\(\int \frac {64-16 x^2+e^x (-256-512 x+64 x^2-128 x^3)}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx\) [3786]

Optimal result

Integrand size = 63, antiderivative size = 24 \[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=\frac {x+x \left (3+x^2\right )}{\left (\frac {1}{4}-e^x\right )^2 x^2} \]

[Out]

(x+(x^2+3)*x)/x^2/(1/4-exp(x))^2

Rubi [F]

\[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=\int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx \]

[In]

Int[(64 - 16*x^2 + E^x*(-256 - 512*x + 64*x^2 - 128*x^3))/(-x^2 + 12*E^x*x^2 - 48*E^(2*x)*x^2 + 64*E^(3*x)*x^2
),x]

[Out]

(16*x)/(1 - 4*E^x)^2 - 64*Defer[Int][1/((-1 + 4*E^x)^2*x^2), x] - 128*Defer[Int][1/((-1 + 4*E^x)^3*x), x] - 12
8*Defer[Int][1/((-1 + 4*E^x)^2*x), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {16 \left (-4+x^2+4 e^x \left (4+8 x-x^2+2 x^3\right )\right )}{\left (1-4 e^x\right )^3 x^2} \, dx \\ & = 16 \int \frac {-4+x^2+4 e^x \left (4+8 x-x^2+2 x^3\right )}{\left (1-4 e^x\right )^3 x^2} \, dx \\ & = 16 \int \left (-\frac {2 \left (4+x^2\right )}{\left (-1+4 e^x\right )^3 x}-\frac {4+8 x-x^2+2 x^3}{\left (-1+4 e^x\right )^2 x^2}\right ) \, dx \\ & = -\left (16 \int \frac {4+8 x-x^2+2 x^3}{\left (-1+4 e^x\right )^2 x^2} \, dx\right )-32 \int \frac {4+x^2}{\left (-1+4 e^x\right )^3 x} \, dx \\ & = -\left (16 \int \left (-\frac {1}{\left (-1+4 e^x\right )^2}+\frac {4}{\left (-1+4 e^x\right )^2 x^2}+\frac {8}{\left (-1+4 e^x\right )^2 x}+\frac {2 x}{\left (-1+4 e^x\right )^2}\right ) \, dx\right )-32 \int \left (\frac {4}{\left (-1+4 e^x\right )^3 x}+\frac {x}{\left (-1+4 e^x\right )^3}\right ) \, dx \\ & = 16 \int \frac {1}{\left (-1+4 e^x\right )^2} \, dx-32 \int \frac {x}{\left (-1+4 e^x\right )^3} \, dx-32 \int \frac {x}{\left (-1+4 e^x\right )^2} \, dx-64 \int \frac {1}{\left (-1+4 e^x\right )^2 x^2} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^3 x} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^2 x} \, dx \\ & = 16 \text {Subst}\left (\int \frac {1}{x (-1+4 x)^2} \, dx,x,e^x\right )+32 \int \frac {x}{\left (-1+4 e^x\right )^2} \, dx+32 \int \frac {x}{-1+4 e^x} \, dx-64 \int \frac {1}{\left (-1+4 e^x\right )^2 x^2} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^3 x} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^2 x} \, dx-128 \int \frac {e^x x}{\left (-1+4 e^x\right )^3} \, dx-128 \int \frac {e^x x}{\left (-1+4 e^x\right )^2} \, dx \\ & = \frac {16 x}{\left (1-4 e^x\right )^2}-\frac {32 x}{1-4 e^x}-16 x^2-16 \int \frac {1}{\left (-1+4 e^x\right )^2} \, dx+16 \text {Subst}\left (\int \left (\frac {1}{x}+\frac {4}{(-1+4 x)^2}-\frac {4}{-1+4 x}\right ) \, dx,x,e^x\right )-32 \int \frac {1}{-1+4 e^x} \, dx-32 \int \frac {x}{-1+4 e^x} \, dx-64 \int \frac {1}{\left (-1+4 e^x\right )^2 x^2} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^3 x} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^2 x} \, dx+128 \int \frac {e^x x}{\left (-1+4 e^x\right )^2} \, dx+128 \int \frac {e^x x}{-1+4 e^x} \, dx \\ & = \frac {16}{1-4 e^x}+16 x+\frac {16 x}{\left (1-4 e^x\right )^2}-16 \log \left (1-4 e^x\right )+32 x \log \left (1-4 e^x\right )-16 \text {Subst}\left (\int \frac {1}{x (-1+4 x)^2} \, dx,x,e^x\right )+32 \int \frac {1}{-1+4 e^x} \, dx-32 \int \log \left (1-4 e^x\right ) \, dx-32 \text {Subst}\left (\int \frac {1}{x (-1+4 x)} \, dx,x,e^x\right )-64 \int \frac {1}{\left (-1+4 e^x\right )^2 x^2} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^3 x} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^2 x} \, dx-128 \int \frac {e^x x}{-1+4 e^x} \, dx \\ & = \frac {16}{1-4 e^x}+16 x+\frac {16 x}{\left (1-4 e^x\right )^2}-16 \log \left (1-4 e^x\right )-16 \text {Subst}\left (\int \left (\frac {1}{x}+\frac {4}{(-1+4 x)^2}-\frac {4}{-1+4 x}\right ) \, dx,x,e^x\right )+32 \int \log \left (1-4 e^x\right ) \, dx+32 \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+32 \text {Subst}\left (\int \frac {1}{x (-1+4 x)} \, dx,x,e^x\right )-32 \text {Subst}\left (\int \frac {\log (1-4 x)}{x} \, dx,x,e^x\right )-64 \int \frac {1}{\left (-1+4 e^x\right )^2 x^2} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^3 x} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^2 x} \, dx-128 \text {Subst}\left (\int \frac {1}{-1+4 x} \, dx,x,e^x\right ) \\ & = 32 x+\frac {16 x}{\left (1-4 e^x\right )^2}-32 \log \left (1-4 e^x\right )+32 \text {Li}_2\left (4 e^x\right )-32 \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+32 \text {Subst}\left (\int \frac {\log (1-4 x)}{x} \, dx,x,e^x\right )-64 \int \frac {1}{\left (-1+4 e^x\right )^2 x^2} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^3 x} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^2 x} \, dx+128 \text {Subst}\left (\int \frac {1}{-1+4 x} \, dx,x,e^x\right ) \\ & = \frac {16 x}{\left (1-4 e^x\right )^2}-64 \int \frac {1}{\left (-1+4 e^x\right )^2 x^2} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^3 x} \, dx-128 \int \frac {1}{\left (-1+4 e^x\right )^2 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=-\frac {16 \left (-4-x^2\right )}{\left (-1+4 e^x\right )^2 x} \]

[In]

Integrate[(64 - 16*x^2 + E^x*(-256 - 512*x + 64*x^2 - 128*x^3))/(-x^2 + 12*E^x*x^2 - 48*E^(2*x)*x^2 + 64*E^(3*
x)*x^2),x]

[Out]

(-16*(-4 - x^2))/((-1 + 4*E^x)^2*x)

Maple [A] (verified)

Time = 495.81 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {16 x^{2}+64}{x \left (4 \,{\mathrm e}^{x}-1\right )^{2}}\)	\(19\)
norman	\(\frac {16 x^{2}+64}{x \left (4 \,{\mathrm e}^{x}-1\right )^{2}}\)	\(20\)
parallelrisch	\(\frac {256 x^{2}+1024}{16 \left (16 \,{\mathrm e}^{2 x}-8 \,{\mathrm e}^{x}+1\right ) x}\)	\(27\)

[In]

int(((-128*x^3+64*x^2-512*x-256)*exp(x)-16*x^2+64)/(64*x^2*exp(x)^3-48*exp(x)^2*x^2+12*exp(x)*x^2-x^2),x,metho
d=_RETURNVERBOSE)

[Out]

16/x*(x^2+4)/(4*exp(x)-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=\frac {16 \, {\left (x^{2} + 4\right )}}{16 \, x e^{\left (2 \, x\right )} - 8 \, x e^{x} + x} \]

[In]

integrate(((-128*x^3+64*x^2-512*x-256)*exp(x)-16*x^2+64)/(64*x^2*exp(x)^3-48*exp(x)^2*x^2+12*exp(x)*x^2-x^2),x
, algorithm="fricas")

[Out]

16*(x^2 + 4)/(16*x*e^(2*x) - 8*x*e^x + x)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=\frac {16 x^{2} + 64}{16 x e^{2 x} - 8 x e^{x} + x} \]

[In]

integrate(((-128*x**3+64*x**2-512*x-256)*exp(x)-16*x**2+64)/(64*x**2*exp(x)**3-48*exp(x)**2*x**2+12*exp(x)*x**
2-x**2),x)

[Out]

(16*x**2 + 64)/(16*x*exp(2*x) - 8*x*exp(x) + x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=\frac {16 \, {\left (x^{2} + 4\right )}}{16 \, x e^{\left (2 \, x\right )} - 8 \, x e^{x} + x} \]

[In]

integrate(((-128*x^3+64*x^2-512*x-256)*exp(x)-16*x^2+64)/(64*x^2*exp(x)^3-48*exp(x)^2*x^2+12*exp(x)*x^2-x^2),x
, algorithm="maxima")

[Out]

16*(x^2 + 4)/(16*x*e^(2*x) - 8*x*e^x + x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=\frac {16 \, {\left (x^{2} + 4\right )}}{16 \, x e^{\left (2 \, x\right )} - 8 \, x e^{x} + x} \]

[In]

integrate(((-128*x^3+64*x^2-512*x-256)*exp(x)-16*x^2+64)/(64*x^2*exp(x)^3-48*exp(x)^2*x^2+12*exp(x)*x^2-x^2),x
, algorithm="giac")

[Out]

16*(x^2 + 4)/(16*x*e^(2*x) - 8*x*e^x + x)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.75 \[ \int \frac {64-16 x^2+e^x \left (-256-512 x+64 x^2-128 x^3\right )}{-x^2+12 e^x x^2-48 e^{2 x} x^2+64 e^{3 x} x^2} \, dx=\frac {16\,\left (x^2+4\right )}{x\,{\left (4\,{\mathrm {e}}^x-1\right )}^2} \]

[In]

int(-(16*x^2 + exp(x)*(512*x - 64*x^2 + 128*x^3 + 256) - 64)/(12*x^2*exp(x) - 48*x^2*exp(2*x) + 64*x^2*exp(3*x
) - x^2),x)

[Out]

(16*(x^2 + 4))/(x*(4*exp(x) - 1)^2)