\(\int \frac {-520+40 e^4-20 x+e^{-x+x^2} (-52+24 x-50 x^2-4 x^3+e^4 (4-2 x+4 x^2))}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx\) [4642]

Optimal result

Integrand size = 87, antiderivative size = 27 \[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\frac {13-e^4+x}{\left (5+\frac {1}{2} e^{(-1+x) x}\right ) x^2} \]

[Out]

(13+x-exp(4))/x^2/(1/2*exp(x*(-1+x))+5)

Rubi [F]

\[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx \]

[In]

Int[(-520 + 40*E^4 - 20*x + E^(-x + x^2)*(-52 + 24*x - 50*x^2 - 4*x^3 + E^4*(4 - 2*x + 4*x^2)))/(100*x^3 + 20*
E^(-x + x^2)*x^3 + E^(-2*x + 2*x^2)*x^3),x]

[Out]

40*Defer[Int][E^(2*x)/(10*E^x + E^x^2)^2, x] - 4*Defer[Int][E^x/(10*E^x + E^x^2), x] - 4*(13 - E^4)*Defer[Int]
[E^x/((10*E^x + E^x^2)*x^3), x] - 20*(13 - E^4)*Defer[Int][E^(2*x)/((10*E^x + E^x^2)^2*x^2), x] + 2*(12 - E^4)
*Defer[Int][E^x/((10*E^x + E^x^2)*x^2), x] + 20*(25 - 2*E^4)*Defer[Int][E^(2*x)/((10*E^x + E^x^2)^2*x), x] - 2
*(25 - 2*E^4)*Defer[Int][E^x/((10*E^x + E^x^2)*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{2 x} \left (-520 \left (1-\frac {e^4}{13}\right )-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )\right )}{\left (10 e^x+e^{x^2}\right )^2 x^3} \, dx \\ & = \int \left (\frac {20 e^{2 x} \left (13-e^4+x\right ) (-1+2 x)}{\left (10 e^x+e^{x^2}\right )^2 x^2}+\frac {2 e^x \left (-2 \left (13-e^4\right )+\left (12-e^4\right ) x-\left (25-2 e^4\right ) x^2-2 x^3\right )}{\left (10 e^x+e^{x^2}\right ) x^3}\right ) \, dx \\ & = 2 \int \frac {e^x \left (-2 \left (13-e^4\right )+\left (12-e^4\right ) x-\left (25-2 e^4\right ) x^2-2 x^3\right )}{\left (10 e^x+e^{x^2}\right ) x^3} \, dx+20 \int \frac {e^{2 x} \left (13-e^4+x\right ) (-1+2 x)}{\left (10 e^x+e^{x^2}\right )^2 x^2} \, dx \\ & = 2 \int \left (-\frac {2 e^x}{10 e^x+e^{x^2}}+\frac {2 e^x \left (-13+e^4\right )}{\left (10 e^x+e^{x^2}\right ) x^3}-\frac {e^x \left (-12+e^4\right )}{\left (10 e^x+e^{x^2}\right ) x^2}+\frac {e^x \left (-25+2 e^4\right )}{\left (10 e^x+e^{x^2}\right ) x}\right ) \, dx+20 \int \left (\frac {2 e^{2 x}}{\left (10 e^x+e^{x^2}\right )^2}+\frac {e^{2 x} \left (-13+e^4\right )}{\left (10 e^x+e^{x^2}\right )^2 x^2}-\frac {e^{2 x} \left (-25+2 e^4\right )}{\left (10 e^x+e^{x^2}\right )^2 x}\right ) \, dx \\ & = -\left (4 \int \frac {e^x}{10 e^x+e^{x^2}} \, dx\right )+40 \int \frac {e^{2 x}}{\left (10 e^x+e^{x^2}\right )^2} \, dx-\left (2 \left (25-2 e^4\right )\right ) \int \frac {e^x}{\left (10 e^x+e^{x^2}\right ) x} \, dx+\left (20 \left (25-2 e^4\right )\right ) \int \frac {e^{2 x}}{\left (10 e^x+e^{x^2}\right )^2 x} \, dx+\left (2 \left (12-e^4\right )\right ) \int \frac {e^x}{\left (10 e^x+e^{x^2}\right ) x^2} \, dx-\left (4 \left (13-e^4\right )\right ) \int \frac {e^x}{\left (10 e^x+e^{x^2}\right ) x^3} \, dx-\left (20 \left (13-e^4\right )\right ) \int \frac {e^{2 x}}{\left (10 e^x+e^{x^2}\right )^2 x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 7.84 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\frac {2 e^x \left (13-e^4+x\right )}{\left (10 e^x+e^{x^2}\right ) x^2} \]

[In]

Integrate[(-520 + 40*E^4 - 20*x + E^(-x + x^2)*(-52 + 24*x - 50*x^2 - 4*x^3 + E^4*(4 - 2*x + 4*x^2)))/(100*x^3
+ 20*E^(-x + x^2)*x^3 + E^(-2*x + 2*x^2)*x^3),x]

[Out]

(2*E^x*(13 - E^4 + x))/((10*E^x + E^x^2)*x^2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.85

method	result	size
risch	\(-\frac {2 \left (-x +{\mathrm e}^{4}-13\right )}{x^{2} \left ({\mathrm e}^{x \left (-1+x \right )}+10\right )}\)	\(23\)
norman	\(\frac {2 x +26-2 \,{\mathrm e}^{4}}{x^{2} \left ({\mathrm e}^{x^{2}-x}+10\right )}\)	\(26\)
parallelrisch	\(-\frac {-26+2 \,{\mathrm e}^{4}-2 x}{\left ({\mathrm e}^{x^{2}-x}+10\right ) x^{2}}\)	\(27\)

[In]

int((((4*x^2-2*x+4)*exp(4)-4*x^3-50*x^2+24*x-52)*exp(x^2-x)+40*exp(4)-20*x-520)/(x^3*exp(x^2-x)^2+20*x^3*exp(x
^2-x)+100*x^3),x,method=_RETURNVERBOSE)

[Out]

-2*(-x+exp(4)-13)/x^2/(exp(x*(-1+x))+10)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\frac {2 \, {\left (x - e^{4} + 13\right )}}{x^{2} e^{\left (x^{2} - x\right )} + 10 \, x^{2}} \]

[In]

integrate((((4*x^2-2*x+4)*exp(4)-4*x^3-50*x^2+24*x-52)*exp(x^2-x)+40*exp(4)-20*x-520)/(x^3*exp(x^2-x)^2+20*x^3
*exp(x^2-x)+100*x^3),x, algorithm="fricas")

[Out]

2*(x - e^4 + 13)/(x^2*e^(x^2 - x) + 10*x^2)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\frac {2 x - 2 e^{4} + 26}{x^{2} e^{x^{2} - x} + 10 x^{2}} \]

[In]

integrate((((4*x**2-2*x+4)*exp(4)-4*x**3-50*x**2+24*x-52)*exp(x**2-x)+40*exp(4)-20*x-520)/(x**3*exp(x**2-x)**2
+20*x**3*exp(x**2-x)+100*x**3),x)

[Out]

(2*x - 2*exp(4) + 26)/(x**2*exp(x**2 - x) + 10*x**2)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\frac {2 \, {\left (x - e^{4} + 13\right )} e^{x}}{x^{2} e^{\left (x^{2}\right )} + 10 \, x^{2} e^{x}} \]

[In]

integrate((((4*x^2-2*x+4)*exp(4)-4*x^3-50*x^2+24*x-52)*exp(x^2-x)+40*exp(4)-20*x-520)/(x^3*exp(x^2-x)^2+20*x^3
*exp(x^2-x)+100*x^3),x, algorithm="maxima")

[Out]

2*(x - e^4 + 13)*e^x/(x^2*e^(x^2) + 10*x^2*e^x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\frac {2 \, {\left (x - e^{4} + 13\right )}}{x^{2} e^{\left (x^{2} - x\right )} + 10 \, x^{2}} \]

[In]

integrate((((4*x^2-2*x+4)*exp(4)-4*x^3-50*x^2+24*x-52)*exp(x^2-x)+40*exp(4)-20*x-520)/(x^3*exp(x^2-x)^2+20*x^3
*exp(x^2-x)+100*x^3),x, algorithm="giac")

[Out]

2*(x - e^4 + 13)/(x^2*e^(x^2 - x) + 10*x^2)

Mupad [B] (verification not implemented)

Time = 11.38 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {-520+40 e^4-20 x+e^{-x+x^2} \left (-52+24 x-50 x^2-4 x^3+e^4 \left (4-2 x+4 x^2\right )\right )}{100 x^3+20 e^{-x+x^2} x^3+e^{-2 x+2 x^2} x^3} \, dx=\frac {2\,\left (x-{\mathrm {e}}^4+13\right )}{x^2\,\left ({\mathrm {e}}^{x^2-x}+10\right )} \]

[In]

int(-(20*x - 40*exp(4) + exp(x^2 - x)*(50*x^2 - exp(4)*(4*x^2 - 2*x + 4) - 24*x + 4*x^3 + 52) + 520)/(x^3*exp(
2*x^2 - 2*x) + 20*x^3*exp(x^2 - x) + 100*x^3),x)

[Out]

(2*(x - exp(4) + 13))/(x^2*(exp(x^2 - x) + 10))