\(\int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx\) [5620]

Optimal result

Integrand size = 29, antiderivative size = 24 \[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=\frac {1}{2} \log \left (\frac {256 e^{20}}{\left (5-e^x+\frac {5}{x}\right )^2}\right ) \]

[Out]

1/2*ln(256*exp(5)^4/(5/x+5-exp(x))^2)

Rubi [F]

\[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=\int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.71 \[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=\log (x)-\log \left (5+5 x-e^x x\right ) \]

[In]

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

[Out]

Log[x] - Log[5 + 5*x - E^x*x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62

method	result	size
risch	\(-\ln \left ({\mathrm e}^{x}-\frac {5 \left (1+x \right )}{x}\right )\)	\(15\)
norman	\(-\ln \left ({\mathrm e}^{x} x -5 x -5\right )+\ln \left (x \right )\)	\(16\)
parallelrisch	\(-\ln \left ({\mathrm e}^{x} x -5 x -5\right )+\ln \left (x \right )\)	\(16\)

[In]

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

[Out]

-ln(exp(x)-5*(1+x)/x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=-\log \left (\frac {x e^{x} - 5 \, x - 5}{x}\right ) \]

[In]

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

[Out]

-log((x*e^x - 5*x - 5)/x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.58 \[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=- \log {\left (e^{x} + \frac {- 5 x - 5}{x} \right )} \]

[In]

integrate((-exp(x)*x**2-5)/(exp(x)*x**2-5*x**2-5*x),x)

[Out]

-log(exp(x) + (-5*x - 5)/x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=-\log \left (\frac {x e^{x} - 5 \, x - 5}{x}\right ) \]

[In]

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

[Out]

-log((x*e^x - 5*x - 5)/x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.62 \[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=-\log \left (x e^{x} - 5 \, x - 5\right ) + \log \left (x\right ) \]

[In]

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

[Out]

-log(x*e^x - 5*x - 5) + log(x)

Mupad [B] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {-5-e^x x^2}{-5 x-5 x^2+e^x x^2} \, dx=\ln \left (x\right )-\ln \left (5\,x-x\,{\mathrm {e}}^x+5\right ) \]

[In]

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

[Out]

log(x) - log(5*x - x*exp(x) + 5)