3.24.19 \(\int \frac {e^{-\frac {1}{4} e^x (16-8 x+x^2)} (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} (200 x^3-350 x^4+175 x^5-25 x^6)+e^{2 x} (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7)+e^x (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8))}{100 x^3-200 x^4+100 x^5} \, dx\)

Optimal. Leaf size=36 \[ \frac {e^{-\frac {1}{4} e^x (-4+x)^2} \left (-e^x+\frac {4}{5 x}+x\right )^2}{-1+x} \]

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Rubi [F]  time = 18.85, 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 {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{100 x^3-200 x^4+100 x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(128 - 192*x - 160*x^3 - 200*x^4 + 100*x^5 + E^(3*x)*(200*x^3 - 350*x^4 + 175*x^5 - 25*x^6) + E^(2*x)*(-32
0*x^2 + 260*x^3 - 480*x^4 + 740*x^5 - 350*x^6 + 50*x^7) + E^x*(-32*x + 256*x^2 + 472*x^3 - 376*x^4 + 280*x^5 -
390*x^6 + 175*x^7 - 25*x^8))/(E^((E^x*(16 - 8*x + x^2))/4)*(100*x^3 - 200*x^4 + 100*x^5)),x]

[Out]

Defer[Int][E^(-1/4*(E^x*(16 - 8*x + x^2))), x] - (3*Defer[Int][E^(x - (E^x*(16 - 8*x + x^2))/4), x])/4 + (19*D
efer[Int][E^(2*x - (E^x*(16 - 8*x + x^2))/4), x])/10 + (5*Defer[Int][E^(3*x - (E^x*(16 - 8*x + x^2))/4), x])/4
- (81*Defer[Int][1/(E^((E^x*(16 - 8*x + x^2))/4)*(-1 + x)^2), x])/25 + (18*Defer[Int][E^(x - (E^x*(16 - 8*x +
x^2))/4)/(-1 + x)^2, x])/5 - Defer[Int][E^(2*x - (E^x*(16 - 8*x + x^2))/4)/(-1 + x)^2, x] - (603*Defer[Int][E
^(x - (E^x*(16 - 8*x + x^2))/4)/(-1 + x), x])/100 + (47*Defer[Int][E^(2*x - (E^x*(16 - 8*x + x^2))/4)/(-1 + x)
, x])/10 - (3*Defer[Int][E^(3*x - (E^x*(16 - 8*x + x^2))/4)/(-1 + x), x])/4 + (32*Defer[Int][1/(E^((E^x*(16 -
8*x + x^2))/4)*x^3), x])/25 + (16*Defer[Int][1/(E^((E^x*(16 - 8*x + x^2))/4)*x^2), x])/25 - (8*Defer[Int][E^(x
- (E^x*(16 - 8*x + x^2))/4)/x^2, x])/25 + (48*Defer[Int][E^(x - (E^x*(16 - 8*x + x^2))/4)/x, x])/25 - (16*Def
er[Int][E^(2*x - (E^x*(16 - 8*x + x^2))/4)/x, x])/5 - (23*Defer[Int][E^(x - (E^x*(16 - 8*x + x^2))/4)*x, x])/2
0 - (5*Defer[Int][E^(2*x - (E^x*(16 - 8*x + x^2))/4)*x, x])/2 - Defer[Int][E^(3*x - (E^x*(16 - 8*x + x^2))/4)*
x, x]/4 + (5*Defer[Int][E^(x - (E^x*(16 - 8*x + x^2))/4)*x^2, x])/4 + Defer[Int][E^(2*x - (E^x*(16 - 8*x + x^2
))/4)*x^2, x]/2 - Defer[Int][E^(x - (E^x*(16 - 8*x + x^2))/4)*x^3, x]/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{x^3 \left (100-200 x+100 x^2\right )} \, dx\\ &=\int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{100 (-1+x)^2 x^3} \, dx\\ &=\frac {1}{100} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (128-192 x-160 x^3-200 x^4+100 x^5+e^{3 x} \left (200 x^3-350 x^4+175 x^5-25 x^6\right )+e^{2 x} \left (-320 x^2+260 x^3-480 x^4+740 x^5-350 x^6+50 x^7\right )+e^x \left (-32 x+256 x^2+472 x^3-376 x^4+280 x^5-390 x^6+175 x^7-25 x^8\right )\right )}{(-1+x)^2 x^3} \, dx\\ &=\frac {1}{100} \int \left (-\frac {160 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2}-\frac {25 e^{3 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} (-4+x) (-2+x)}{-1+x}+\frac {128 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^3}-\frac {192 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^2}-\frac {200 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x}{(-1+x)^2}+\frac {100 e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x^2}{(-1+x)^2}+\frac {10 e^{2 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (-32+26 x-48 x^2+74 x^3-35 x^4+5 x^5\right )}{(-1+x)^2 x}-\frac {e^{x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (32-256 x-472 x^2+376 x^3-280 x^4+390 x^5-175 x^6+25 x^7\right )}{(-1+x)^2 x^2}\right ) \, dx\\ &=-\left (\frac {1}{100} \int \frac {e^{x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (32-256 x-472 x^2+376 x^3-280 x^4+390 x^5-175 x^6+25 x^7\right )}{(-1+x)^2 x^2} \, dx\right )+\frac {1}{10} \int \frac {e^{2 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} \left (-32+26 x-48 x^2+74 x^3-35 x^4+5 x^5\right )}{(-1+x)^2 x} \, dx-\frac {1}{4} \int \frac {e^{3 x-\frac {1}{4} e^x \left (16-8 x+x^2\right )} (-4+x) (-2+x)}{-1+x} \, dx+\frac {32}{25} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^3} \, dx-\frac {8}{5} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2} \, dx-\frac {48}{25} \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )}}{(-1+x)^2 x^2} \, dx-2 \int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x}{(-1+x)^2} \, dx+\int \frac {e^{-\frac {1}{4} e^x \left (16-8 x+x^2\right )} x^2}{(-1+x)^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 41, normalized size = 1.14 \begin {gather*} \frac {e^{-\frac {1}{4} e^x (-4+x)^2} \left (4-5 e^x x+5 x^2\right )^2}{25 (-1+x) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(128 - 192*x - 160*x^3 - 200*x^4 + 100*x^5 + E^(3*x)*(200*x^3 - 350*x^4 + 175*x^5 - 25*x^6) + E^(2*x
)*(-320*x^2 + 260*x^3 - 480*x^4 + 740*x^5 - 350*x^6 + 50*x^7) + E^x*(-32*x + 256*x^2 + 472*x^3 - 376*x^4 + 280
*x^5 - 390*x^6 + 175*x^7 - 25*x^8))/(E^((E^x*(16 - 8*x + x^2))/4)*(100*x^3 - 200*x^4 + 100*x^5)),x]

[Out]

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

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fricas [A]  time = 0.86, size = 60, normalized size = 1.67 \begin {gather*} \frac {{\left (25 \, x^{4} + 25 \, x^{2} e^{\left (2 \, x\right )} + 40 \, x^{2} - 10 \, {\left (5 \, x^{3} + 4 \, x\right )} e^{x} + 16\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} - 8 \, x + 16\right )} e^{x}\right )}}{25 \, {\left (x^{3} - x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^6+175*x^5-350*x^4+200*x^3)*exp(x)^3+(50*x^7-350*x^6+740*x^5-480*x^4+260*x^3-320*x^2)*exp(x)^
2+(-25*x^8+175*x^7-390*x^6+280*x^5-376*x^4+472*x^3+256*x^2-32*x)*exp(x)+100*x^5-200*x^4-160*x^3-192*x+128)/(10
0*x^5-200*x^4+100*x^3)/exp(1/4*(x^2-8*x+16)*exp(x)),x, algorithm="fricas")

[Out]

1/25*(25*x^4 + 25*x^2*e^(2*x) + 40*x^2 - 10*(5*x^3 + 4*x)*e^x + 16)*e^(-1/4*(x^2 - 8*x + 16)*e^x)/(x^3 - x^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (100 \, x^{5} - 200 \, x^{4} - 160 \, x^{3} - 25 \, {\left (x^{6} - 7 \, x^{5} + 14 \, x^{4} - 8 \, x^{3}\right )} e^{\left (3 \, x\right )} + 10 \, {\left (5 \, x^{7} - 35 \, x^{6} + 74 \, x^{5} - 48 \, x^{4} + 26 \, x^{3} - 32 \, x^{2}\right )} e^{\left (2 \, x\right )} - {\left (25 \, x^{8} - 175 \, x^{7} + 390 \, x^{6} - 280 \, x^{5} + 376 \, x^{4} - 472 \, x^{3} - 256 \, x^{2} + 32 \, x\right )} e^{x} - 192 \, x + 128\right )} e^{\left (-\frac {1}{4} \, {\left (x^{2} - 8 \, x + 16\right )} e^{x}\right )}}{100 \, {\left (x^{5} - 2 \, x^{4} + x^{3}\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^6+175*x^5-350*x^4+200*x^3)*exp(x)^3+(50*x^7-350*x^6+740*x^5-480*x^4+260*x^3-320*x^2)*exp(x)^
2+(-25*x^8+175*x^7-390*x^6+280*x^5-376*x^4+472*x^3+256*x^2-32*x)*exp(x)+100*x^5-200*x^4-160*x^3-192*x+128)/(10
0*x^5-200*x^4+100*x^3)/exp(1/4*(x^2-8*x+16)*exp(x)),x, algorithm="giac")

[Out]

integrate(1/100*(100*x^5 - 200*x^4 - 160*x^3 - 25*(x^6 - 7*x^5 + 14*x^4 - 8*x^3)*e^(3*x) + 10*(5*x^7 - 35*x^6
+ 74*x^5 - 48*x^4 + 26*x^3 - 32*x^2)*e^(2*x) - (25*x^8 - 175*x^7 + 390*x^6 - 280*x^5 + 376*x^4 - 472*x^3 - 256
*x^2 + 32*x)*e^x - 192*x + 128)*e^(-1/4*(x^2 - 8*x + 16)*e^x)/(x^5 - 2*x^4 + x^3), x)

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maple [A]  time = 0.06, size = 54, normalized size = 1.50

method	result	size
risch	\(\frac {\left (25 x^{4}-50 \,{\mathrm e}^{x} x^{3}+25 \,{\mathrm e}^{2 x} x^{2}+40 x^{2}-40 \,{\mathrm e}^{x} x +16\right ) {\mathrm e}^{-\frac {\left (x -4\right )^{2} {\mathrm e}^{x}}{4}}}{25 x^{2} \left (x -1\right )}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-25*x^6+175*x^5-350*x^4+200*x^3)*exp(x)^3+(50*x^7-350*x^6+740*x^5-480*x^4+260*x^3-320*x^2)*exp(x)^2+(-25
*x^8+175*x^7-390*x^6+280*x^5-376*x^4+472*x^3+256*x^2-32*x)*exp(x)+100*x^5-200*x^4-160*x^3-192*x+128)/(100*x^5-
200*x^4+100*x^3)/exp(1/4*(x^2-8*x+16)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

1/25/x^2*(25*x^4-50*exp(x)*x^3+25*exp(2*x)*x^2+40*x^2-40*exp(x)*x+16)/(x-1)*exp(-1/4*(x-4)^2*exp(x))

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maxima [B]  time = 0.86, size = 65, normalized size = 1.81 \begin {gather*} \frac {{\left (25 \, x^{4} + 25 \, x^{2} e^{\left (2 \, x\right )} + 40 \, x^{2} - 10 \, {\left (5 \, x^{3} + 4 \, x\right )} e^{x} + 16\right )} e^{\left (-\frac {1}{4} \, x^{2} e^{x} + 2 \, x e^{x} - 4 \, e^{x}\right )}}{25 \, {\left (x^{3} - x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x^6+175*x^5-350*x^4+200*x^3)*exp(x)^3+(50*x^7-350*x^6+740*x^5-480*x^4+260*x^3-320*x^2)*exp(x)^
2+(-25*x^8+175*x^7-390*x^6+280*x^5-376*x^4+472*x^3+256*x^2-32*x)*exp(x)+100*x^5-200*x^4-160*x^3-192*x+128)/(10
0*x^5-200*x^4+100*x^3)/exp(1/4*(x^2-8*x+16)*exp(x)),x, algorithm="maxima")

[Out]

1/25*(25*x^4 + 25*x^2*e^(2*x) + 40*x^2 - 10*(5*x^3 + 4*x)*e^x + 16)*e^(-1/4*x^2*e^x + 2*x*e^x - 4*e^x)/(x^3 -
x^2)

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mupad [B]  time = 1.77, size = 61, normalized size = 1.69 \begin {gather*} -\frac {{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x-\frac {x^2\,{\mathrm {e}}^x}{4}-4\,{\mathrm {e}}^x}\,\left (x^2\,{\mathrm {e}}^{2\,x}-2\,x^3\,{\mathrm {e}}^x-\frac {8\,x\,{\mathrm {e}}^x}{5}+\frac {8\,x^2}{5}+x^4+\frac {16}{25}\right )}{x^2-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(exp(x)*(x^2 - 8*x + 16))/4)*(192*x + exp(2*x)*(320*x^2 - 260*x^3 + 480*x^4 - 740*x^5 + 350*x^6 - 5
0*x^7) - exp(3*x)*(200*x^3 - 350*x^4 + 175*x^5 - 25*x^6) + 160*x^3 + 200*x^4 - 100*x^5 + exp(x)*(32*x - 256*x^
2 - 472*x^3 + 376*x^4 - 280*x^5 + 390*x^6 - 175*x^7 + 25*x^8) - 128))/(100*x^3 - 200*x^4 + 100*x^5),x)

[Out]

-(exp(2*x*exp(x) - (x^2*exp(x))/4 - 4*exp(x))*(x^2*exp(2*x) - 2*x^3*exp(x) - (8*x*exp(x))/5 + (8*x^2)/5 + x^4
+ 16/25))/(x^2 - x^3)

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sympy [B]  time = 0.51, size = 61, normalized size = 1.69 \begin {gather*} \frac {\left (25 x^{4} - 50 x^{3} e^{x} + 25 x^{2} e^{2 x} + 40 x^{2} - 40 x e^{x} + 16\right ) e^{- \left (\frac {x^{2}}{4} - 2 x + 4\right ) e^{x}}}{25 x^{3} - 25 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-25*x**6+175*x**5-350*x**4+200*x**3)*exp(x)**3+(50*x**7-350*x**6+740*x**5-480*x**4+260*x**3-320*x*
*2)*exp(x)**2+(-25*x**8+175*x**7-390*x**6+280*x**5-376*x**4+472*x**3+256*x**2-32*x)*exp(x)+100*x**5-200*x**4-1
60*x**3-192*x+128)/(100*x**5-200*x**4+100*x**3)/exp(1/4*(x**2-8*x+16)*exp(x)),x)

[Out]

(25*x**4 - 50*x**3*exp(x) + 25*x**2*exp(2*x) + 40*x**2 - 40*x*exp(x) + 16)*exp(-(x**2/4 - 2*x + 4)*exp(x))/(25
*x**3 - 25*x**2)

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