\(\int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx\) [2809]

Optimal result

Integrand size = 66, antiderivative size = 30 \[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=\log \left (4 \left (1+e^2 \left (4+e^{2 x}-e^{3+x}\right )+\frac {3}{2 x}\right )\right ) \]

[Out]

ln(4*(4+exp(x)^2-exp(3+x))*exp(2)+6/x+4)

Rubi [F]

\[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=\int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx \]

[In]

Int[(3 + 2*E^(5 + x)*x^2 - 4*E^(2 + 2*x)*x^2)/(-3*x - 2*x^2 - 8*E^2*x^2 + 2*E^(5 + x)*x^2 - 2*E^(2 + 2*x)*x^2)
,x]

[Out]

2*x + 6*Defer[Int][(-3 + 2*E^(5 + x)*x - 2*E^(2 + 2*x)*x - 2*(1 + 4*E^2)*x)^(-1), x] + 3*Defer[Int][1/(x*(-3 +
2*E^(5 + x)*x - 2*E^(2 + 2*x)*x - 2*(1 + 4*E^2)*x)), x] - 4*(1 + 4*E^2)*Defer[Int][x/(3 - 2*E^(5 + x)*x + 2*E
^(2 + 2*x)*x + 2*(1 + 4*E^2)*x), x] + 2*Defer[Int][(E^(5 + x)*x)/(3 - 2*E^(5 + x)*x + 2*E^(2 + 2*x)*x + 2*(1 +
4*E^2)*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x+2 e^{5+x} x^2-2 e^{2+2 x} x^2+\left (-2-8 e^2\right ) x^2} \, dx \\ & = \int \left (2+\frac {-3-6 x+2 e^{5+x} x^2-4 \left (1+4 e^2\right ) x^2}{x \left (3-2 e^{5+x} x+2 e^{2+2 x} x+2 \left (1+4 e^2\right ) x\right )}\right ) \, dx \\ & = 2 x+\int \frac {-3-6 x+2 e^{5+x} x^2-4 \left (1+4 e^2\right ) x^2}{x \left (3-2 e^{5+x} x+2 e^{2+2 x} x+2 \left (1+4 e^2\right ) x\right )} \, dx \\ & = 2 x+\int \left (\frac {6}{-3+2 e^{5+x} x-2 e^{2+2 x} x-2 \left (1+4 e^2\right ) x}+\frac {3}{x \left (-3+2 e^{5+x} x-2 e^{2+2 x} x-2 \left (1+4 e^2\right ) x\right )}+\frac {2 e^{5+x} x}{3-2 e^{5+x} x+2 e^{2+2 x} x+2 \left (1+4 e^2\right ) x}+\frac {4 \left (-1-4 e^2\right ) x}{3-2 e^{5+x} x+2 e^{2+2 x} x+2 \left (1+4 e^2\right ) x}\right ) \, dx \\ & = 2 x+2 \int \frac {e^{5+x} x}{3-2 e^{5+x} x+2 e^{2+2 x} x+2 \left (1+4 e^2\right ) x} \, dx+3 \int \frac {1}{x \left (-3+2 e^{5+x} x-2 e^{2+2 x} x-2 \left (1+4 e^2\right ) x\right )} \, dx+6 \int \frac {1}{-3+2 e^{5+x} x-2 e^{2+2 x} x-2 \left (1+4 e^2\right ) x} \, dx-\left (4 \left (1+4 e^2\right )\right ) \int \frac {x}{3-2 e^{5+x} x+2 e^{2+2 x} x+2 \left (1+4 e^2\right ) x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.11 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=-\log (x)+\log \left (3+2 x+8 e^2 x-2 e^{5+x} x+2 e^{2+2 x} x\right ) \]

[In]

Integrate[(3 + 2*E^(5 + x)*x^2 - 4*E^(2 + 2*x)*x^2)/(-3*x - 2*x^2 - 8*E^2*x^2 + 2*E^(5 + x)*x^2 - 2*E^(2 + 2*x
)*x^2),x]

[Out]

-Log[x] + Log[3 + 2*x + 8*E^2*x - 2*E^(5 + x)*x + 2*E^(2 + 2*x)*x]

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00

method	result	size
risch	\(\ln \left ({\mathrm e}^{2 x}-{\mathrm e}^{3+x}+\frac {\left (8 \,{\mathrm e}^{2} x +2 x +3\right ) {\mathrm e}^{-2}}{2 x}\right )\)	\(30\)
norman	\(-\ln \left (x \right )+\ln \left (2 \,{\mathrm e}^{2} {\mathrm e}^{x} {\mathrm e}^{3} x -2 x \,{\mathrm e}^{2} {\mathrm e}^{2 x}-8 \,{\mathrm e}^{2} x -2 x -3\right )\)	\(35\)
parallelrisch	\(-\ln \left (x \right )+\ln \left (\frac {\left (2 x \,{\mathrm e}^{2} {\mathrm e}^{2 x}-2 x \,{\mathrm e}^{2} {\mathrm e}^{3+x}+8 \,{\mathrm e}^{2} x +2 x +3\right ) {\mathrm e}^{-2}}{2}\right )\)	\(41\)

[In]

int((2*x^2*exp(2)*exp(3+x)-4*x^2*exp(2)*exp(x)^2+3)/(2*x^2*exp(2)*exp(3+x)-2*x^2*exp(2)*exp(x)^2-8*x^2*exp(2)-
2*x^2-3*x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=\log \left (\frac {8 \, x e^{10} + {\left (2 \, x + 3\right )} e^{8} + 2 \, x e^{\left (2 \, x + 10\right )} - 2 \, x e^{\left (x + 13\right )}}{x}\right ) \]

[In]

integrate((2*x^2*exp(2)*exp(3+x)-4*x^2*exp(2)*exp(x)^2+3)/(2*x^2*exp(2)*exp(3+x)-2*x^2*exp(2)*exp(x)^2-8*x^2*e
xp(2)-2*x^2-3*x),x, algorithm="fricas")

[Out]

log((8*x*e^10 + (2*x + 3)*e^8 + 2*x*e^(2*x + 10) - 2*x*e^(x + 13))/x)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20 \[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=\log {\left (e^{2 x} - e^{3} \sqrt {e^{2 x}} + \frac {2 x + 8 x e^{2} + 3}{2 x e^{2}} \right )} \]

[In]

integrate((2*x**2*exp(2)*exp(3+x)-4*x**2*exp(2)*exp(x)**2+3)/(2*x**2*exp(2)*exp(3+x)-2*x**2*exp(2)*exp(x)**2-8
*x**2*exp(2)-2*x**2-3*x),x)

[Out]

log(exp(2*x) - exp(3)*sqrt(exp(2*x)) + (2*x + 8*x*exp(2) + 3)*exp(-2)/(2*x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.17 \[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=\log \left (\frac {{\left (2 \, x {\left (4 \, e^{2} + 1\right )} + 2 \, x e^{\left (2 \, x + 2\right )} - 2 \, x e^{\left (x + 5\right )} + 3\right )} e^{\left (-2\right )}}{2 \, x}\right ) \]

[In]

integrate((2*x^2*exp(2)*exp(3+x)-4*x^2*exp(2)*exp(x)^2+3)/(2*x^2*exp(2)*exp(3+x)-2*x^2*exp(2)*exp(x)^2-8*x^2*e
xp(2)-2*x^2-3*x),x, algorithm="maxima")

[Out]

log(1/2*(2*x*(4*e^2 + 1) + 2*x*e^(2*x + 2) - 2*x*e^(x + 5) + 3)*e^(-2)/x)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).

Time = 0.31 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.10 \[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=\log \left (8 \, {\left (x + 5\right )} e^{10} + 2 \, {\left (x + 5\right )} e^{8} + 2 \, {\left (x + 5\right )} e^{\left (2 \, x + 10\right )} - 2 \, {\left (x + 5\right )} e^{\left (x + 13\right )} - 40 \, e^{10} - 7 \, e^{8} - 10 \, e^{\left (2 \, x + 10\right )} + 10 \, e^{\left (x + 13\right )}\right ) - \log \left (x\right ) \]

[In]

integrate((2*x^2*exp(2)*exp(3+x)-4*x^2*exp(2)*exp(x)^2+3)/(2*x^2*exp(2)*exp(3+x)-2*x^2*exp(2)*exp(x)^2-8*x^2*e
xp(2)-2*x^2-3*x),x, algorithm="giac")

[Out]

log(8*(x + 5)*e^10 + 2*(x + 5)*e^8 + 2*(x + 5)*e^(2*x + 10) - 2*(x + 5)*e^(x + 13) - 40*e^10 - 7*e^8 - 10*e^(2
*x + 10) + 10*e^(x + 13)) - log(x)

Mupad [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {3+2 e^{5+x} x^2-4 e^{2+2 x} x^2}{-3 x-2 x^2-8 e^2 x^2+2 e^{5+x} x^2-2 e^{2+2 x} x^2} \, dx=\ln \left (3\,{\mathrm {e}}^8+2\,x\,{\mathrm {e}}^8+8\,x\,{\mathrm {e}}^{10}-2\,x\,{\mathrm {e}}^{13}\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{10}\right )-\ln \left (x\right ) \]

[In]

int(-(2*x^2*exp(x + 3)*exp(2) - 4*x^2*exp(2*x)*exp(2) + 3)/(3*x + 8*x^2*exp(2) + 2*x^2 + 2*x^2*exp(2*x)*exp(2)
- 2*x^2*exp(x + 3)*exp(2)),x)

[Out]

log(3*exp(8) + 2*x*exp(8) + 8*x*exp(10) - 2*x*exp(13)*exp(x) + 2*x*exp(2*x)*exp(10)) - log(x)