3.63.8 \(\int \frac {100 e^{\frac {3 (8+2 x)}{x^2}} x^2+e^{\frac {2 (8+2 x)}{x^2}} (400+50 x+25 x^2)}{-8 x^5+e^{\frac {3 (8+2 x)}{x^2}} (8 e^3 x^2-96 e^2 x^3+384 e x^4-512 x^5)+e^{\frac {2 (8+2 x)}{x^2}} (-24 e^2 x^3+192 e x^4-384 x^5)+e^{\frac {8+2 x}{x^2}} (24 e x^4-96 x^5)} \, dx\)

Optimal. Leaf size=30 \[ \frac {1}{16 \left (-x+\frac {1}{5} \left (e+x-e^{-\frac {2 (4+x)}{x^2}} x\right )\right )^2} \]

________________________________________________________________________________________

Rubi [F]  time = 7.15, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {100 e^{\frac {3 (8+2 x)}{x^2}} x^2+e^{\frac {2 (8+2 x)}{x^2}} \left (400+50 x+25 x^2\right )}{-8 x^5+e^{\frac {3 (8+2 x)}{x^2}} \left (8 e^3 x^2-96 e^2 x^3+384 e x^4-512 x^5\right )+e^{\frac {2 (8+2 x)}{x^2}} \left (-24 e^2 x^3+192 e x^4-384 x^5\right )+e^{\frac {8+2 x}{x^2}} \left (24 e x^4-96 x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(100*E^((3*(8 + 2*x))/x^2)*x^2 + E^((2*(8 + 2*x))/x^2)*(400 + 50*x + 25*x^2))/(-8*x^5 + E^((3*(8 + 2*x))/x
^2)*(8*E^3*x^2 - 96*E^2*x^3 + 384*E*x^4 - 512*x^5) + E^((2*(8 + 2*x))/x^2)*(-24*E^2*x^3 + 192*E*x^4 - 384*x^5)
+ E^((8 + 2*x)/x^2)*(24*E*x^4 - 96*x^5)),x]

[Out]

(25*Defer[Int][E^(1 + (4*(4 + x))/x^2)/((E - 4*x)*(E^(1 + 8/x^2 + 2/x) - x - 4*E^(8/x^2 + 2/x)*x)^3), x])/8 +
(25*Defer[Int][E^((4*(4 + x))/x^2)/((E - 4*x)*(E^(1 + 8/x^2 + 2/x) - x - 4*E^(8/x^2 + 2/x)*x)^2), x])/2 - 50*D
efer[Int][E^((4*(4 + x))/x^2)/(x^2*(-E^(1 + 8/x^2 + 2/x) + x + 4*E^(8/x^2 + 2/x)*x)^3), x] - (25*Defer[Int][E^
((4*(4 + x))/x^2)/(x*(-E^(1 + 8/x^2 + 2/x) + x + 4*E^(8/x^2 + 2/x)*x)^3), x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 e^{\frac {4 (4+x)}{x^2}} \left (16+2 x+\left (1+4 e^{\frac {2 (4+x)}{x^2}}\right ) x^2\right )}{8 x^2 \left (e^{\frac {8+2 x+x^2}{x^2}}-x-4 e^{\frac {2 (4+x)}{x^2}} x\right )^3} \, dx\\ &=\frac {25}{8} \int \frac {e^{\frac {4 (4+x)}{x^2}} \left (16+2 x+\left (1+4 e^{\frac {2 (4+x)}{x^2}}\right ) x^2\right )}{x^2 \left (e^{\frac {8+2 x+x^2}{x^2}}-x-4 e^{\frac {2 (4+x)}{x^2}} x\right )^3} \, dx\\ &=\frac {25}{8} \int \left (\frac {4 e^{\frac {4 (4+x)}{x^2}}}{(e-4 x) \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^2}+\frac {e^{\frac {4 (4+x)}{x^2}} \left (16 e-2 (32-e) x-(8-e) x^2\right )}{(e-4 x) x^2 \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3}\right ) \, dx\\ &=\frac {25}{8} \int \frac {e^{\frac {4 (4+x)}{x^2}} \left (16 e-2 (32-e) x-(8-e) x^2\right )}{(e-4 x) x^2 \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3} \, dx+\frac {25}{2} \int \frac {e^{\frac {4 (4+x)}{x^2}}}{(e-4 x) \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^2} \, dx\\ &=\frac {25}{8} \int \left (\frac {e^{1+\frac {4 (4+x)}{x^2}}}{(e-4 x) \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3}-\frac {16 e^{\frac {4 (4+x)}{x^2}}}{x^2 \left (-e^{1+\frac {8}{x^2}+\frac {2}{x}}+x+4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3}-\frac {2 e^{\frac {4 (4+x)}{x^2}}}{x \left (-e^{1+\frac {8}{x^2}+\frac {2}{x}}+x+4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3}\right ) \, dx+\frac {25}{2} \int \frac {e^{\frac {4 (4+x)}{x^2}}}{(e-4 x) \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^2} \, dx\\ &=\frac {25}{8} \int \frac {e^{1+\frac {4 (4+x)}{x^2}}}{(e-4 x) \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3} \, dx-\frac {25}{4} \int \frac {e^{\frac {4 (4+x)}{x^2}}}{x \left (-e^{1+\frac {8}{x^2}+\frac {2}{x}}+x+4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3} \, dx+\frac {25}{2} \int \frac {e^{\frac {4 (4+x)}{x^2}}}{(e-4 x) \left (e^{1+\frac {8}{x^2}+\frac {2}{x}}-x-4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^2} \, dx-50 \int \frac {e^{\frac {4 (4+x)}{x^2}}}{x^2 \left (-e^{1+\frac {8}{x^2}+\frac {2}{x}}+x+4 e^{\frac {8}{x^2}+\frac {2}{x}} x\right )^3} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 1.68, size = 47, normalized size = 1.57 \begin {gather*} \frac {25 e^{\frac {4 (4+x)}{x^2}}}{16 \left (-e^{\frac {8+2 x+x^2}{x^2}}+x+4 e^{\frac {2 (4+x)}{x^2}} x\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(100*E^((3*(8 + 2*x))/x^2)*x^2 + E^((2*(8 + 2*x))/x^2)*(400 + 50*x + 25*x^2))/(-8*x^5 + E^((3*(8 + 2
*x))/x^2)*(8*E^3*x^2 - 96*E^2*x^3 + 384*E*x^4 - 512*x^5) + E^((2*(8 + 2*x))/x^2)*(-24*E^2*x^3 + 192*E*x^4 - 38
4*x^5) + E^((8 + 2*x)/x^2)*(24*E*x^4 - 96*x^5)),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.55, size = 62, normalized size = 2.07 \begin {gather*} \frac {25 \, e^{\left (\frac {4 \, {\left (x + 4\right )}}{x^{2}}\right )}}{16 \, {\left (x^{2} + {\left (16 \, x^{2} - 8 \, x e + e^{2}\right )} e^{\left (\frac {4 \, {\left (x + 4\right )}}{x^{2}}\right )} + 2 \, {\left (4 \, x^{2} - x e\right )} e^{\left (\frac {2 \, {\left (x + 4\right )}}{x^{2}}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x^2*exp((2*x+8)/x^2)^3+(25*x^2+50*x+400)*exp((2*x+8)/x^2)^2)/((8*x^2*exp(1)^3-96*x^3*exp(1)^2+3
84*x^4*exp(1)-512*x^5)*exp((2*x+8)/x^2)^3+(-24*x^3*exp(1)^2+192*x^4*exp(1)-384*x^5)*exp((2*x+8)/x^2)^2+(24*x^4
*exp(1)-96*x^5)*exp((2*x+8)/x^2)-8*x^5),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.31, size = 100, normalized size = 3.33 \begin {gather*} \frac {25 \, e^{\left (\frac {4 \, {\left (x + 4\right )}}{x^{2}}\right )}}{16 \, {\left (16 \, x^{2} e^{\left (\frac {4 \, {\left (x + 4\right )}}{x^{2}}\right )} + 8 \, x^{2} e^{\left (\frac {2 \, {\left (x + 4\right )}}{x^{2}}\right )} + x^{2} - 8 \, x e^{\left (\frac {x^{2} + 2 \, x + 8}{x^{2}} + \frac {2 \, {\left (x + 4\right )}}{x^{2}}\right )} - 2 \, x e^{\left (\frac {x^{2} + 2 \, x + 8}{x^{2}}\right )} + e^{\left (\frac {2 \, {\left (x^{2} + 2 \, x + 8\right )}}{x^{2}}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x^2*exp((2*x+8)/x^2)^3+(25*x^2+50*x+400)*exp((2*x+8)/x^2)^2)/((8*x^2*exp(1)^3-96*x^3*exp(1)^2+3
84*x^4*exp(1)-512*x^5)*exp((2*x+8)/x^2)^3+(-24*x^3*exp(1)^2+192*x^4*exp(1)-384*x^5)*exp((2*x+8)/x^2)^2+(24*x^4
*exp(1)-96*x^5)*exp((2*x+8)/x^2)-8*x^5),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.54, size = 47, normalized size = 1.57

method	result	size
norman	\(\frac {25 \,{\mathrm e}^{\frac {4 x +16}{x^{2}}}}{16 \left ({\mathrm e} \,{\mathrm e}^{\frac {2 x +8}{x^{2}}}-4 x \,{\mathrm e}^{\frac {2 x +8}{x^{2}}}-x \right )^{2}}\)	\(47\)
risch	\(\frac {25}{16 \left ({\mathrm e}^{2}-8 x \,{\mathrm e}+16 x^{2}\right )}+\frac {25 \left (2 \,{\mathrm e}^{\frac {x^{2}+2 x +8}{x^{2}}}-8 x \,{\mathrm e}^{\frac {2 x +8}{x^{2}}}-x \right ) x}{16 \left ({\mathrm e}^{2}-8 x \,{\mathrm e}+16 x^{2}\right ) \left ({\mathrm e}^{\frac {x^{2}+2 x +8}{x^{2}}}-4 x \,{\mathrm e}^{\frac {2 x +8}{x^{2}}}-x \right )^{2}}\)	\(99\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((100*x^2*exp((2*x+8)/x^2)^3+(25*x^2+50*x+400)*exp((2*x+8)/x^2)^2)/((8*x^2*exp(1)^3-96*x^3*exp(1)^2+384*x^4
*exp(1)-512*x^5)*exp((2*x+8)/x^2)^3+(-24*x^3*exp(1)^2+192*x^4*exp(1)-384*x^5)*exp((2*x+8)/x^2)^2+(24*x^4*exp(1
)-96*x^5)*exp((2*x+8)/x^2)-8*x^5),x,method=_RETURNVERBOSE)

[Out]

25/16*exp((2*x+8)/x^2)^2/(exp(1)*exp((2*x+8)/x^2)-4*x*exp((2*x+8)/x^2)-x)^2

________________________________________________________________________________________

maxima [B]  time = 0.62, size = 71, normalized size = 2.37 \begin {gather*} \frac {25 \, e^{\left (\frac {4}{x} + \frac {16}{x^{2}}\right )}}{16 \, {\left (x^{2} + {\left (16 \, x^{2} - 8 \, x e + e^{2}\right )} e^{\left (\frac {4}{x} + \frac {16}{x^{2}}\right )} + 2 \, {\left (4 \, x^{2} - x e\right )} e^{\left (\frac {2}{x} + \frac {8}{x^{2}}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x^2*exp((2*x+8)/x^2)^3+(25*x^2+50*x+400)*exp((2*x+8)/x^2)^2)/((8*x^2*exp(1)^3-96*x^3*exp(1)^2+3
84*x^4*exp(1)-512*x^5)*exp((2*x+8)/x^2)^3+(-24*x^3*exp(1)^2+192*x^4*exp(1)-384*x^5)*exp((2*x+8)/x^2)^2+(24*x^4
*exp(1)-96*x^5)*exp((2*x+8)/x^2)-8*x^5),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int -\frac {100\,x^2\,{\mathrm {e}}^{\frac {3\,\left (2\,x+8\right )}{x^2}}+{\mathrm {e}}^{\frac {2\,\left (2\,x+8\right )}{x^2}}\,\left (25\,x^2+50\,x+400\right )}{{\mathrm {e}}^{\frac {2\,\left (2\,x+8\right )}{x^2}}\,\left (384\,x^5-192\,\mathrm {e}\,x^4+24\,{\mathrm {e}}^2\,x^3\right )-{\mathrm {e}}^{\frac {2\,x+8}{x^2}}\,\left (24\,x^4\,\mathrm {e}-96\,x^5\right )-{\mathrm {e}}^{\frac {3\,\left (2\,x+8\right )}{x^2}}\,\left (-512\,x^5+384\,\mathrm {e}\,x^4-96\,{\mathrm {e}}^2\,x^3+8\,{\mathrm {e}}^3\,x^2\right )+8\,x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(100*x^2*exp((3*(2*x + 8))/x^2) + exp((2*(2*x + 8))/x^2)*(50*x + 25*x^2 + 400))/(exp((2*(2*x + 8))/x^2)*(
24*x^3*exp(2) - 192*x^4*exp(1) + 384*x^5) - exp((2*x + 8)/x^2)*(24*x^4*exp(1) - 96*x^5) - exp((3*(2*x + 8))/x^
2)*(8*x^2*exp(3) - 96*x^3*exp(2) + 384*x^4*exp(1) - 512*x^5) + 8*x^5),x)

[Out]

int(-(100*x^2*exp((3*(2*x + 8))/x^2) + exp((2*(2*x + 8))/x^2)*(50*x + 25*x^2 + 400))/(exp((2*(2*x + 8))/x^2)*(
24*x^3*exp(2) - 192*x^4*exp(1) + 384*x^5) - exp((2*x + 8)/x^2)*(24*x^4*exp(1) - 96*x^5) - exp((3*(2*x + 8))/x^
2)*(8*x^2*exp(3) - 96*x^3*exp(2) + 384*x^4*exp(1) - 512*x^5) + 8*x^5), x)

________________________________________________________________________________________

sympy [B]  time = 0.56, size = 150, normalized size = 5.00 \begin {gather*} \frac {- 25 x^{2} + \left (- 200 x^{2} + 50 e x\right ) e^{\frac {2 x + 8}{x^{2}}}}{256 x^{4} - 128 e x^{3} + 16 x^{2} e^{2} + \left (2048 x^{4} - 1536 e x^{3} + 384 x^{2} e^{2} - 32 x e^{3}\right ) e^{\frac {2 x + 8}{x^{2}}} + \left (4096 x^{4} - 4096 e x^{3} + 1536 x^{2} e^{2} - 256 x e^{3} + 16 e^{4}\right ) e^{\frac {2 \left (2 x + 8\right )}{x^{2}}}} + \frac {25}{256 x^{2} - 128 e x + 16 e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((100*x**2*exp((2*x+8)/x**2)**3+(25*x**2+50*x+400)*exp((2*x+8)/x**2)**2)/((8*x**2*exp(1)**3-96*x**3*e
xp(1)**2+384*x**4*exp(1)-512*x**5)*exp((2*x+8)/x**2)**3+(-24*x**3*exp(1)**2+192*x**4*exp(1)-384*x**5)*exp((2*x
+8)/x**2)**2+(24*x**4*exp(1)-96*x**5)*exp((2*x+8)/x**2)-8*x**5),x)

[Out]

(-25*x**2 + (-200*x**2 + 50*E*x)*exp((2*x + 8)/x**2))/(256*x**4 - 128*E*x**3 + 16*x**2*exp(2) + (2048*x**4 - 1
536*E*x**3 + 384*x**2*exp(2) - 32*x*exp(3))*exp((2*x + 8)/x**2) + (4096*x**4 - 4096*E*x**3 + 1536*x**2*exp(2)
- 256*x*exp(3) + 16*exp(4))*exp(2*(2*x + 8)/x**2)) + 25/(256*x**2 - 128*E*x + 16*exp(2))

________________________________________________________________________________________