\(\int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} (960 e^x x-200 x^2+e^x (-1040 x+960 x^2) \log (x)-8 e^{2 x} \log ^2(x))}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx\) [2989]

Optimal result

Integrand size = 87, antiderivative size = 25 \[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=5+\frac {4 e^{-\frac {48 x}{x+\frac {1}{5} e^x \log (x)}}}{x^2} \]

[Out]

5+4/x^2/exp(3*x/(1/5*exp(x)*ln(x)+x))^16

Rubi [F]

\[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=\int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx \]

[In]

Int[(960*E^x*x - 200*x^2 + E^x*(-1040*x + 960*x^2)*Log[x] - 8*E^(2*x)*Log[x]^2)/(E^((240*x)/(5*x + E^x*Log[x])
)*(25*x^5 + 10*E^x*x^4*Log[x] + E^(2*x)*x^3*Log[x]^2)),x]

[Out]

-8*Defer[Int][1/(E^((240*x)/(5*x + E^x*Log[x]))*x^3), x] - 4800*Defer[Int][1/(E^((240*x)/(5*x + E^x*Log[x]))*(
5*x + E^x*Log[x])^2), x] + 4800*Defer[Int][1/(E^((240*x)/(5*x + E^x*Log[x]))*x*(5*x + E^x*Log[x])^2), x] - 480
0*Defer[Int][1/(E^((240*x)/(5*x + E^x*Log[x]))*x*Log[x]*(5*x + E^x*Log[x])^2), x] - 960*Defer[Int][1/(E^((240*
x)/(5*x + E^x*Log[x]))*x^2*(5*x + E^x*Log[x])), x] + 960*Defer[Int][1/(E^((240*x)/(5*x + E^x*Log[x]))*x*(5*x +
E^x*Log[x])), x] + 960*Defer[Int][1/(E^((240*x)/(5*x + E^x*Log[x]))*x^2*Log[x]*(5*x + E^x*Log[x])), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{x^3 \left (5 x+e^x \log (x)\right )^2} \, dx \\ & = \int \left (-\frac {8 e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^3}-\frac {4800 e^{-\frac {240 x}{5 x+e^x \log (x)}} (1-\log (x)+x \log (x))}{x \log (x) \left (5 x+e^x \log (x)\right )^2}+\frac {960 e^{-\frac {240 x}{5 x+e^x \log (x)}} (1-\log (x)+x \log (x))}{x^2 \log (x) \left (5 x+e^x \log (x)\right )}\right ) \, dx \\ & = -\left (8 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^3} \, dx\right )+960 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} (1-\log (x)+x \log (x))}{x^2 \log (x) \left (5 x+e^x \log (x)\right )} \, dx-4800 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} (1-\log (x)+x \log (x))}{x \log (x) \left (5 x+e^x \log (x)\right )^2} \, dx \\ & = -\left (8 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^3} \, dx\right )+960 \int \left (-\frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^2 \left (5 x+e^x \log (x)\right )}+\frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x \left (5 x+e^x \log (x)\right )}+\frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^2 \log (x) \left (5 x+e^x \log (x)\right )}\right ) \, dx-4800 \int \left (\frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{\left (5 x+e^x \log (x)\right )^2}-\frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x \left (5 x+e^x \log (x)\right )^2}+\frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x \log (x) \left (5 x+e^x \log (x)\right )^2}\right ) \, dx \\ & = -\left (8 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^3} \, dx\right )-960 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^2 \left (5 x+e^x \log (x)\right )} \, dx+960 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x \left (5 x+e^x \log (x)\right )} \, dx+960 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^2 \log (x) \left (5 x+e^x \log (x)\right )} \, dx-4800 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{\left (5 x+e^x \log (x)\right )^2} \, dx+4800 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x \left (5 x+e^x \log (x)\right )^2} \, dx-4800 \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x \log (x) \left (5 x+e^x \log (x)\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.50 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=\frac {4 e^{-\frac {240 x}{5 x+e^x \log (x)}}}{x^2} \]

[In]

Integrate[(960*E^x*x - 200*x^2 + E^x*(-1040*x + 960*x^2)*Log[x] - 8*E^(2*x)*Log[x]^2)/(E^((240*x)/(5*x + E^x*L
og[x]))*(25*x^5 + 10*E^x*x^4*Log[x] + E^(2*x)*x^3*Log[x]^2)),x]

[Out]

4/(E^((240*x)/(5*x + E^x*Log[x]))*x^2)

Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84

\[\frac {4 \,{\mathrm e}^{-\frac {240 x}{{\mathrm e}^{x} \ln \left (x \right )+5 x}}}{x^{2}}\]

[In]

int((-8*exp(x)^2*ln(x)^2+(960*x^2-1040*x)*exp(x)*ln(x)+960*exp(x)*x-200*x^2)/(x^3*exp(x)^2*ln(x)^2+10*x^4*exp(
x)*ln(x)+25*x^5)/exp(15*x/(exp(x)*ln(x)+5*x))^16,x)

[Out]

4/x^2*exp(-240*x/(exp(x)*ln(x)+5*x))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=\frac {4 \, e^{\left (-\frac {240 \, x}{e^{x} \log \left (x\right ) + 5 \, x}\right )}}{x^{2}} \]

[In]

integrate((-8*exp(x)^2*log(x)^2+(960*x^2-1040*x)*exp(x)*log(x)+960*exp(x)*x-200*x^2)/(x^3*exp(x)^2*log(x)^2+10
*x^4*exp(x)*log(x)+25*x^5)/exp(15*x/(exp(x)*log(x)+5*x))^16,x, algorithm="fricas")

[Out]

4*e^(-240*x/(e^x*log(x) + 5*x))/x^2

Sympy [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=\frac {4 e^{- \frac {240 x}{5 x + e^{x} \log {\left (x \right )}}}}{x^{2}} \]

[In]

integrate((-8*exp(x)**2*ln(x)**2+(960*x**2-1040*x)*exp(x)*ln(x)+960*exp(x)*x-200*x**2)/(x**3*exp(x)**2*ln(x)**
2+10*x**4*exp(x)*ln(x)+25*x**5)/exp(15*x/(exp(x)*ln(x)+5*x))**16,x)

[Out]

4*exp(-240*x/(5*x + exp(x)*log(x)))/x**2

Maxima [F]

\[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=\int { \frac {8 \, {\left (10 \, {\left (12 \, x^{2} - 13 \, x\right )} e^{x} \log \left (x\right ) - e^{\left (2 \, x\right )} \log \left (x\right )^{2} - 25 \, x^{2} + 120 \, x e^{x}\right )} e^{\left (-\frac {240 \, x}{e^{x} \log \left (x\right ) + 5 \, x}\right )}}{10 \, x^{4} e^{x} \log \left (x\right ) + x^{3} e^{\left (2 \, x\right )} \log \left (x\right )^{2} + 25 \, x^{5}} \,d x } \]

[In]

integrate((-8*exp(x)^2*log(x)^2+(960*x^2-1040*x)*exp(x)*log(x)+960*exp(x)*x-200*x^2)/(x^3*exp(x)^2*log(x)^2+10
*x^4*exp(x)*log(x)+25*x^5)/exp(15*x/(exp(x)*log(x)+5*x))^16,x, algorithm="maxima")

[Out]

8*integrate((10*(12*x^2 - 13*x)*e^x*log(x) - e^(2*x)*log(x)^2 - 25*x^2 + 120*x*e^x)*e^(-240*x/(e^x*log(x) + 5*
x))/(10*x^4*e^x*log(x) + x^3*e^(2*x)*log(x)^2 + 25*x^5), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=\frac {4 \, e^{\left (-x + \frac {x e^{x} \log \left (x\right ) + 5 \, x^{2} - 240 \, x}{e^{x} \log \left (x\right ) + 5 \, x}\right )}}{x^{2}} \]

[In]

integrate((-8*exp(x)^2*log(x)^2+(960*x^2-1040*x)*exp(x)*log(x)+960*exp(x)*x-200*x^2)/(x^3*exp(x)^2*log(x)^2+10
*x^4*exp(x)*log(x)+25*x^5)/exp(15*x/(exp(x)*log(x)+5*x))^16,x, algorithm="giac")

[Out]

4*e^(-x + (x*e^x*log(x) + 5*x^2 - 240*x)/(e^x*log(x) + 5*x))/x^2

Mupad [B] (verification not implemented)

Time = 9.94 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{-\frac {240 x}{5 x+e^x \log (x)}} \left (960 e^x x-200 x^2+e^x \left (-1040 x+960 x^2\right ) \log (x)-8 e^{2 x} \log ^2(x)\right )}{25 x^5+10 e^x x^4 \log (x)+e^{2 x} x^3 \log ^2(x)} \, dx=\frac {4\,{\mathrm {e}}^{-\frac {240\,x}{5\,x+{\mathrm {e}}^x\,\ln \left (x\right )}}}{x^2} \]

[In]

int(-(exp(-(240*x)/(5*x + exp(x)*log(x)))*(200*x^2 - 960*x*exp(x) + 8*exp(2*x)*log(x)^2 + exp(x)*log(x)*(1040*
x - 960*x^2)))/(25*x^5 + x^3*exp(2*x)*log(x)^2 + 10*x^4*exp(x)*log(x)),x)

[Out]

(4*exp(-(240*x)/(5*x + exp(x)*log(x))))/x^2