\(\int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} (-1+3 x-100 x^2)}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx\) [7418]

Optimal result

Integrand size = 62, antiderivative size = 23 \[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=2-x+25 x^2+\frac {x}{-e^{2+x}+x} \]

[Out]

x/(x-exp(2+x))-x+25*x^2+2

Rubi [F]

\[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=\int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx \]

[In]

Int[(-x^2 + 50*x^3 + E^(4 + 2*x)*(-1 + 50*x) + E^(2 + x)*(-1 + 3*x - 100*x^2))/(E^(4 + 2*x) - 2*E^(2 + x)*x +
x^2),x]

[Out]

-x + 25*x^2 - Defer[Int][(E^(2 + x) - x)^(-1), x] - Defer[Int][x/(E^(2 + x) - x)^2, x] + Defer[Int][x/(E^(2 +
x) - x), x] + Defer[Int][x^2/(E^(2 + x) - x)^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{\left (e^{2+x}-x\right )^2} \, dx \\ & = \int \left (-1+\frac {-1+x}{e^{2+x}-x}+50 x+\frac {(-1+x) x}{\left (e^{2+x}-x\right )^2}\right ) \, dx \\ & = -x+25 x^2+\int \frac {-1+x}{e^{2+x}-x} \, dx+\int \frac {(-1+x) x}{\left (e^{2+x}-x\right )^2} \, dx \\ & = -x+25 x^2+\int \left (-\frac {1}{e^{2+x}-x}+\frac {x}{e^{2+x}-x}\right ) \, dx+\int \left (-\frac {x}{\left (e^{2+x}-x\right )^2}+\frac {x^2}{\left (e^{2+x}-x\right )^2}\right ) \, dx \\ & = -x+25 x^2-\int \frac {1}{e^{2+x}-x} \, dx-\int \frac {x}{\left (e^{2+x}-x\right )^2} \, dx+\int \frac {x}{e^{2+x}-x} \, dx+\int \frac {x^2}{\left (e^{2+x}-x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=x \left (-1-\frac {1}{e^{2+x}-x}+25 x\right ) \]

[In]

Integrate[(-x^2 + 50*x^3 + E^(4 + 2*x)*(-1 + 50*x) + E^(2 + x)*(-1 + 3*x - 100*x^2))/(E^(4 + 2*x) - 2*E^(2 + x
)*x + x^2),x]

[Out]

x*(-1 - (E^(2 + x) - x)^(-1) + 25*x)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96

method	result	size
risch	\(25 x^{2}-x +\frac {x}{x -{\mathrm e}^{2+x}}\)	\(22\)
parallelrisch	\(\frac {25 x^{3}-25 x^{2} {\mathrm e}^{2+x}-x^{2}+x \,{\mathrm e}^{2+x}+x}{x -{\mathrm e}^{2+x}}\)	\(39\)
norman	\(\frac {x \,{\mathrm e}^{2+x}+{\mathrm e}^{2+x}-x^{2}+25 x^{3}-25 x^{2} {\mathrm e}^{2+x}}{x -{\mathrm e}^{2+x}}\)	\(42\)

[In]

int(((50*x-1)*exp(2+x)^2+(-100*x^2+3*x-1)*exp(2+x)+50*x^3-x^2)/(exp(2+x)^2-2*x*exp(2+x)+x^2),x,method=_RETURNV
ERBOSE)

[Out]

25*x^2-x+x/(x-exp(2+x))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=\frac {25 \, x^{3} - x^{2} - {\left (25 \, x^{2} - x\right )} e^{\left (x + 2\right )} + x}{x - e^{\left (x + 2\right )}} \]

[In]

integrate(((50*x-1)*exp(2+x)^2+(-100*x^2+3*x-1)*exp(2+x)+50*x^3-x^2)/(exp(2+x)^2-2*x*exp(2+x)+x^2),x, algorith
m="fricas")

[Out]

(25*x^3 - x^2 - (25*x^2 - x)*e^(x + 2) + x)/(x - e^(x + 2))

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.61 \[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=25 x^{2} - x - \frac {x}{- x + e^{x + 2}} \]

[In]

integrate(((50*x-1)*exp(2+x)**2+(-100*x**2+3*x-1)*exp(2+x)+50*x**3-x**2)/(exp(2+x)**2-2*x*exp(2+x)+x**2),x)

[Out]

25*x**2 - x - x/(-x + exp(x + 2))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=\frac {25 \, x^{3} - x^{2} - {\left (25 \, x^{2} e^{2} - x e^{2}\right )} e^{x} + x}{x - e^{\left (x + 2\right )}} \]

[In]

integrate(((50*x-1)*exp(2+x)^2+(-100*x^2+3*x-1)*exp(2+x)+50*x^3-x^2)/(exp(2+x)^2-2*x*exp(2+x)+x^2),x, algorith
m="maxima")

[Out]

(25*x^3 - x^2 - (25*x^2*e^2 - x*e^2)*e^x + x)/(x - e^(x + 2))

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.65 \[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=\frac {25 \, x^{3} - 25 \, x^{2} e^{\left (x + 2\right )} - x^{2} + x e^{\left (x + 2\right )} + x}{x - e^{\left (x + 2\right )}} \]

[In]

integrate(((50*x-1)*exp(2+x)^2+(-100*x^2+3*x-1)*exp(2+x)+50*x^3-x^2)/(exp(2+x)^2-2*x*exp(2+x)+x^2),x, algorith
m="giac")

[Out]

(25*x^3 - 25*x^2*e^(x + 2) - x^2 + x*e^(x + 2) + x)/(x - e^(x + 2))

Mupad [B] (verification not implemented)

Time = 13.13 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-x^2+50 x^3+e^{4+2 x} (-1+50 x)+e^{2+x} \left (-1+3 x-100 x^2\right )}{e^{4+2 x}-2 e^{2+x} x+x^2} \, dx=\frac {{\mathrm {e}}^{x+2}}{x-{\mathrm {e}}^{x+2}}-x+25\,x^2 \]

[In]

int(-(exp(x + 2)*(100*x^2 - 3*x + 1) - exp(2*x + 4)*(50*x - 1) + x^2 - 50*x^3)/(exp(2*x + 4) - 2*x*exp(x + 2)
+ x^2),x)

[Out]

exp(x + 2)/(x - exp(x + 2)) - x + 25*x^2