3.13.76 \(\int \frac {e^{\frac {3-x+3 \log (\frac {-20-5 x-2 x^2}{6 x})}{\log (\frac {-20-5 x-2 x^2}{6 x})}} (60-20 x-6 x^2+2 x^3+(-20 x-5 x^2-2 x^3) \log (\frac {-20-5 x-2 x^2}{6 x}))}{(20 x+5 x^2+2 x^3) \log ^2(\frac {-20-5 x-2 x^2}{6 x})} \, dx\) [1276]

3.13.76.1 Optimal result
3.13.76.2 Mathematica [A] (verified)
3.13.76.3 Rubi [F]
3.13.76.4 Maple [A] (verified)
3.13.76.5 Fricas [A] (verification not implemented)
3.13.76.6 Sympy [F(-2)]
3.13.76.7 Maxima [F]
3.13.76.8 Giac [A] (verification not implemented)
3.13.76.9 Mupad [B] (verification not implemented)

3.13.76.1 Optimal result

Integrand size = 133, antiderivative size = 31 \[ \int \frac {e^{\frac {3-x+3 \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}{\log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}} \left (60-20 x-6 x^2+2 x^3+\left (-20 x-5 x^2-2 x^3\right ) \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )\right )}{\left (20 x+5 x^2+2 x^3\right ) \log ^2\left (\frac {-20-5 x-2 x^2}{6 x}\right )} \, dx=e^{3+\frac {3-x}{\log \left (-\frac {3}{2}+\frac {1}{3} \left (2-\frac {10}{x}-x\right )\right )}} \]

output
exp((-x+3)/ln(-5/6-10/3/x-1/3*x)+3)
 
3.13.76.2 Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{\frac {3-x+3 \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}{\log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}} \left (60-20 x-6 x^2+2 x^3+\left (-20 x-5 x^2-2 x^3\right ) \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )\right )}{\left (20 x+5 x^2+2 x^3\right ) \log ^2\left (\frac {-20-5 x-2 x^2}{6 x}\right )} \, dx=e^{3+\frac {3-x}{\log \left (\frac {1}{6} \left (-5-\frac {20}{x}-2 x\right )\right )}} \]

input
Integrate[(E^((3 - x + 3*Log[(-20 - 5*x - 2*x^2)/(6*x)])/Log[(-20 - 5*x - 
2*x^2)/(6*x)])*(60 - 20*x - 6*x^2 + 2*x^3 + (-20*x - 5*x^2 - 2*x^3)*Log[(- 
20 - 5*x - 2*x^2)/(6*x)]))/((20*x + 5*x^2 + 2*x^3)*Log[(-20 - 5*x - 2*x^2) 
/(6*x)]^2),x]
 
output
E^(3 + (3 - x)/Log[(-5 - 20/x - 2*x)/6])
 
3.13.76.3 Rubi [F]

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\left (2 x^3-6 x^2+\left (-2 x^3-5 x^2-20 x\right ) \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )-20 x+60\right ) \exp \left (\frac {3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )-x+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\left (2 x^3+5 x^2+20 x\right ) \log ^2\left (\frac {-2 x^2-5 x-20}{6 x}\right )} \, dx\)

\(\Big \downarrow \) 2026

\(\displaystyle \int \frac {\left (2 x^3-6 x^2+\left (-2 x^3-5 x^2-20 x\right ) \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )-20 x+60\right ) \exp \left (\frac {3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )-x+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{x \left (2 x^2+5 x+20\right ) \log ^2\left (\frac {-2 x^2-5 x-20}{6 x}\right )}dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {2 \left (x^3-3 x^2-10 x+30\right ) \exp \left (\frac {3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )-x+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{x \left (2 x^2+5 x+20\right ) \log ^2\left (\frac {1}{6} \left (-2 x-\frac {20}{x}-5\right )\right )}-\frac {\exp \left (\frac {3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )-x+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\log \left (\frac {1}{6} \left (-2 x-\frac {20}{x}-5\right )\right )}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \int \frac {\exp \left (\frac {-x+3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\log ^2\left (\frac {1}{6} \left (-2 x-5-\frac {20}{x}\right )\right )}dx-\frac {44}{3} i \sqrt {\frac {5}{3}} \int \frac {\exp \left (\frac {-x+3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\left (-4 x+3 i \sqrt {15}-5\right ) \log ^2\left (\frac {1}{6} \left (-2 x-5-\frac {20}{x}\right )\right )}dx+3 \int \frac {\exp \left (\frac {-x+3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{x \log ^2\left (\frac {1}{6} \left (-2 x-5-\frac {20}{x}\right )\right )}dx-\frac {17}{9} \left (9+i \sqrt {15}\right ) \int \frac {\exp \left (\frac {-x+3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\left (4 x-3 i \sqrt {15}+5\right ) \log ^2\left (\frac {1}{6} \left (-2 x-5-\frac {20}{x}\right )\right )}dx-\frac {17}{9} \left (9-i \sqrt {15}\right ) \int \frac {\exp \left (\frac {-x+3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\left (4 x+3 i \sqrt {15}+5\right ) \log ^2\left (\frac {1}{6} \left (-2 x-5-\frac {20}{x}\right )\right )}dx-\frac {44}{3} i \sqrt {\frac {5}{3}} \int \frac {\exp \left (\frac {-x+3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\left (4 x+3 i \sqrt {15}+5\right ) \log ^2\left (\frac {1}{6} \left (-2 x-5-\frac {20}{x}\right )\right )}dx-\int \frac {\exp \left (\frac {-x+3 \log \left (\frac {-2 x^2-5 x-20}{6 x}\right )+3}{\log \left (\frac {-2 x^2-5 x-20}{6 x}\right )}\right )}{\log \left (\frac {1}{6} \left (-2 x-5-\frac {20}{x}\right )\right )}dx\)

input
Int[(E^((3 - x + 3*Log[(-20 - 5*x - 2*x^2)/(6*x)])/Log[(-20 - 5*x - 2*x^2) 
/(6*x)])*(60 - 20*x - 6*x^2 + 2*x^3 + (-20*x - 5*x^2 - 2*x^3)*Log[(-20 - 5 
*x - 2*x^2)/(6*x)]))/((20*x + 5*x^2 + 2*x^3)*Log[(-20 - 5*x - 2*x^2)/(6*x) 
]^2),x]
 
output
$Aborted
 

3.13.76.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2026
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p 
*r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ 
erQ[p] &&  !MonomialQ[Px, x] && (ILtQ[p, 0] ||  !PolyQ[u, x])
 

rule 7279
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 
3.13.76.4 Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.39

method result size
risch \({\mathrm e}^{-\frac {-3 \ln \left (\frac {-2 x^{2}-5 x -20}{6 x}\right )-3+x}{\ln \left (\frac {-2 x^{2}-5 x -20}{6 x}\right )}}\) \(43\)
parallelrisch \({\mathrm e}^{\frac {3 \ln \left (-\frac {2 x^{2}+5 x +20}{6 x}\right )+3-x}{\ln \left (-\frac {2 x^{2}+5 x +20}{6 x}\right )}}\) \(44\)

input
int(((-2*x^3-5*x^2-20*x)*ln(1/6*(-2*x^2-5*x-20)/x)+2*x^3-6*x^2-20*x+60)*ex 
p((3*ln(1/6*(-2*x^2-5*x-20)/x)+3-x)/ln(1/6*(-2*x^2-5*x-20)/x))/(2*x^3+5*x^ 
2+20*x)/ln(1/6*(-2*x^2-5*x-20)/x)^2,x,method=_RETURNVERBOSE)
 
output
exp(-(-3*ln(1/6*(-2*x^2-5*x-20)/x)-3+x)/ln(1/6*(-2*x^2-5*x-20)/x))
 
3.13.76.5 Fricas [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.35 \[ \int \frac {e^{\frac {3-x+3 \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}{\log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}} \left (60-20 x-6 x^2+2 x^3+\left (-20 x-5 x^2-2 x^3\right ) \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )\right )}{\left (20 x+5 x^2+2 x^3\right ) \log ^2\left (\frac {-20-5 x-2 x^2}{6 x}\right )} \, dx=e^{\left (-\frac {x - 3 \, \log \left (-\frac {2 \, x^{2} + 5 \, x + 20}{6 \, x}\right ) - 3}{\log \left (-\frac {2 \, x^{2} + 5 \, x + 20}{6 \, x}\right )}\right )} \]

input
integrate(((-2*x^3-5*x^2-20*x)*log(1/6*(-2*x^2-5*x-20)/x)+2*x^3-6*x^2-20*x 
+60)*exp((3*log(1/6*(-2*x^2-5*x-20)/x)+3-x)/log(1/6*(-2*x^2-5*x-20)/x))/(2 
*x^3+5*x^2+20*x)/log(1/6*(-2*x^2-5*x-20)/x)^2,x, algorithm=\
 
output
e^(-(x - 3*log(-1/6*(2*x^2 + 5*x + 20)/x) - 3)/log(-1/6*(2*x^2 + 5*x + 20) 
/x))
 
3.13.76.6 Sympy [F(-2)]

Exception generated. \[ \int \frac {e^{\frac {3-x+3 \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}{\log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}} \left (60-20 x-6 x^2+2 x^3+\left (-20 x-5 x^2-2 x^3\right ) \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )\right )}{\left (20 x+5 x^2+2 x^3\right ) \log ^2\left (\frac {-20-5 x-2 x^2}{6 x}\right )} \, dx=\text {Exception raised: TypeError} \]

input
integrate(((-2*x**3-5*x**2-20*x)*ln(1/6*(-2*x**2-5*x-20)/x)+2*x**3-6*x**2- 
20*x+60)*exp((3*ln(1/6*(-2*x**2-5*x-20)/x)+3-x)/ln(1/6*(-2*x**2-5*x-20)/x) 
)/(2*x**3+5*x**2+20*x)/ln(1/6*(-2*x**2-5*x-20)/x)**2,x)
 
output
Exception raised: TypeError >> '>' not supported between instances of 'Pol 
y' and 'int'
 
3.13.76.7 Maxima [F]

\[ \int \frac {e^{\frac {3-x+3 \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}{\log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}} \left (60-20 x-6 x^2+2 x^3+\left (-20 x-5 x^2-2 x^3\right ) \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )\right )}{\left (20 x+5 x^2+2 x^3\right ) \log ^2\left (\frac {-20-5 x-2 x^2}{6 x}\right )} \, dx=\int { \frac {{\left (2 \, x^{3} - 6 \, x^{2} - {\left (2 \, x^{3} + 5 \, x^{2} + 20 \, x\right )} \log \left (-\frac {2 \, x^{2} + 5 \, x + 20}{6 \, x}\right ) - 20 \, x + 60\right )} e^{\left (-\frac {x - 3 \, \log \left (-\frac {2 \, x^{2} + 5 \, x + 20}{6 \, x}\right ) - 3}{\log \left (-\frac {2 \, x^{2} + 5 \, x + 20}{6 \, x}\right )}\right )}}{{\left (2 \, x^{3} + 5 \, x^{2} + 20 \, x\right )} \log \left (-\frac {2 \, x^{2} + 5 \, x + 20}{6 \, x}\right )^{2}} \,d x } \]

input
integrate(((-2*x^3-5*x^2-20*x)*log(1/6*(-2*x^2-5*x-20)/x)+2*x^3-6*x^2-20*x 
+60)*exp((3*log(1/6*(-2*x^2-5*x-20)/x)+3-x)/log(1/6*(-2*x^2-5*x-20)/x))/(2 
*x^3+5*x^2+20*x)/log(1/6*(-2*x^2-5*x-20)/x)^2,x, algorithm=\
 
output
integrate((2*x^3 - 6*x^2 - (2*x^3 + 5*x^2 + 20*x)*log(-1/6*(2*x^2 + 5*x + 
20)/x) - 20*x + 60)*e^(-(x - 3*log(-1/6*(2*x^2 + 5*x + 20)/x) - 3)/log(-1/ 
6*(2*x^2 + 5*x + 20)/x))/((2*x^3 + 5*x^2 + 20*x)*log(-1/6*(2*x^2 + 5*x + 2 
0)/x)^2), x)
 
3.13.76.8 Giac [A] (verification not implemented)

Time = 1.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int \frac {e^{\frac {3-x+3 \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}{\log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}} \left (60-20 x-6 x^2+2 x^3+\left (-20 x-5 x^2-2 x^3\right ) \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )\right )}{\left (20 x+5 x^2+2 x^3\right ) \log ^2\left (\frac {-20-5 x-2 x^2}{6 x}\right )} \, dx=e^{\left (-\frac {x}{\log \left (-\frac {1}{3} \, x - \frac {10}{3 \, x} - \frac {5}{6}\right )} + \frac {3}{\log \left (-\frac {1}{3} \, x - \frac {10}{3 \, x} - \frac {5}{6}\right )} + 3\right )} \]

input
integrate(((-2*x^3-5*x^2-20*x)*log(1/6*(-2*x^2-5*x-20)/x)+2*x^3-6*x^2-20*x 
+60)*exp((3*log(1/6*(-2*x^2-5*x-20)/x)+3-x)/log(1/6*(-2*x^2-5*x-20)/x))/(2 
*x^3+5*x^2+20*x)/log(1/6*(-2*x^2-5*x-20)/x)^2,x, algorithm=\
 
output
e^(-x/log(-1/3*x - 10/3/x - 5/6) + 3/log(-1/3*x - 10/3/x - 5/6) + 3)
 
3.13.76.9 Mupad [B] (verification not implemented)

Time = 9.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.26 \[ \int \frac {e^{\frac {3-x+3 \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}{\log \left (\frac {-20-5 x-2 x^2}{6 x}\right )}} \left (60-20 x-6 x^2+2 x^3+\left (-20 x-5 x^2-2 x^3\right ) \log \left (\frac {-20-5 x-2 x^2}{6 x}\right )\right )}{\left (20 x+5 x^2+2 x^3\right ) \log ^2\left (\frac {-20-5 x-2 x^2}{6 x}\right )} \, dx={\mathrm {e}}^{\frac {3}{\ln \left (-\frac {2\,x^2+5\,x+20}{6\,x}\right )}-\frac {x}{\ln \left (-\frac {2\,x^2+5\,x+20}{6\,x}\right )}}\,{\left (-\frac {1}{6\,x}\right )}^{\frac {3}{\ln \left (-\frac {2\,x^2+5\,x+20}{6\,x}\right )}}\,{\left (2\,x^2+5\,x+20\right )}^{\frac {3}{\ln \left (-\frac {2\,x^2+5\,x+20}{6\,x}\right )}} \]

input
int(-(exp((3*log(-((5*x)/6 + x^2/3 + 10/3)/x) - x + 3)/log(-((5*x)/6 + x^2 
/3 + 10/3)/x))*(20*x + log(-((5*x)/6 + x^2/3 + 10/3)/x)*(20*x + 5*x^2 + 2* 
x^3) + 6*x^2 - 2*x^3 - 60))/(log(-((5*x)/6 + x^2/3 + 10/3)/x)^2*(20*x + 5* 
x^2 + 2*x^3)),x)
 
output
exp(3/log(-(5*x + 2*x^2 + 20)/(6*x)) - x/log(-(5*x + 2*x^2 + 20)/(6*x)))*( 
-1/(6*x))^(3/log(-(5*x + 2*x^2 + 20)/(6*x)))*(5*x + 2*x^2 + 20)^(3/log(-(5 
*x + 2*x^2 + 20)/(6*x)))