\(\int \frac {8 x+e^x x (-3-3 x-x^2)}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx\) [4618]

Optimal result

Integrand size = 43, antiderivative size = 13 \[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\frac {3+x}{8+e^x x} \]

[Out]

(3+x)/(8+exp(x+ln(x)))

Rubi [F]

\[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx \]

[In]

Int[(8*x + E^x*x*(-3 - 3*x - x^2))/(64*x + 16*E^x*x^2 + E^(2*x)*x^3),x]

[Out]

32*Defer[Int][(8 + E^x*x)^(-2), x] + 24*Defer[Int][1/(x*(8 + E^x*x)^2), x] + 8*Defer[Int][x/(8 + E^x*x)^2, x]
- 3*Defer[Int][(8 + E^x*x)^(-1), x] - 3*Defer[Int][1/(x*(8 + E^x*x)), x] - Defer[Int][x/(8 + E^x*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {8-e^x \left (3+3 x+x^2\right )}{\left (8+e^x x\right )^2} \, dx \\ & = \int \left (-\frac {3+3 x+x^2}{x \left (8+e^x x\right )}+\frac {8 \left (3+4 x+x^2\right )}{x \left (8+e^x x\right )^2}\right ) \, dx \\ & = 8 \int \frac {3+4 x+x^2}{x \left (8+e^x x\right )^2} \, dx-\int \frac {3+3 x+x^2}{x \left (8+e^x x\right )} \, dx \\ & = 8 \int \left (\frac {4}{\left (8+e^x x\right )^2}+\frac {3}{x \left (8+e^x x\right )^2}+\frac {x}{\left (8+e^x x\right )^2}\right ) \, dx-\int \left (\frac {3}{8+e^x x}+\frac {3}{x \left (8+e^x x\right )}+\frac {x}{8+e^x x}\right ) \, dx \\ & = -\left (3 \int \frac {1}{8+e^x x} \, dx\right )-3 \int \frac {1}{x \left (8+e^x x\right )} \, dx+8 \int \frac {x}{\left (8+e^x x\right )^2} \, dx+24 \int \frac {1}{x \left (8+e^x x\right )^2} \, dx+32 \int \frac {1}{\left (8+e^x x\right )^2} \, dx-\int \frac {x}{8+e^x x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\frac {3+x}{8+e^x x} \]

[In]

Integrate[(8*x + E^x*x*(-3 - 3*x - x^2))/(64*x + 16*E^x*x^2 + E^(2*x)*x^3),x]

[Out]

(3 + x)/(8 + E^x*x)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {3+x}{8+{\mathrm e}^{x} x}\)	\(13\)
norman	\(\frac {3+x}{8+{\mathrm e}^{x +\ln \left (x \right )}}\)	\(14\)
parallelrisch	\(\frac {3+x}{8+{\mathrm e}^{x +\ln \left (x \right )}}\)	\(14\)

[In]

int(((-x^2-3*x-3)*exp(x+ln(x))+8*x)/(x*exp(x+ln(x))^2+16*exp(x+ln(x))*x+64*x),x,method=_RETURNVERBOSE)

[Out]

(3+x)/(8+exp(x)*x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\frac {x + 3}{e^{\left (x + \log \left (x\right )\right )} + 8} \]

[In]

integrate(((-x^2-3*x-3)*exp(x+log(x))+8*x)/(x*exp(x+log(x))^2+16*exp(x+log(x))*x+64*x),x, algorithm="fricas")

[Out]

(x + 3)/(e^(x + log(x)) + 8)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.62 \[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\frac {x + 3}{x e^{x} + 8} \]

[In]

integrate(((-x**2-3*x-3)*exp(x+ln(x))+8*x)/(x*exp(x+ln(x))**2+16*exp(x+ln(x))*x+64*x),x)

[Out]

(x + 3)/(x*exp(x) + 8)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\frac {x + 3}{x e^{x} + 8} \]

[In]

integrate(((-x^2-3*x-3)*exp(x+log(x))+8*x)/(x*exp(x+log(x))^2+16*exp(x+log(x))*x+64*x),x, algorithm="maxima")

[Out]

(x + 3)/(x*e^x + 8)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\frac {x + 3}{x e^{x} + 8} \]

[In]

integrate(((-x^2-3*x-3)*exp(x+log(x))+8*x)/(x*exp(x+log(x))^2+16*exp(x+log(x))*x+64*x),x, algorithm="giac")

[Out]

(x + 3)/(x*e^x + 8)

Mupad [B] (verification not implemented)

Time = 10.42 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {8 x+e^x x \left (-3-3 x-x^2\right )}{64 x+16 e^x x^2+e^{2 x} x^3} \, dx=\frac {x+3}{x\,{\mathrm {e}}^x+8} \]

[In]

int((8*x - exp(x + log(x))*(3*x + x^2 + 3))/(64*x + x*exp(2*x + 2*log(x)) + 16*x*exp(x + log(x))),x)

[Out]

(x + 3)/(x*exp(x) + 8)