\(\int \frac {-4 x-x^3+e^{-5+2 x} (4+2 x^3)}{e^{-5+2 x} x^3-x^4} \, dx\) [6060]

Optimal result

Integrand size = 44, antiderivative size = 26 \[ \int \frac {-4 x-x^3+e^{-5+2 x} \left (4+2 x^3\right )}{e^{-5+2 x} x^3-x^4} \, dx=9-\frac {2}{x^2}+\log \left (4 e^{2 e} \left (e^{-5+2 x}-x\right )\right ) \]

[Out]

9-2/x^2+ln(4*exp(exp(1))^2*(exp(-5+2*x)-x))

Rubi [F]

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

[In]

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

[Out]

-2/x^2 + 2*x - E^5*Defer[Int][(E^(2*x) - E^5*x)^(-1), x] - 2*E^5*Defer[Int][x/(-E^(2*x) + E^5*x), x]

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

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {-4 x-x^3+e^{-5+2 x} \left (4+2 x^3\right )}{e^{-5+2 x} x^3-x^4} \, dx=-\frac {2}{x^2}+\log \left (e^{2 x}-e^5 x\right ) \]

[In]

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

[Out]

-2/x^2 + Log[E^(2*x) - E^5*x]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69

method	result	size
norman	\(-\frac {2}{x^{2}}+\ln \left (x -{\mathrm e}^{-5+2 x}\right )\)	\(18\)
risch	\(-\frac {2}{x^{2}}+5+\ln \left ({\mathrm e}^{-5+2 x}-x \right )\)	\(19\)
parallelrisch	\(\frac {\ln \left (x -{\mathrm e}^{-5+2 x}\right ) x^{2}-2}{x^{2}}\)	\(22\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-4 x-x^3+e^{-5+2 x} \left (4+2 x^3\right )}{e^{-5+2 x} x^3-x^4} \, dx=\frac {x^{2} \log \left (-x + e^{\left (2 \, x - 5\right )}\right ) - 2}{x^{2}} \]

[In]

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

[Out]

(x^2*log(-x + e^(2*x - 5)) - 2)/x^2

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.54 \[ \int \frac {-4 x-x^3+e^{-5+2 x} \left (4+2 x^3\right )}{e^{-5+2 x} x^3-x^4} \, dx=\log {\left (- x + e^{2 x - 5} \right )} - \frac {2}{x^{2}} \]

[In]

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

[Out]

log(-x + exp(2*x - 5)) - 2/x**2

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-4 x-x^3+e^{-5+2 x} \left (4+2 x^3\right )}{e^{-5+2 x} x^3-x^4} \, dx=-\frac {2}{x^{2}} + \log \left (-x e^{5} + e^{\left (2 \, x\right )}\right ) \]

[In]

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

[Out]

-2/x^2 + log(-x*e^5 + e^(2*x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {-4 x-x^3+e^{-5+2 x} \left (4+2 x^3\right )}{e^{-5+2 x} x^3-x^4} \, dx=\frac {x^{2} \log \left (-x e^{5} + e^{\left (2 \, x\right )}\right ) - 2}{x^{2}} \]

[In]

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

[Out]

(x^2*log(-x*e^5 + e^(2*x)) - 2)/x^2

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {-4 x-x^3+e^{-5+2 x} \left (4+2 x^3\right )}{e^{-5+2 x} x^3-x^4} \, dx=\ln \left (x-{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-5}\right )-\frac {2}{x^2} \]

[In]

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

[Out]

log(x - exp(2*x)*exp(-5)) - 2/x^2