\(\int \frac {e^{2 x} (-200 x^7+50 e^4 x^7-25 x^8)+e^x (-120 x^2-80 x^3-10 x^4+e^4 (30 x^2+10 x^3))}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx\) [2196]

Optimal result

Integrand size = 84, antiderivative size = 27 \[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=\frac {4-e^4+x}{-\frac {2 e^{-x}}{5 x^3}+x^2} \]

[Out]

(4-exp(4)+x)/(x^2-2/5/x^3/exp(x))

Rubi [F]

\[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=\int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx \]

[In]

Int[(E^(2*x)*(-200*x^7 + 50*E^4*x^7 - 25*x^8) + E^x*(-120*x^2 - 80*x^3 - 10*x^4 + E^4*(30*x^2 + 10*x^3)))/(4 -
20*E^x*x^5 + 25*E^(2*x)*x^10),x]

[Out]

-50*(4 - E^4)*Defer[Int][(E^x*x^2)/(-2 + 5*E^x*x^5)^2, x] - 10*(9 - E^4)*Defer[Int][(E^x*x^3)/(-2 + 5*E^x*x^5)
^2, x] - 10*Defer[Int][(E^x*x^4)/(-2 + 5*E^x*x^5)^2, x] - 10*(4 - E^4)*Defer[Int][(E^x*x^2)/(-2 + 5*E^x*x^5),
x] - 5*Defer[Int][(E^x*x^3)/(-2 + 5*E^x*x^5), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e^x x^2 \left (10 e^{4+x} x^5+2 e^4 (3+x)-5 e^x x^5 (8+x)-2 \left (12+8 x+x^2\right )\right )}{\left (2-5 e^x x^5\right )^2} \, dx \\ & = 5 \int \frac {e^x x^2 \left (10 e^{4+x} x^5+2 e^4 (3+x)-5 e^x x^5 (8+x)-2 \left (12+8 x+x^2\right )\right )}{\left (2-5 e^x x^5\right )^2} \, dx \\ & = 5 \int \left (-\frac {2 e^x x^2 (5+x) \left (4-e^4+x\right )}{\left (-2+5 e^x x^5\right )^2}-\frac {e^x x^2 \left (8-2 e^4+x\right )}{-2+5 e^x x^5}\right ) \, dx \\ & = -\left (5 \int \frac {e^x x^2 \left (8-2 e^4+x\right )}{-2+5 e^x x^5} \, dx\right )-10 \int \frac {e^x x^2 (5+x) \left (4-e^4+x\right )}{\left (-2+5 e^x x^5\right )^2} \, dx \\ & = -\left (5 \int \left (-\frac {2 e^x \left (-4+e^4\right ) x^2}{-2+5 e^x x^5}+\frac {e^x x^3}{-2+5 e^x x^5}\right ) \, dx\right )-10 \int \left (-\frac {5 e^x \left (-4+e^4\right ) x^2}{\left (-2+5 e^x x^5\right )^2}-\frac {e^x \left (-9+e^4\right ) x^3}{\left (-2+5 e^x x^5\right )^2}+\frac {e^x x^4}{\left (-2+5 e^x x^5\right )^2}\right ) \, dx \\ & = -\left (5 \int \frac {e^x x^3}{-2+5 e^x x^5} \, dx\right )-10 \int \frac {e^x x^4}{\left (-2+5 e^x x^5\right )^2} \, dx-\left (10 \left (4-e^4\right )\right ) \int \frac {e^x x^2}{-2+5 e^x x^5} \, dx-\left (50 \left (4-e^4\right )\right ) \int \frac {e^x x^2}{\left (-2+5 e^x x^5\right )^2} \, dx-\left (10 \left (9-e^4\right )\right ) \int \frac {e^x x^3}{\left (-2+5 e^x x^5\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 9.67 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=-\frac {5 e^x \left (-4+e^4-x\right ) x^3}{-2+5 e^x x^5} \]

[In]

Integrate[(E^(2*x)*(-200*x^7 + 50*E^4*x^7 - 25*x^8) + E^x*(-120*x^2 - 80*x^3 - 10*x^4 + E^4*(30*x^2 + 10*x^3))
)/(4 - 20*E^x*x^5 + 25*E^(2*x)*x^10),x]

[Out]

(-5*E^x*(-4 + E^4 - x)*x^3)/(-2 + 5*E^x*x^5)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22

method	result	size
norman	\(\frac {\left (20-5 \,{\mathrm e}^{4}\right ) x^{3} {\mathrm e}^{x}+5 \,{\mathrm e}^{x} x^{4}}{5 x^{5} {\mathrm e}^{x}-2}\)	\(33\)
risch	\(\frac {4-{\mathrm e}^{4}+x}{x^{2}}-\frac {2 \left ({\mathrm e}^{4}-x -4\right )}{x^{2} \left (5 x^{5} {\mathrm e}^{x}-2\right )}\)	\(36\)
parallelrisch	\(-\frac {25 \,{\mathrm e}^{x} x^{3} {\mathrm e}^{4}-25 \,{\mathrm e}^{x} x^{4}-100 \,{\mathrm e}^{x} x^{3}}{5 \left (5 x^{5} {\mathrm e}^{x}-2\right )}\)	\(38\)

[In]

int(((50*x^7*exp(4)-25*x^8-200*x^7)*exp(x)^2+((10*x^3+30*x^2)*exp(4)-10*x^4-80*x^3-120*x^2)*exp(x))/(25*x^10*e
xp(x)^2-20*x^5*exp(x)+4),x,method=_RETURNVERBOSE)

[Out]

((20-5*exp(4))*x^3*exp(x)+5*exp(x)*x^4)/(5*x^5*exp(x)-2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=\frac {5 \, {\left (x^{4} - x^{3} e^{4} + 4 \, x^{3}\right )} e^{x}}{5 \, x^{5} e^{x} - 2} \]

[In]

integrate(((50*x^7*exp(4)-25*x^8-200*x^7)*exp(x)^2+((10*x^3+30*x^2)*exp(4)-10*x^4-80*x^3-120*x^2)*exp(x))/(25*
x^10*exp(x)^2-20*x^5*exp(x)+4),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=\frac {2 x - 2 e^{4} + 8}{5 x^{7} e^{x} - 2 x^{2}} - \frac {- x - 4 + e^{4}}{x^{2}} \]

[In]

integrate(((50*x**7*exp(4)-25*x**8-200*x**7)*exp(x)**2+((10*x**3+30*x**2)*exp(4)-10*x**4-80*x**3-120*x**2)*exp
(x))/(25*x**10*exp(x)**2-20*x**5*exp(x)+4),x)

[Out]

(2*x - 2*exp(4) + 8)/(5*x**7*exp(x) - 2*x**2) - (-x - 4 + exp(4))/x**2

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04 \[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=\frac {5 \, {\left (x^{4} - x^{3} {\left (e^{4} - 4\right )}\right )} e^{x}}{5 \, x^{5} e^{x} - 2} \]

[In]

integrate(((50*x^7*exp(4)-25*x^8-200*x^7)*exp(x)^2+((10*x^3+30*x^2)*exp(4)-10*x^4-80*x^3-120*x^2)*exp(x))/(25*
x^10*exp(x)^2-20*x^5*exp(x)+4),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=\frac {5 \, {\left (x^{4} e^{x} - x^{3} e^{\left (x + 4\right )} + 4 \, x^{3} e^{x}\right )}}{5 \, x^{5} e^{x} - 2} \]

[In]

integrate(((50*x^7*exp(4)-25*x^8-200*x^7)*exp(x)^2+((10*x^3+30*x^2)*exp(4)-10*x^4-80*x^3-120*x^2)*exp(x))/(25*
x^10*exp(x)^2-20*x^5*exp(x)+4),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.91 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {e^{2 x} \left (-200 x^7+50 e^4 x^7-25 x^8\right )+e^x \left (-120 x^2-80 x^3-10 x^4+e^4 \left (30 x^2+10 x^3\right )\right )}{4-20 e^x x^5+25 e^{2 x} x^{10}} \, dx=\frac {5\,x^3\,{\mathrm {e}}^x\,\left (x-{\mathrm {e}}^4+4\right )}{5\,x^5\,{\mathrm {e}}^x-2} \]

[In]

int(-(exp(2*x)*(200*x^7 - 50*x^7*exp(4) + 25*x^8) + exp(x)*(120*x^2 - exp(4)*(30*x^2 + 10*x^3) + 80*x^3 + 10*x
^4))/(25*x^10*exp(2*x) - 20*x^5*exp(x) + 4),x)

[Out]

(5*x^3*exp(x)*(x - exp(4) + 4))/(5*x^5*exp(x) - 2)