\(\int \frac {e^2 (160-10 x^2)+e^{-1+x} (-32-64 x-30 x^2-4 x^3)}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx\) [6729]

Optimal result

Integrand size = 76, antiderivative size = 28 \[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=-2+\frac {2 e^2 (4+x)^2}{25 \left (5-e^{-3+x}\right )^2 x} \]

[Out]

2/25/x/(5-exp(-3+x))^2*exp(1)^2*(4+x)^2-2

Rubi [F]

\[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=\int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx \]

[In]

Int[(E^2*(160 - 10*x^2) + E^(-1 + x)*(-32 - 64*x - 30*x^2 - 4*x^3))/(-3125*x^2 + 1875*E^(-3 + x)*x^2 - 375*E^(
-6 + 2*x)*x^2 + 25*E^(-9 + 3*x)*x^2),x]

[Out]

(16*E^8)/(25*(5*E^3 - E^x)^2) + (2*E^8*x)/(25*(5*E^3 - E^x)^2) - (32*E^8*Defer[Int][1/((-5*E^3 + E^x)^2*x^2),
x])/25 - (64*E^11*Defer[Int][1/((-5*E^3 + E^x)^3*x), x])/5 - (64*E^8*Defer[Int][1/((-5*E^3 + E^x)^2*x), x])/25

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 e^8 (4+x) \left (5 e^3 (-4+x)+e^x \left (4+7 x+2 x^2\right )\right )}{25 \left (5 e^3-e^x\right )^3 x^2} \, dx \\ & = \frac {1}{25} \left (2 e^8\right ) \int \frac {(4+x) \left (5 e^3 (-4+x)+e^x \left (4+7 x+2 x^2\right )\right )}{\left (5 e^3-e^x\right )^3 x^2} \, dx \\ & = \frac {1}{25} \left (2 e^8\right ) \int \left (\frac {10 e^3 (4+x)^2}{\left (5 e^3-e^x\right )^3 x}-\frac {16+32 x+15 x^2+2 x^3}{\left (-5 e^3+e^x\right )^2 x^2}\right ) \, dx \\ & = -\left (\frac {1}{25} \left (2 e^8\right ) \int \frac {16+32 x+15 x^2+2 x^3}{\left (-5 e^3+e^x\right )^2 x^2} \, dx\right )+\frac {1}{5} \left (4 e^{11}\right ) \int \frac {(4+x)^2}{\left (5 e^3-e^x\right )^3 x} \, dx \\ & = -\left (\frac {1}{25} \left (2 e^8\right ) \int \left (\frac {15}{\left (-5 e^3+e^x\right )^2}+\frac {16}{\left (-5 e^3+e^x\right )^2 x^2}+\frac {32}{\left (-5 e^3+e^x\right )^2 x}+\frac {2 x}{\left (-5 e^3+e^x\right )^2}\right ) \, dx\right )+\frac {1}{5} \left (4 e^{11}\right ) \int \left (-\frac {8}{\left (-5 e^3+e^x\right )^3}-\frac {16}{\left (-5 e^3+e^x\right )^3 x}-\frac {x}{\left (-5 e^3+e^x\right )^3}\right ) \, dx \\ & = -\left (\frac {1}{25} \left (4 e^8\right ) \int \frac {x}{\left (-5 e^3+e^x\right )^2} \, dx\right )-\frac {1}{5} \left (6 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (4 e^{11}\right ) \int \frac {x}{\left (-5 e^3+e^x\right )^3} \, dx-\frac {1}{5} \left (32 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx \\ & = -\left (\frac {1}{125} \left (4 e^5\right ) \int \frac {e^x x}{\left (-5 e^3+e^x\right )^2} \, dx\right )+\frac {1}{125} \left (4 e^5\right ) \int \frac {x}{-5 e^3+e^x} \, dx-\frac {1}{25} \left (4 e^8\right ) \int \frac {e^x x}{\left (-5 e^3+e^x\right )^3} \, dx+\frac {1}{25} \left (4 e^8\right ) \int \frac {x}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{5} \left (6 e^8\right ) \text {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )^2} \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (32 e^{11}\right ) \text {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )^3} \, dx,x,e^x\right )-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx \\ & = \frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {4 e^5 x}{125 \left (5 e^3-e^x\right )}-\frac {2 e^2 x^2}{625}+\frac {1}{625} \left (4 e^2\right ) \int \frac {e^x x}{-5 e^3+e^x} \, dx-\frac {1}{125} \left (4 e^5\right ) \int \frac {1}{-5 e^3+e^x} \, dx+\frac {1}{125} \left (4 e^5\right ) \int \frac {e^x x}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{125} \left (4 e^5\right ) \int \frac {x}{-5 e^3+e^x} \, dx-\frac {1}{25} \left (2 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2} \, dx-\frac {1}{5} \left (6 e^8\right ) \text {Subst}\left (\int \left (\frac {1}{5 e^3 \left (5 e^3-x\right )^2}+\frac {1}{25 e^6 \left (5 e^3-x\right )}+\frac {1}{25 e^6 x}\right ) \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (32 e^{11}\right ) \text {Subst}\left (\int \left (-\frac {1}{5 e^3 \left (5 e^3-x\right )^3}-\frac {1}{25 e^6 \left (5 e^3-x\right )^2}-\frac {1}{125 e^9 \left (5 e^3-x\right )}-\frac {1}{125 e^9 x}\right ) \, dx,x,e^x\right )-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx \\ & = \frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {2 e^5}{125 \left (5 e^3-e^x\right )}+\frac {2 e^2 x}{625}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}+\frac {4}{625} e^2 x \log \left (1-\frac {e^{-3+x}}{5}\right )-\frac {2}{625} e^2 \log \left (5 e^3-e^x\right )-\frac {1}{625} \left (4 e^2\right ) \int \frac {e^x x}{-5 e^3+e^x} \, dx-\frac {1}{625} \left (4 e^2\right ) \int \log \left (1-\frac {e^{-3+x}}{5}\right ) \, dx+\frac {1}{125} \left (4 e^5\right ) \int \frac {1}{-5 e^3+e^x} \, dx-\frac {1}{125} \left (4 e^5\right ) \text {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )} \, dx,x,e^x\right )-\frac {1}{25} \left (2 e^8\right ) \text {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )^2} \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx \\ & = \frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {2 e^5}{125 \left (5 e^3-e^x\right )}+\frac {2 e^2 x}{625}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {2}{625} e^2 \log \left (5 e^3-e^x\right )+\frac {1}{625} \left (4 e^2\right ) \int \log \left (1-\frac {e^{-3+x}}{5}\right ) \, dx+\frac {1}{625} \left (4 e^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\frac {1}{625} \left (4 e^2\right ) \text {Subst}\left (\int \frac {1}{-5 e^3+x} \, dx,x,e^x\right )-\frac {1}{625} \left (4 e^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right )}{x} \, dx,x,e^{-3+x}\right )+\frac {1}{125} \left (4 e^5\right ) \text {Subst}\left (\int \frac {1}{x \left (-5 e^3+x\right )} \, dx,x,e^x\right )-\frac {1}{25} \left (2 e^8\right ) \text {Subst}\left (\int \left (\frac {1}{5 e^3 \left (5 e^3-x\right )^2}+\frac {1}{25 e^6 \left (5 e^3-x\right )}+\frac {1}{25 e^6 x}\right ) \, dx,x,e^x\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx \\ & = \frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {4 e^2 x}{625}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {4}{625} e^2 \log \left (5 e^3-e^x\right )+\frac {4}{625} e^2 \text {Li}_2\left (\frac {e^{-3+x}}{5}\right )-\frac {1}{625} \left (4 e^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {1}{625} \left (4 e^2\right ) \text {Subst}\left (\int \frac {1}{-5 e^3+x} \, dx,x,e^x\right )+\frac {1}{625} \left (4 e^2\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right )}{x} \, dx,x,e^{-3+x}\right )-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx \\ & = \frac {16 e^8}{25 \left (5 e^3-e^x\right )^2}+\frac {2 e^8 x}{25 \left (5 e^3-e^x\right )^2}-\frac {1}{25} \left (32 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x^2} \, dx-\frac {1}{25} \left (64 e^8\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^2 x} \, dx-\frac {1}{5} \left (64 e^{11}\right ) \int \frac {1}{\left (-5 e^3+e^x\right )^3 x} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.97 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=\frac {2 e^8 (4+x)^2}{25 \left (-5 e^3+e^x\right )^2 x} \]

[In]

Integrate[(E^2*(160 - 10*x^2) + E^(-1 + x)*(-32 - 64*x - 30*x^2 - 4*x^3))/(-3125*x^2 + 1875*E^(-3 + x)*x^2 - 3
75*E^(-6 + 2*x)*x^2 + 25*E^(-9 + 3*x)*x^2),x]

[Out]

(2*E^8*(4 + x)^2)/(25*(-5*E^3 + E^x)^2*x)

Maple [A] (verified)

Time = 7.57 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{2} \left (x^{2}+8 x +16\right )}{25 x \left ({\mathrm e}^{-3+x}-5\right )^{2}}\)	\(24\)
norman	\(\frac {\frac {16 \,{\mathrm e}^{2} x}{25}+\frac {32 \,{\mathrm e}^{2}}{25}+\frac {2 x^{2} {\mathrm e}^{2}}{25}}{x \left ({\mathrm e}^{-3+x}-5\right )^{2}}\)	\(36\)
parallelrisch	\(\frac {2 x^{2} {\mathrm e}^{2}+16 \,{\mathrm e}^{2} x +32 \,{\mathrm e}^{2}}{25 x \left ({\mathrm e}^{2 x -6}-10 \,{\mathrm e}^{-3+x}+25\right )}\)	\(45\)

[In]

int(((-4*x^3-30*x^2-64*x-32)*exp(1)^2*exp(-3+x)+(-10*x^2+160)*exp(1)^2)/(25*x^2*exp(-3+x)^3-375*x^2*exp(-3+x)^
2+1875*x^2*exp(-3+x)-3125*x^2),x,method=_RETURNVERBOSE)

[Out]

2/25/x*exp(2)*(x^2+8*x+16)/(exp(-3+x)-5)^2

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.25 \[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=\frac {2 \, {\left (x^{2} + 8 \, x + 16\right )} e^{6}}{25 \, {\left (25 \, x e^{4} + x e^{\left (2 \, x - 2\right )} - 10 \, x e^{\left (x + 1\right )}\right )}} \]

[In]

integrate(((-4*x^3-30*x^2-64*x-32)*exp(1)^2*exp(-3+x)+(-10*x^2+160)*exp(1)^2)/(25*x^2*exp(-3+x)^3-375*x^2*exp(
-3+x)^2+1875*x^2*exp(-3+x)-3125*x^2),x, algorithm="fricas")

[Out]

2/25*(x^2 + 8*x + 16)*e^6/(25*x*e^4 + x*e^(2*x - 2) - 10*x*e^(x + 1))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.46 \[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=\frac {2 x^{2} e^{2} + 16 x e^{2} + 32 e^{2}}{- 250 x e^{x - 3} + 25 x e^{2 x - 6} + 625 x} \]

[In]

integrate(((-4*x**3-30*x**2-64*x-32)*exp(1)**2*exp(-3+x)+(-10*x**2+160)*exp(1)**2)/(25*x**2*exp(-3+x)**3-375*x
**2*exp(-3+x)**2+1875*x**2*exp(-3+x)-3125*x**2),x)

[Out]

(2*x**2*exp(2) + 16*x*exp(2) + 32*exp(2))/(-250*x*exp(x - 3) + 25*x*exp(2*x - 6) + 625*x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=\frac {2 \, {\left (x^{2} e^{8} + 8 \, x e^{8} + 16 \, e^{8}\right )}}{25 \, {\left (25 \, x e^{6} + x e^{\left (2 \, x\right )} - 10 \, x e^{\left (x + 3\right )}\right )}} \]

[In]

integrate(((-4*x^3-30*x^2-64*x-32)*exp(1)^2*exp(-3+x)+(-10*x^2+160)*exp(1)^2)/(25*x^2*exp(-3+x)^3-375*x^2*exp(
-3+x)^2+1875*x^2*exp(-3+x)-3125*x^2),x, algorithm="maxima")

[Out]

2/25*(x^2*e^8 + 8*x*e^8 + 16*e^8)/(25*x*e^6 + x*e^(2*x) - 10*x*e^(x + 3))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (22) = 44\).

Time = 0.28 (sec) , antiderivative size = 61, normalized size of antiderivative = 2.18 \[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=\frac {2 \, {\left ({\left (x - 3\right )}^{2} e^{2} + 14 \, {\left (x - 3\right )} e^{2} + 49 \, e^{2}\right )}}{25 \, {\left ({\left (x - 3\right )} e^{\left (2 \, x - 6\right )} - 10 \, {\left (x - 3\right )} e^{\left (x - 3\right )} + 25 \, x + 3 \, e^{\left (2 \, x - 6\right )} - 30 \, e^{\left (x - 3\right )}\right )}} \]

[In]

integrate(((-4*x^3-30*x^2-64*x-32)*exp(1)^2*exp(-3+x)+(-10*x^2+160)*exp(1)^2)/(25*x^2*exp(-3+x)^3-375*x^2*exp(
-3+x)^2+1875*x^2*exp(-3+x)-3125*x^2),x, algorithm="giac")

[Out]

2/25*((x - 3)^2*e^2 + 14*(x - 3)*e^2 + 49*e^2)/((x - 3)*e^(2*x - 6) - 10*(x - 3)*e^(x - 3) + 25*x + 3*e^(2*x -
6) - 30*e^(x - 3))

Mupad [B] (verification not implemented)

Time = 15.81 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.43 \[ \int \frac {e^2 \left (160-10 x^2\right )+e^{-1+x} \left (-32-64 x-30 x^2-4 x^3\right )}{-3125 x^2+1875 e^{-3+x} x^2-375 e^{-6+2 x} x^2+25 e^{-9+3 x} x^2} \, dx=\frac {2\,\left ({\mathrm {e}}^2\,x^3+8\,{\mathrm {e}}^2\,x^2+16\,{\mathrm {e}}^2\,x\right )}{25\,x^2\,\left ({\mathrm {e}}^{2\,x-6}-10\,{\mathrm {e}}^{x-3}+25\right )} \]

[In]

int(-(exp(2)*(10*x^2 - 160) + exp(x - 3)*exp(2)*(64*x + 30*x^2 + 4*x^3 + 32))/(1875*x^2*exp(x - 3) - 375*x^2*e
xp(2*x - 6) + 25*x^2*exp(3*x - 9) - 3125*x^2),x)

[Out]

(2*(16*x*exp(2) + 8*x^2*exp(2) + x^3*exp(2)))/(25*x^2*(exp(2*x - 6) - 10*exp(x - 3) + 25))