}{9} \int \frac {e^x (-9+9 x)-45 \left (1-\frac {\log (3)}{5}\right )+e^{\frac {1}{9} \left (16 x+8 x^2+x^3\right )} \left (-80-80 x-15 x^2+e^x \left (-7-16 x-3 x^2\right )+\left (16+16 x+3 x^2\right ) \log (3)\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx\\ &=\frac {1}{9} \int \left (\frac {9 \left (-e^x+e^x x-5 \left (1-\frac {\log (3)}{5}\right )\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}+\frac {e^{\frac {1}{9} x (4+x)^2} \left (-7 e^x-16 e^x x-3 e^x x^2-80 \left (1-\frac {\log (3)}{5}\right )-80 x \left (1-\frac {\log (3)}{5}\right )-15 x^2 \left (1-\frac {\log (3)}{5}\right )\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}\right ) \, dx\\ &=\frac {1}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2} \left (-7 e^x-16 e^x x-3 e^x x^2-80 \left (1-\frac {\log (3)}{5}\right )-80 x \left (1-\frac {\log (3)}{5}\right )-15 x^2 \left (1-\frac {\log (3)}{5}\right )\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx+\int \frac {-e^x+e^x x-5 \left (1-\frac {\log (3)}{5}\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx\\ &=\frac {1}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2} \left (-e^x \left (7+16 x+3 x^2\right )+\left (16+16 x+3 x^2\right ) (-5+\log (3))\right )}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx+\int \left (\frac {-1+x}{e^x+5 \left (1-\frac {\log (3)}{5}\right )}+\frac {x (-5+\log (3))}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}\right ) \, dx\\ &=\frac {1}{9} \int \left (\frac {e^{\frac {1}{9} x (4+x)^2} \left (-7-16 x-3 x^2\right )}{e^x+5 \left (1-\frac {\log (3)}{5}\right )}+\frac {9 e^{\frac {1}{9} x (4+x)^2} (-5+\log (3))}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2}\right ) \, dx+(-5+\log (3)) \int \frac {x}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx+\int \frac {-1+x}{e^x+5 \left (1-\frac {\log (3)}{5}\right )} \, dx\\ &=\frac {(1-x)^2}{2 (5-\log (3))}+\frac {1}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2} \left (-7-16 x-3 x^2\right )}{e^x+5 \left (1-\frac {\log (3)}{5}\right )} \, dx-\frac {\int \frac {e^x (-1+x)}{e^x+5 \left (1-\frac {\log (3)}{5}\right )} \, dx}{5-\log (3)}+(-5+\log (3)) \int \frac {e^{\frac {1}{9} x (4+x)^2}}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx+\int \frac {e^x x}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx-\int \frac {x}{e^x+5 \left (1-\frac {\log (3)}{5}\right )} \, dx\\ &=\frac {(1-x)^2}{2 (5-\log (3))}-\frac {x^2}{2 (5-\log (3))}-\frac {x}{5+e^x-\log (3)}+\frac {(1-x) \log \left (1+\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}+\frac {1}{9} \int \left (\frac {7 e^{\frac {1}{9} x (4+x)^2}}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )}+\frac {16 e^{\frac {1}{9} x (4+x)^2} x}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )}+\frac {3 e^{\frac {1}{9} x (4+x)^2} x^2}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )}\right ) \, dx+\frac {\int \frac {e^x x}{e^x+5 \left (1-\frac {\log (3)}{5}\right )} \, dx}{5-\log (3)}+\frac {\int \log \left (1+\frac {e^x}{5 \left (1-\frac {\log (3)}{5}\right )}\right ) \, dx}{5-\log (3)}+(-5+\log (3)) \int \frac {e^{\frac {1}{9} x (4+x)^2}}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx+\int \frac {1}{e^x+5 \left (1-\frac {\log (3)}{5}\right )} \, dx\\ &=\frac {(1-x)^2}{2 (5-\log (3))}-\frac {x^2}{2 (5-\log (3))}-\frac {x}{5+e^x-\log (3)}+\frac {(1-x) \log \left (1+\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}+\frac {x \log \left (1+\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}+\frac {1}{3} \int \frac {e^{\frac {1}{9} x (4+x)^2} x^2}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+\frac {7}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2}}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+\frac {16}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2} x}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx-\frac {\int \log \left (1+\frac {e^x}{5 \left (1-\frac {\log (3)}{5}\right )}\right ) \, dx}{5-\log (3)}+\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5 \left (1-\frac {\log (3)}{5}\right )}\right )}{x} \, dx,x,e^x\right )}{5-\log (3)}+(-5+\log (3)) \int \frac {e^{\frac {1}{9} x (4+x)^2}}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx+\operatorname {Subst}\left (\int \frac {1}{x (5+x-\log (3))} \, dx,x,e^x\right )\\ &=\frac {(1-x)^2}{2 (5-\log (3))}-\frac {x^2}{2 (5-\log (3))}-\frac {x}{5+e^x-\log (3)}+\frac {(1-x) \log \left (1+\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}+\frac {x \log \left (1+\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}-\frac {\text {Li}_2\left (-\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}+\frac {1}{3} \int \frac {e^{\frac {1}{9} x (4+x)^2} x^2}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+\frac {7}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2}}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+\frac {16}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2} x}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+\frac {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{5-\log (3)}-\frac {\operatorname {Subst}\left (\int \frac {1}{5+x-\log (3)} \, dx,x,e^x\right )}{5-\log (3)}-\frac {\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{5 \left (1-\frac {\log (3)}{5}\right )}\right )}{x} \, dx,x,e^x\right )}{5-\log (3)}+(-5+\log (3)) \int \frac {e^{\frac {1}{9} x (4+x)^2}}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx\\ &=\frac {(1-x)^2}{2 (5-\log (3))}+\frac {x}{5-\log (3)}-\frac {x^2}{2 (5-\log (3))}-\frac {x}{5+e^x-\log (3)}+\frac {(1-x) \log \left (1+\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}+\frac {x \log \left (1+\frac {e^x}{5-\log (3)}\right )}{5-\log (3)}-\frac {\log \left (5+e^x-\log (3)\right )}{5-\log (3)}+\frac {1}{3} \int \frac {e^{\frac {1}{9} x (4+x)^2} x^2}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+\frac {7}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2}}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+\frac {16}{9} \int \frac {e^{\frac {1}{9} x (4+x)^2} x}{-e^x-5 \left (1-\frac {\log (3)}{5}\right )} \, dx+(-5+\log (3)) \int \frac {e^{\frac {1}{9} x (4+x)^2}}{\left (e^x+5 \left (1-\frac {\log (3)}{5}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 2.15, size = 27, normalized size = 0.93 \begin {gather*} -\frac {e^{\frac {1}{9} x (4+x)^2}+x}{5+e^x-\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-45 + E^x*(-9 + 9*x) + 9*Log[3] + E^((16*x + 8*x^2 + x^3)/9)*(-80 - 80*x - 15*x^2 + E^x*(-7 - 16*x
- 3*x^2) + (16 + 16*x + 3*x^2)*Log[3]))/(225 + 9*E^(2*x) + E^x*(90 - 18*Log[3]) - 90*Log[3] + 9*Log[3]^2),x]

[Out]

-((E^((x*(4 + x)^2)/9) + x)/(5 + E^x - Log[3]))

________________________________________________________________________________________

fricas [A]  time = 0.65, size = 29, normalized size = 1.00 \begin {gather*} -\frac {x + e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )}}{e^{x} - \log \relax (3) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*log(3)-15*x^2-80*x-80)*exp(1/9*x^3+8/9*x^2+16/9*x)+(9*x-9)*
exp(x)+9*log(3)-45)/(9*exp(x)^2+(-18*log(3)+90)*exp(x)+9*log(3)^2-90*log(3)+225),x, algorithm="fricas")

[Out]

-(x + e^(1/9*x^3 + 8/9*x^2 + 16/9*x))/(e^x - log(3) + 5)

________________________________________________________________________________________

giac [B]  time = 0.25, size = 140, normalized size = 4.83 \begin {gather*} -\frac {x \log \relax (3) + e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )} \log \relax (3) - e^{x} \log \left (e^{x} - \log \relax (3) + 5\right ) + \log \relax (3) \log \left (e^{x} - \log \relax (3) + 5\right ) + e^{x} \log \left (-e^{x} + \log \relax (3) - 5\right ) - \log \relax (3) \log \left (-e^{x} + \log \relax (3) - 5\right ) - 5 \, x - 5 \, e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )} - 5 \, \log \left (e^{x} - \log \relax (3) + 5\right ) + 5 \, \log \left (-e^{x} + \log \relax (3) - 5\right )}{e^{x} \log \relax (3) - \log \relax (3)^{2} - 5 \, e^{x} + 10 \, \log \relax (3) - 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*log(3)-15*x^2-80*x-80)*exp(1/9*x^3+8/9*x^2+16/9*x)+(9*x-9)*
exp(x)+9*log(3)-45)/(9*exp(x)^2+(-18*log(3)+90)*exp(x)+9*log(3)^2-90*log(3)+225),x, algorithm="giac")

[Out]

-(x*log(3) + e^(1/9*x^3 + 8/9*x^2 + 16/9*x)*log(3) - e^x*log(e^x - log(3) + 5) + log(3)*log(e^x - log(3) + 5)
+ e^x*log(-e^x + log(3) - 5) - log(3)*log(-e^x + log(3) - 5) - 5*x - 5*e^(1/9*x^3 + 8/9*x^2 + 16/9*x) - 5*log(
e^x - log(3) + 5) + 5*log(-e^x + log(3) - 5))/(e^x*log(3) - log(3)^2 - 5*e^x + 10*log(3) - 25)

________________________________________________________________________________________

maple [A]  time = 0.26, size = 29, normalized size = 1.00

method	result	size
norman	\(\frac {x +{\mathrm e}^{\frac {1}{9} x^{3}+\frac {8}{9} x^{2}+\frac {16}{9} x}}{\ln \relax (3)-{\mathrm e}^{x}-5}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*ln(3)-15*x^2-80*x-80)*exp(1/9*x^3+8/9*x^2+16/9*x)+(9*x-9)*exp(x)+
9*ln(3)-45)/(9*exp(x)^2+(-18*ln(3)+90)*exp(x)+9*ln(3)^2-90*ln(3)+225),x,method=_RETURNVERBOSE)

[Out]

(x+exp(1/9*x^3+8/9*x^2+16/9*x))/(ln(3)-exp(x)-5)

________________________________________________________________________________________

maxima [B]  time = 0.55, size = 205, normalized size = 7.07 \begin {gather*} {\left (\frac {x}{\log \relax (3)^{2} - 10 \, \log \relax (3) + 25} - \frac {\log \left (e^{x} - \log \relax (3) + 5\right )}{\log \relax (3)^{2} - 10 \, \log \relax (3) + 25} - \frac {1}{{\left (\log \relax (3) - 5\right )} e^{x} - \log \relax (3)^{2} + 10 \, \log \relax (3) - 25}\right )} \log \relax (3) - \frac {x e^{x}}{{\left (\log \relax (3) - 5\right )} e^{x} - \log \relax (3)^{2} + 10 \, \log \relax (3) - 25} - \frac {5 \, x}{\log \relax (3)^{2} - 10 \, \log \relax (3) + 25} - \frac {e^{\left (\frac {1}{9} \, x^{3} + \frac {8}{9} \, x^{2} + \frac {16}{9} \, x\right )}}{e^{x} - \log \relax (3) + 5} + \frac {5 \, \log \left (e^{x} - \log \relax (3) + 5\right )}{\log \relax (3)^{2} - 10 \, \log \relax (3) + 25} + \frac {\log \left (e^{x} - \log \relax (3) + 5\right )}{\log \relax (3) - 5} + \frac {5}{{\left (\log \relax (3) - 5\right )} e^{x} - \log \relax (3)^{2} + 10 \, \log \relax (3) - 25} + \frac {1}{e^{x} - \log \relax (3) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x^2-16*x-7)*exp(x)+(3*x^2+16*x+16)*log(3)-15*x^2-80*x-80)*exp(1/9*x^3+8/9*x^2+16/9*x)+(9*x-9)*
exp(x)+9*log(3)-45)/(9*exp(x)^2+(-18*log(3)+90)*exp(x)+9*log(3)^2-90*log(3)+225),x, algorithm="maxima")

[Out]

(x/(log(3)^2 - 10*log(3) + 25) - log(e^x - log(3) + 5)/(log(3)^2 - 10*log(3) + 25) - 1/((log(3) - 5)*e^x - log
(3)^2 + 10*log(3) - 25))*log(3) - x*e^x/((log(3) - 5)*e^x - log(3)^2 + 10*log(3) - 25) - 5*x/(log(3)^2 - 10*lo
g(3) + 25) - e^(1/9*x^3 + 8/9*x^2 + 16/9*x)/(e^x - log(3) + 5) + 5*log(e^x - log(3) + 5)/(log(3)^2 - 10*log(3)
+ 25) + log(e^x - log(3) + 5)/(log(3) - 5) + 5/((log(3) - 5)*e^x - log(3)^2 + 10*log(3) - 25) + 1/(e^x - log(
3) + 5)

________________________________________________________________________________________

mupad [B]  time = 6.63, size = 225, normalized size = 7.76 \begin {gather*} \frac {5\,\ln \left ({\mathrm {e}}^x-\ln \relax (3)+5\right )}{{\ln \relax (3)}^2-10\,\ln \relax (3)+25}+\frac {\frac {{\mathrm {e}}^x}{\ln \relax (3)-5}-\frac {x\,{\mathrm {e}}^x}{\ln \relax (3)-5}}{{\mathrm {e}}^x-\ln \relax (3)+5}+\frac {\frac {5\,x}{\ln \relax (3)-5}+\frac {5\,{\mathrm {e}}^x}{{\left (\ln \relax (3)-5\right )}^2}-\frac {5\,x\,{\mathrm {e}}^x}{{\ln \relax (3)}^2-10\,\ln \relax (3)+25}}{{\mathrm {e}}^x-\ln \relax (3)+5}-\frac {{\mathrm {e}}^{\frac {x^3}{9}+\frac {8\,x^2}{9}+\frac {16\,x}{9}}}{{\mathrm {e}}^x-\ln \relax (3)+5}-\frac {\frac {x\,\ln \relax (3)}{\ln \relax (3)-5}+\frac {{\mathrm {e}}^x\,\ln \relax (3)}{{\left (\ln \relax (3)-5\right )}^2}-\frac {x\,{\mathrm {e}}^x\,\ln \relax (3)}{{\ln \relax (3)}^2-10\,\ln \relax (3)+25}}{{\mathrm {e}}^x-\ln \relax (3)+5}+\frac {\ln \left ({\mathrm {e}}^x-\ln \relax (3)+5\right )}{\ln \relax (3)-5}-\frac {\ln \relax (3)\,\ln \left ({\mathrm {e}}^x-\ln \relax (3)+5\right )}{{\ln \relax (3)}^2-10\,\ln \relax (3)+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*log(3) + exp(x)*(9*x - 9) - exp((16*x)/9 + (8*x^2)/9 + x^3/9)*(80*x - log(3)*(16*x + 3*x^2 + 16) + exp(
x)*(16*x + 3*x^2 + 7) + 15*x^2 + 80) - 45)/(9*exp(2*x) - 90*log(3) - exp(x)*(18*log(3) - 90) + 9*log(3)^2 + 22
5),x)

[Out]

(5*log(exp(x) - log(3) + 5))/(log(3)^2 - 10*log(3) + 25) + (exp(x)/(log(3) - 5) - (x*exp(x))/(log(3) - 5))/(ex
p(x) - log(3) + 5) + ((5*x)/(log(3) - 5) + (5*exp(x))/(log(3) - 5)^2 - (5*x*exp(x))/(log(3)^2 - 10*log(3) + 25
))/(exp(x) - log(3) + 5) - exp((16*x)/9 + (8*x^2)/9 + x^3/9)/(exp(x) - log(3) + 5) - ((x*log(3))/(log(3) - 5)
+ (exp(x)*log(3))/(log(3) - 5)^2 - (x*exp(x)*log(3))/(log(3)^2 - 10*log(3) + 25))/(exp(x) - log(3) + 5) + log(
exp(x) - log(3) + 5)/(log(3) - 5) - (log(3)*log(exp(x) - log(3) + 5))/(log(3)^2 - 10*log(3) + 25)

________________________________________________________________________________________

sympy [A]  time = 0.32, size = 37, normalized size = 1.28 \begin {gather*} - \frac {x}{e^{x} - \log {\relax (3 )} + 5} - \frac {e^{\frac {x^{3}}{9} + \frac {8 x^{2}}{9} + \frac {16 x}{9}}}{e^{x} - \log {\relax (3 )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x**2-16*x-7)*exp(x)+(3*x**2+16*x+16)*ln(3)-15*x**2-80*x-80)*exp(1/9*x**3+8/9*x**2+16/9*x)+(9*x
-9)*exp(x)+9*ln(3)-45)/(9*exp(x)**2+(-18*ln(3)+90)*exp(x)+9*ln(3)**2-90*ln(3)+225),x)

[Out]

-x/(exp(x) - log(3) + 5) - exp(x**3/9 + 8*x**2/9 + 16*x/9)/(exp(x) - log(3) + 5)

________________________________________________________________________________________