\(\int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} (2 x-x^2-2 e^8 x^3)}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx\) [4481]

Optimal result

Integrand size = 92, antiderivative size = 26 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {e^{-3+x}}{x \left (5-e^{-e^8 x^2} x\right )} \]

[Out]

exp(x)/exp(3)/x/(5-x/exp(x^2*exp(4)^2))

Rubi [F]

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

[In]

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

[Out]

-Defer[Int][E^(-3 + x + E^8*x^2)/((5*E^(E^8*x^2) - x)*x^2), x] + Defer[Int][E^(-3 + x + E^8*x^2)/((5*E^(E^8*x^
2) - x)*x), x] - 2*Defer[Int][(E^(5 + x + E^8*x^2)*x)/(5*E^(E^8*x^2) - x)^2, x] + Defer[Int][E^(-3 + x + E^8*x
^2)/(x*(-5*E^(E^8*x^2) + x)^2), x]

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

Mathematica [A] (verified)

Time = 3.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {e^{-3+x+e^8 x^2}}{\left (5 e^{e^8 x^2}-x\right ) x} \]

[In]

Integrate[(E^(x + 2*E^8*x^2)*(-5 + 5*x) + E^(x + E^8*x^2)*(2*x - x^2 - 2*E^8*x^3))/(25*E^(3 + 2*E^8*x^2)*x^2 -
10*E^(3 + E^8*x^2)*x^3 + E^3*x^4),x]

[Out]

E^(-3 + x + E^8*x^2)/((5*E^(E^8*x^2) - x)*x)

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {{\mathrm e}^{-3+x +x^{2} {\mathrm e}^{8}}}{\left (-x +5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right ) x}\)	\(30\)
norman	\(-\frac {{\mathrm e}^{x} {\mathrm e}^{-3} {\mathrm e}^{x^{2} {\mathrm e}^{8}}}{x \left (x -5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right )}\)	\(36\)
parallelrisch	\(-\frac {{\mathrm e}^{x} {\mathrm e}^{-3} {\mathrm e}^{x^{2} {\mathrm e}^{8}}}{x \left (x -5 \,{\mathrm e}^{x^{2} {\mathrm e}^{8}}\right )}\)	\(36\)

[In]

int(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(3)*exp
(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=-\frac {e^{\left (2 \, x^{2} e^{8} + 2 \, x\right )}}{x^{2} e^{\left (x^{2} e^{8} + x + 3\right )} - 5 \, x e^{\left (2 \, x^{2} e^{8} + x + 3\right )}} \]

[In]

integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(
3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="fricas")

[Out]

-e^(2*x^2*e^8 + 2*x)/(x^2*e^(x^2*e^8 + x + 3) - 5*x*e^(2*x^2*e^8 + x + 3))

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {e^{x}}{- 5 x e^{3} + 25 e^{3} e^{x^{2} e^{8}}} + \frac {e^{x}}{5 x e^{3}} \]

[In]

integrate(((5*x-5)*exp(x)*exp(x**2*exp(4)**2)**2+(-2*x**3*exp(4)**2-x**2+2*x)*exp(x)*exp(x**2*exp(4)**2))/(25*
x**2*exp(3)*exp(x**2*exp(4)**2)**2-10*x**3*exp(3)*exp(x**2*exp(4)**2)+x**4*exp(3)),x)

[Out]

exp(x)/(-5*x*exp(3) + 25*exp(3)*exp(x**2*exp(8))) + exp(-3)*exp(x)/(5*x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=-\frac {e^{\left (x^{2} e^{8} + x\right )}}{x^{2} e^{3} - 5 \, x e^{\left (x^{2} e^{8} + 3\right )}} \]

[In]

integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(
3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="maxima")

[Out]

-e^(x^2*e^8 + x)/(x^2*e^3 - 5*x*e^(x^2*e^8 + 3))

Giac [F]

\[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\int { \frac {5 \, {\left (x - 1\right )} e^{\left (2 \, x^{2} e^{8} + x\right )} - {\left (2 \, x^{3} e^{8} + x^{2} - 2 \, x\right )} e^{\left (x^{2} e^{8} + x\right )}}{x^{4} e^{3} - 10 \, x^{3} e^{\left (x^{2} e^{8} + 3\right )} + 25 \, x^{2} e^{\left (2 \, x^{2} e^{8} + 3\right )}} \,d x } \]

[In]

integrate(((5*x-5)*exp(x)*exp(x^2*exp(4)^2)^2+(-2*x^3*exp(4)^2-x^2+2*x)*exp(x)*exp(x^2*exp(4)^2))/(25*x^2*exp(
3)*exp(x^2*exp(4)^2)^2-10*x^3*exp(3)*exp(x^2*exp(4)^2)+x^4*exp(3)),x, algorithm="giac")

[Out]

integrate((5*(x - 1)*e^(2*x^2*e^8 + x) - (2*x^3*e^8 + x^2 - 2*x)*e^(x^2*e^8 + x))/(x^4*e^3 - 10*x^3*e^(x^2*e^8
+ 3) + 25*x^2*e^(2*x^2*e^8 + 3)), x)

Mupad [B] (verification not implemented)

Time = 10.50 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.92 \[ \int \frac {e^{x+2 e^8 x^2} (-5+5 x)+e^{x+e^8 x^2} \left (2 x-x^2-2 e^8 x^3\right )}{25 e^{3+2 e^8 x^2} x^2-10 e^{3+e^8 x^2} x^3+e^3 x^4} \, dx=\frac {{\mathrm {e}}^{x-3}}{5\,x}+\frac {{\mathrm {e}}^{-3}\,\left ({\mathrm {e}}^x-2\,x^2\,{\mathrm {e}}^{x+8}\right )}{5\,\left (x-5\,{\mathrm {e}}^{x^2\,{\mathrm {e}}^8}\right )\,\left (2\,x^2\,{\mathrm {e}}^8-1\right )} \]

[In]

int((exp(2*x^2*exp(8))*exp(x)*(5*x - 5) - exp(x^2*exp(8))*exp(x)*(2*x^3*exp(8) - 2*x + x^2))/(x^4*exp(3) - 10*
x^3*exp(x^2*exp(8))*exp(3) + 25*x^2*exp(2*x^2*exp(8))*exp(3)),x)

[Out]

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