\(\int \frac {80+16 x^2+e^{2 x} x^4+e^x (40 x+22 x^2+10 x^3)}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx\) [9978]

Optimal result

Integrand size = 60, antiderivative size = 20 \[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=1+x+\frac {-2-\frac {20}{x}}{4+e^x x} \]

[Out]

(-2-20/x)/(exp(x)*x+4)+x+1

Rubi [F]

\[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=\int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx \]

[In]

Int[(80 + 16*x^2 + E^(2*x)*x^4 + E^x*(40*x + 22*x^2 + 10*x^3))/(16*x^2 + 8*E^x*x^3 + E^(2*x)*x^4),x]

[Out]

x - 8*Defer[Int][(4 + E^x*x)^(-2), x] - 80*Defer[Int][1/(x^2*(4 + E^x*x)^2), x] - 88*Defer[Int][1/(x*(4 + E^x*
x)^2), x] + 2*Defer[Int][(4 + E^x*x)^(-1), x] + 40*Defer[Int][1/(x^2*(4 + E^x*x)), x] + 22*Defer[Int][1/(x*(4
+ E^x*x)), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{x^2 \left (4+e^x x\right )^2} \, dx \\ & = \int \left (1-\frac {8 \left (10+11 x+x^2\right )}{x^2 \left (4+e^x x\right )^2}+\frac {2 \left (20+11 x+x^2\right )}{x^2 \left (4+e^x x\right )}\right ) \, dx \\ & = x+2 \int \frac {20+11 x+x^2}{x^2 \left (4+e^x x\right )} \, dx-8 \int \frac {10+11 x+x^2}{x^2 \left (4+e^x x\right )^2} \, dx \\ & = x+2 \int \left (\frac {1}{4+e^x x}+\frac {20}{x^2 \left (4+e^x x\right )}+\frac {11}{x \left (4+e^x x\right )}\right ) \, dx-8 \int \left (\frac {1}{\left (4+e^x x\right )^2}+\frac {10}{x^2 \left (4+e^x x\right )^2}+\frac {11}{x \left (4+e^x x\right )^2}\right ) \, dx \\ & = x+2 \int \frac {1}{4+e^x x} \, dx-8 \int \frac {1}{\left (4+e^x x\right )^2} \, dx+22 \int \frac {1}{x \left (4+e^x x\right )} \, dx+40 \int \frac {1}{x^2 \left (4+e^x x\right )} \, dx-80 \int \frac {1}{x^2 \left (4+e^x x\right )^2} \, dx-88 \int \frac {1}{x \left (4+e^x x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 2.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=x-\frac {2 (10+x)}{x \left (4+e^x x\right )} \]

[In]

Integrate[(80 + 16*x^2 + E^(2*x)*x^4 + E^x*(40*x + 22*x^2 + 10*x^3))/(16*x^2 + 8*E^x*x^3 + E^(2*x)*x^4),x]

[Out]

x - (2*(10 + x))/(x*(4 + E^x*x))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
risch	\(x -\frac {2 \left (x +10\right )}{x \left ({\mathrm e}^{x} x +4\right )}\)	\(19\)
norman	\(\frac {-20-2 x +{\mathrm e}^{x} x^{3}+4 x^{2}}{x \left ({\mathrm e}^{x} x +4\right )}\)	\(29\)
parallelrisch	\(\frac {-20-2 x +{\mathrm e}^{x} x^{3}+4 x^{2}}{x \left ({\mathrm e}^{x} x +4\right )}\)	\(29\)

[In]

int((exp(x)^2*x^4+(10*x^3+22*x^2+40*x)*exp(x)+16*x^2+80)/(exp(x)^2*x^4+8*exp(x)*x^3+16*x^2),x,method=_RETURNVE
RBOSE)

[Out]

x-2*(x+10)/x/(exp(x)*x+4)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=\frac {x^{3} e^{x} + 4 \, x^{2} - 2 \, x - 20}{x^{2} e^{x} + 4 \, x} \]

[In]

integrate((exp(x)^2*x^4+(10*x^3+22*x^2+40*x)*exp(x)+16*x^2+80)/(exp(x)^2*x^4+8*exp(x)*x^3+16*x^2),x, algorithm
="fricas")

[Out]

(x^3*e^x + 4*x^2 - 2*x - 20)/(x^2*e^x + 4*x)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=x + \frac {- 2 x - 20}{x^{2} e^{x} + 4 x} \]

[In]

integrate((exp(x)**2*x**4+(10*x**3+22*x**2+40*x)*exp(x)+16*x**2+80)/(exp(x)**2*x**4+8*exp(x)*x**3+16*x**2),x)

[Out]

x + (-2*x - 20)/(x**2*exp(x) + 4*x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=\frac {x^{3} e^{x} + 4 \, x^{2} - 2 \, x - 20}{x^{2} e^{x} + 4 \, x} \]

[In]

integrate((exp(x)^2*x^4+(10*x^3+22*x^2+40*x)*exp(x)+16*x^2+80)/(exp(x)^2*x^4+8*exp(x)*x^3+16*x^2),x, algorithm
="maxima")

[Out]

(x^3*e^x + 4*x^2 - 2*x - 20)/(x^2*e^x + 4*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=\frac {x^{3} e^{x} + 4 \, x^{2} - 2 \, x - 20}{x^{2} e^{x} + 4 \, x} \]

[In]

integrate((exp(x)^2*x^4+(10*x^3+22*x^2+40*x)*exp(x)+16*x^2+80)/(exp(x)^2*x^4+8*exp(x)*x^3+16*x^2),x, algorithm
="giac")

[Out]

(x^3*e^x + 4*x^2 - 2*x - 20)/(x^2*e^x + 4*x)

Mupad [B] (verification not implemented)

Time = 16.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {80+16 x^2+e^{2 x} x^4+e^x \left (40 x+22 x^2+10 x^3\right )}{16 x^2+8 e^x x^3+e^{2 x} x^4} \, dx=x-\frac {2\,x+20}{x\,\left (x\,{\mathrm {e}}^x+4\right )} \]

[In]

int((x^4*exp(2*x) + 16*x^2 + exp(x)*(40*x + 22*x^2 + 10*x^3) + 80)/(8*x^3*exp(x) + x^4*exp(2*x) + 16*x^2),x)

[Out]

x - (2*x + 20)/(x*(x*exp(x) + 4))