3.80.79 \(\int \frac {1}{5} e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} (80 x^2+e^{25-10 x+x^2} (8+4 x))} (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3) \, dx\)

Optimal. Leaf size=26 \[ e^{4 \left (4 e^{-(-5+x)^2} x^2+\frac {2+x}{5}\right )} \]

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Rubi [F]  time = 2.24, 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 {1}{5} \exp \left (-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )\right ) \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-25 + 10*x - x^2 + (E^(-25 + 10*x - x^2)*(80*x^2 + E^(25 - 10*x + x^2)*(8 + 4*x)))/5)*(4*E^(25 - 10*x
+ x^2) + 160*x + 800*x^2 - 160*x^3))/5,x]

[Out]

(4*Defer[Int][E^((4*(2 + x + (20*x^2)/E^(-5 + x)^2))/5), x])/5 + 32*Defer[Int][E^(-117/5 + (54*x)/5 + (-1 + 16
/E^(-5 + x)^2)*x^2)*x, x] + 160*Defer[Int][E^(-117/5 + (54*x)/5 + (-1 + 16/E^(-5 + x)^2)*x^2)*x^2, x] - 32*Def
er[Int][E^(-117/5 + (54*x)/5 + (-1 + 16/E^(-5 + x)^2)*x^2)*x^3, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int \exp \left (-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )\right ) \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx\\ &=\frac {1}{5} \int 4 \exp \left (-\frac {117}{5}+\frac {4 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) \left (e^{25+x^2}-40 e^{10 x} x \left (-1-5 x+x^2\right )\right ) \, dx\\ &=\frac {4}{5} \int \exp \left (-\frac {117}{5}+\frac {4 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) \left (e^{25+x^2}-40 e^{10 x} x \left (-1-5 x+x^2\right )\right ) \, dx\\ &=\frac {4}{5} \int \left (\exp \left (\frac {8}{5}+\frac {4 x}{5}+x^2+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right )-40 \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x \left (-1-5 x+x^2\right )\right ) \, dx\\ &=\frac {4}{5} \int \exp \left (\frac {8}{5}+\frac {4 x}{5}+x^2+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) \, dx-32 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x \left (-1-5 x+x^2\right ) \, dx\\ &=\frac {4}{5} \int e^{\frac {4}{5} \left (2+x+20 e^{-(-5+x)^2} x^2\right )} \, dx-32 \int \left (-\exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x-5 \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^2+\exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^3\right ) \, dx\\ &=\frac {4}{5} \int e^{\frac {4}{5} \left (2+x+20 e^{-(-5+x)^2} x^2\right )} \, dx+32 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x \, dx-32 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^3 \, dx+160 \int \exp \left (-\frac {117}{5}+\frac {54 x}{5}+\left (-1+16 e^{-(-5+x)^2}\right ) x^2\right ) x^2 \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 1.25, size = 83, normalized size = 3.19 \begin {gather*} \frac {1}{5} \int e^{-25+10 x-x^2+\frac {1}{5} e^{-25+10 x-x^2} \left (80 x^2+e^{25-10 x+x^2} (8+4 x)\right )} \left (4 e^{25-10 x+x^2}+160 x+800 x^2-160 x^3\right ) \, dx \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-25 + 10*x - x^2 + (E^(-25 + 10*x - x^2)*(80*x^2 + E^(25 - 10*x + x^2)*(8 + 4*x)))/5)*(4*E^(25 -
10*x + x^2) + 160*x + 800*x^2 - 160*x^3))/5,x]

[Out]

Integrate[E^(-25 + 10*x - x^2 + (E^(-25 + 10*x - x^2)*(80*x^2 + E^(25 - 10*x + x^2)*(8 + 4*x)))/5)*(4*E^(25 -
10*x + x^2) + 160*x + 800*x^2 - 160*x^3), x]/5

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fricas [B]  time = 0.59, size = 49, normalized size = 1.88 \begin {gather*} e^{\left (x^{2} + \frac {1}{5} \, {\left (80 \, x^{2} - {\left (5 \, x^{2} - 54 \, x + 117\right )} e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + 10 \, x - 25\right )} - 10 \, x + 25\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*exp(x^2-10*x+25)-160*x^3+800*x^2+160*x)*exp(1/5*((4*x+8)*exp(x^2-10*x+25)+80*x^2)/exp(x^2-10*
x+25))/exp(x^2-10*x+25),x, algorithm="fricas")

[Out]

e^(x^2 + 1/5*(80*x^2 - (5*x^2 - 54*x + 117)*e^(x^2 - 10*x + 25))*e^(-x^2 + 10*x - 25) - 10*x + 25)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4}{5} \, {\left (40 \, x^{3} - 200 \, x^{2} - 40 \, x - e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + \frac {4}{5} \, {\left (20 \, x^{2} + {\left (x + 2\right )} e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + 10 \, x - 25\right )} + 10 \, x - 25\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*exp(x^2-10*x+25)-160*x^3+800*x^2+160*x)*exp(1/5*((4*x+8)*exp(x^2-10*x+25)+80*x^2)/exp(x^2-10*
x+25))/exp(x^2-10*x+25),x, algorithm="giac")

[Out]

integrate(-4/5*(40*x^3 - 200*x^2 - 40*x - e^(x^2 - 10*x + 25))*e^(-x^2 + 4/5*(20*x^2 + (x + 2)*e^(x^2 - 10*x +
25))*e^(-x^2 + 10*x - 25) + 10*x - 25), x)

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maple [A]  time = 0.08, size = 34, normalized size = 1.31

method	result	size
risch	\({\mathrm e}^{\frac {4 \left ({\mathrm e}^{\left (x -5\right )^{2}} x +20 x^{2}+2 \,{\mathrm e}^{\left (x -5\right )^{2}}\right ) {\mathrm e}^{-\left (x -5\right )^{2}}}{5}}\)	\(34\)
norman	\({\mathrm e}^{\frac {\left (\left (4 x +8\right ) {\mathrm e}^{x^{2}-10 x +25}+80 x^{2}\right ) {\mathrm e}^{-x^{2}+10 x -25}}{5}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*(4*exp(x^2-10*x+25)-160*x^3+800*x^2+160*x)*exp(1/5*((4*x+8)*exp(x^2-10*x+25)+80*x^2)/exp(x^2-10*x+25))
/exp(x^2-10*x+25),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {4}{5} \, \int {\left (40 \, x^{3} - 200 \, x^{2} - 40 \, x - e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + \frac {4}{5} \, {\left (20 \, x^{2} + {\left (x + 2\right )} e^{\left (x^{2} - 10 \, x + 25\right )}\right )} e^{\left (-x^{2} + 10 \, x - 25\right )} + 10 \, x - 25\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*exp(x^2-10*x+25)-160*x^3+800*x^2+160*x)*exp(1/5*((4*x+8)*exp(x^2-10*x+25)+80*x^2)/exp(x^2-10*
x+25))/exp(x^2-10*x+25),x, algorithm="maxima")

[Out]

-4/5*integrate((40*x^3 - 200*x^2 - 40*x - e^(x^2 - 10*x + 25))*e^(-x^2 + 4/5*(20*x^2 + (x + 2)*e^(x^2 - 10*x +
25))*e^(-x^2 + 10*x - 25) + 10*x - 25), x)

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mupad [B]  time = 5.61, size = 25, normalized size = 0.96 \begin {gather*} {\mathrm {e}}^{\frac {4\,x}{5}}\,{\mathrm {e}}^{8/5}\,{\mathrm {e}}^{16\,x^2\,{\mathrm {e}}^{10\,x}\,{\mathrm {e}}^{-25}\,{\mathrm {e}}^{-x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(10*x - x^2 - 25)*((exp(x^2 - 10*x + 25)*(4*x + 8))/5 + 16*x^2))*exp(10*x - x^2 - 25)*(160*x + 4*e
xp(x^2 - 10*x + 25) + 800*x^2 - 160*x^3))/5,x)

[Out]

exp((4*x)/5)*exp(8/5)*exp(16*x^2*exp(10*x)*exp(-25)*exp(-x^2))

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sympy [A]  time = 0.44, size = 32, normalized size = 1.23 \begin {gather*} e^{\left (16 x^{2} + \frac {\left (4 x + 8\right ) e^{x^{2} - 10 x + 25}}{5}\right ) e^{- x^{2} + 10 x - 25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*(4*exp(x**2-10*x+25)-160*x**3+800*x**2+160*x)*exp(1/5*((4*x+8)*exp(x**2-10*x+25)+80*x**2)/exp(x*
*2-10*x+25))/exp(x**2-10*x+25),x)

[Out]

exp((16*x**2 + (4*x + 8)*exp(x**2 - 10*x + 25)/5)*exp(-x**2 + 10*x - 25))

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