\(\int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx\) [3109]

Optimal result

Integrand size = 29, antiderivative size = 20 \[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=\log \left (8-\frac {1}{2} e^{-3-x} x (1+2 x)\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=\int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx \]

[In]

Int[(-1 - 3*x + 2*x^2)/(16*E^(3 + x) - x - 2*x^2),x]

[Out]

-Defer[Int][(16*E^(3 + x) - x - 2*x^2)^(-1), x] + 3*Defer[Int][x/(-16*E^(3 + x) + x + 2*x^2), x] - 2*Defer[Int
][x^2/(-16*E^(3 + x) + x + 2*x^2), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {1}{16 e^{3+x}-x-2 x^2}+\frac {3 x}{-16 e^{3+x}+x+2 x^2}-\frac {2 x^2}{-16 e^{3+x}+x+2 x^2}\right ) \, dx \\ & = -\left (2 \int \frac {x^2}{-16 e^{3+x}+x+2 x^2} \, dx\right )+3 \int \frac {x}{-16 e^{3+x}+x+2 x^2} \, dx-\int \frac {1}{16 e^{3+x}-x-2 x^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=-x+\log \left (16 e^{3+x}-x-2 x^2\right ) \]

[In]

Integrate[(-1 - 3*x + 2*x^2)/(16*E^(3 + x) - x - 2*x^2),x]

[Out]

-x + Log[16*E^(3 + x) - x - 2*x^2]

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
risch	\(-x +\ln \left ({\mathrm e}^{x}-\frac {\left (1+2 x \right ) x \,{\mathrm e}^{-3}}{16}\right )\)	\(19\)
parallelrisch	\(-x +\ln \left (-8 \,{\mathrm e}^{x} {\mathrm e}^{3}+x^{2}+\frac {x}{2}\right )\)	\(19\)
norman	\(-x +\ln \left (16 \,{\mathrm e}^{x} {\mathrm e}^{3}-2 x^{2}-x \right )\)	\(21\)

[In]

int((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x,method=_RETURNVERBOSE)

[Out]

-x+ln(exp(x)-1/16*(1+2*x)*x*exp(-3))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=-x + \log \left (-2 \, x^{2} - x + 16 \, e^{\left (x + 3\right )}\right ) \]

[In]

integrate((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x, algorithm="fricas")

[Out]

-x + log(-2*x^2 - x + 16*e^(x + 3))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=- x + \log {\left (\frac {- 2 x^{2} - x}{16 e^{3}} + e^{x} \right )} \]

[In]

integrate((2*x**2-3*x-1)/(16*exp(x)*exp(3)-2*x**2-x),x)

[Out]

-x + log((-2*x**2 - x)*exp(-3)/16 + exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=-x + \log \left (-\frac {1}{16} \, {\left (2 \, x^{2} + x - 16 \, e^{\left (x + 3\right )}\right )} e^{\left (-3\right )}\right ) \]

[In]

integrate((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x, algorithm="maxima")

[Out]

-x + log(-1/16*(2*x^2 + x - 16*e^(x + 3))*e^(-3))

Giac [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=-x + \log \left (2 \, x^{2} + x - 16 \, e^{\left (x + 3\right )}\right ) \]

[In]

integrate((2*x^2-3*x-1)/(16*exp(x)*exp(3)-2*x^2-x),x, algorithm="giac")

[Out]

-x + log(2*x^2 + x - 16*e^(x + 3))

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {-1-3 x+2 x^2}{16 e^{3+x}-x-2 x^2} \, dx=\ln \left (x-16\,{\mathrm {e}}^{x+3}+2\,x^2\right )-x \]

[In]

int((3*x - 2*x^2 + 1)/(x - 16*exp(3)*exp(x) + 2*x^2),x)

[Out]

log(x - 16*exp(x + 3) + 2*x^2) - x