\(\int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x (80 e^{-3+x} x+(-4-8 x) \log (6))}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx\) [5057]

Optimal result

Integrand size = 77, antiderivative size = 29 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-4 x+\frac {4 e^{3-x} \log (6)}{\left (e^x-\frac {10}{x}\right ) x} \]

[Out]

4*ln(6)/x/(exp(x)-10/x)/exp(-3+x)-4*x

Rubi [F]

\[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=\int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx \]

[In]

Int[(-400*E^(-3 + x) - 4*E^(-3 + 3*x)*x^2 + 40*Log[6] + E^x*(80*E^(-3 + x)*x + (-4 - 8*x)*Log[6]))/(100*E^(-3
+ x) - 20*E^(-3 + 2*x)*x + E^(-3 + 3*x)*x^2),x]

[Out]

-4*x - 40*Log[6]*Defer[Int][E^(3 - x)/(-10 + E^x*x)^2, x] - 40*Log[6]*Defer[Int][E^(3 - x)/(x*(-10 + E^x*x)^2)
, x] - 8*Log[6]*Defer[Int][E^(3 - x)/(-10 + E^x*x), x] - 4*Log[6]*Defer[Int][E^(3 - x)/(x*(-10 + E^x*x)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{3-x} \left (-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )\right )}{\left (10-e^x x\right )^2} \, dx \\ & = \int \left (-4-\frac {40 e^{3-x} (1+x) \log (6)}{x \left (-10+e^x x\right )^2}-\frac {4 e^{3-x} (1+2 x) \log (6)}{x \left (-10+e^x x\right )}\right ) \, dx \\ & = -4 x-(4 \log (6)) \int \frac {e^{3-x} (1+2 x)}{x \left (-10+e^x x\right )} \, dx-(40 \log (6)) \int \frac {e^{3-x} (1+x)}{x \left (-10+e^x x\right )^2} \, dx \\ & = -4 x-(4 \log (6)) \int \left (\frac {2 e^{3-x}}{-10+e^x x}+\frac {e^{3-x}}{x \left (-10+e^x x\right )}\right ) \, dx-(40 \log (6)) \int \left (\frac {e^{3-x}}{\left (-10+e^x x\right )^2}+\frac {e^{3-x}}{x \left (-10+e^x x\right )^2}\right ) \, dx \\ & = -4 x-(4 \log (6)) \int \frac {e^{3-x}}{x \left (-10+e^x x\right )} \, dx-(8 \log (6)) \int \frac {e^{3-x}}{-10+e^x x} \, dx-(40 \log (6)) \int \frac {e^{3-x}}{\left (-10+e^x x\right )^2} \, dx-(40 \log (6)) \int \frac {e^{3-x}}{x \left (-10+e^x x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.40 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \left (10 x-e^x x^2+e^{3-x} \log (6)\right )}{10-e^x x} \]

[In]

Integrate[(-400*E^(-3 + x) - 4*E^(-3 + 3*x)*x^2 + 40*Log[6] + E^x*(80*E^(-3 + x)*x + (-4 - 8*x)*Log[6]))/(100*
E^(-3 + x) - 20*E^(-3 + 2*x)*x + E^(-3 + 3*x)*x^2),x]

[Out]

(-4*(10*x - E^x*x^2 + E^(3 - x)*Log[6]))/(10 - E^x*x)

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21

method	result	size
norman	\(\frac {\left (4 \,{\mathrm e}^{3} \ln \left (6\right )+40 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-x}}{{\mathrm e}^{x} x -10}\)	\(35\)
parallelrisch	\(\frac {\left (-4 \,{\mathrm e}^{-3+x} x^{2} {\mathrm e}^{x}+40 x \,{\mathrm e}^{-3+x}+4 \ln \left (6\right )\right ) {\mathrm e}^{-x +3}}{{\mathrm e}^{x} x -10}\)	\(39\)
risch	\(-4 x -\frac {2 \,{\mathrm e}^{-x +3} \ln \left (2\right )}{5}-\frac {2 \,{\mathrm e}^{-x +3} \ln \left (3\right )}{5}+\frac {2 x \left (\ln \left (2\right )+\ln \left (3\right )\right ) {\mathrm e}^{3}}{5 \left ({\mathrm e}^{x} x -10\right )}\)	\(43\)

[In]

int((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*ln(6))*exp(x)-400*exp(-3+x)+40*ln(6))/(x^2*exp(-3+x)*e
xp(x)^2-20*x*exp(-3+x)*exp(x)+100*exp(-3+x)),x,method=_RETURNVERBOSE)

[Out]

(4*exp(3)*ln(6)+40*exp(x)*x-4*exp(x)^2*x^2)/exp(x)/(exp(x)*x-10)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - 10 \, x e^{x} - e^{3} \log \left (6\right )\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \]

[In]

integrate((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*log(6))*exp(x)-400*exp(-3+x)+40*log(6))/(x^2*exp
(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+100*exp(-3+x)),x, algorithm="fricas")

[Out]

-4*(x^2*e^(2*x) - 10*x*e^x - e^3*log(6))/(x*e^(2*x) - 10*e^x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=- 4 x + \frac {2 x e^{3} \log {\left (6 \right )}}{5 x e^{x} - 50} - \frac {2 e^{3} e^{- x} \log {\left (6 \right )}}{5} \]

[In]

integrate((-4*x**2*exp(-3+x)*exp(x)**2+(80*x*exp(-3+x)+(-8*x-4)*ln(6))*exp(x)-400*exp(-3+x)+40*ln(6))/(x**2*ex
p(-3+x)*exp(x)**2-20*x*exp(-3+x)*exp(x)+100*exp(-3+x)),x)

[Out]

-4*x + 2*x*exp(3)*log(6)/(5*x*exp(x) - 50) - 2*exp(3)*exp(-x)*log(6)/5

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - {\left (\log \left (3\right ) + \log \left (2\right )\right )} e^{3} - 10 \, x e^{x}\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \]

[In]

integrate((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*log(6))*exp(x)-400*exp(-3+x)+40*log(6))/(x^2*exp
(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+100*exp(-3+x)),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (29) = 58\).

Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.28 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left ({\left (x - 3\right )}^{2} e^{\left (2 \, x - 3\right )} + 3 \, {\left (x - 3\right )} e^{\left (2 \, x - 3\right )} - 10 \, {\left (x - 3\right )} e^{\left (x - 3\right )} - \log \left (6\right )\right )}}{{\left (x - 3\right )} e^{\left (2 \, x - 3\right )} + 3 \, e^{\left (2 \, x - 3\right )} - 10 \, e^{\left (x - 3\right )}} \]

[In]

integrate((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*log(6))*exp(x)-400*exp(-3+x)+40*log(6))/(x^2*exp
(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+100*exp(-3+x)),x, algorithm="giac")

[Out]

-4*((x - 3)^2*e^(2*x - 3) + 3*(x - 3)*e^(2*x - 3) - 10*(x - 3)*e^(x - 3) - log(6))/((x - 3)*e^(2*x - 3) + 3*e^
(2*x - 3) - 10*e^(x - 3))

Mupad [B] (verification not implemented)

Time = 11.89 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=\frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,\ln \left (6\right )}{x\,{\mathrm {e}}^x-10}-4\,x \]

[In]

int(-(400*exp(x - 3) - 40*log(6) + exp(x)*(log(6)*(8*x + 4) - 80*x*exp(x - 3)) + 4*x^2*exp(2*x)*exp(x - 3))/(1
00*exp(x - 3) + x^2*exp(2*x)*exp(x - 3) - 20*x*exp(x - 3)*exp(x)),x)

[Out]

(4*exp(-x)*exp(3)*log(6))/(x*exp(x) - 10) - 4*x